course phy201 Multiple lessons. I finally finished my Army training and got settled in my new duty station, so look out!
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17:59:54 prin phy and gen phy problem 4.02 net force 265 N on bike and rider accelerates at 2.30 m/s^2, mass of bike and rider
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RESPONSE --> 115kg b/c 265/2.3=115
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17:59:59 A force Fnet acting on mass m results in acceleration a, where a = Fnet / m. We are given Fnet and a, so we can solve the equation to find m. Multiplying both sides by m we get a * m = Fnet / m * m so a * m = Fnet. Dividing both sides of this equation by a we have m = Fnet / a = 265 N / (2.30 m/s^2) = 115 (kg m/s^2) / (m/s^2) = 115 kg.
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RESPONSE --> ok
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18:03:52 prin phy and gen phy problem 4.07 force to accelerate 7 g pellet to 125 m/s in .7 m barrel
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RESPONSE --> b/c 125/2=62.5, .7/62.5=.011, 125/.011=11400, 11400*.007=80N
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18:03:56 ** The initial velocity of the bullet is zero and the final velocity is 175 m/s. If we assume uniform acceleration (not necessarily the case but not a bad first approximation) the average velocity is (0 + 125 m/s) / 2 = 62.5 m/s and the time required for the trip down the barrel is .7 m / (62.5 m/s) = .011 sec, approx.. Acceleration is therefore rate of velocity change = `dv / `dt = (125 m/s - 0 m/s) / (.11 sec) = 11000 m/s^2, approx.. The force on the bullet is therefore F = m a = .007 kg * 11000 m/s^2 = 77 N approx. **
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RESPONSE --> ok
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18:08:38 gen phy 4.08. breaking strength 22 N, accel 2.5 m/s^2 breaks line. What can we say about the mass of the fish?
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RESPONSE --> m>1.8kg b/c 22=m(9.8+2.5)=m*12.3, 22/12.3=1.8
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18:08:42 The fish is being pulled upward by the tension, downward by gravity. The net force on the fish is therefore equal to the tension in the line, minus the force exerted by gravity. In symbols, Fnet = T - m g, where m is the mass of the fish. To accelerate a fish of mass m upward at 2.5 m/s^2 the net force must be Fnet = m a = m * 2.5 m/s^2. Combined with the preceding we have the condition m * 2.5 m/s^2 = T - m g so that to provide this force we require T = m * 2.5 m/s^2 + m g = m * 2.5 m/s^2 + m * 9.8 m/s^2 = m * 12.3 m/s^2. We know that the line breaks, so the tension must exceed the 22 N breaking strength of the line. So T > 22 N. Thus m * 12.3 m/s^2 > 22 N. Solving this inequality for m we get m > 22 N / (12.3 m/s^2) = 22 kg m/s^2 / (12.3 m/s^2) = 1.8 kg. The fish has a mass exceeding 1.8 kg.
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RESPONSE --> ok
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18:17:37 univ phy 4.38*parachutist 55 kg with parachute, upward 620 N force. What are the weight and acceleration of parachutist?
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RESPONSE --> W=539N b/c 55*9.8=539, a=1.47m/s b/c 620-539=81, 81/55=1.47
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18:18:09 Describe the free body diagram you drew.
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RESPONSE --> down arrow for weight, up arrow for upward force, net force is difference between the two
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18:19:05 The weight of the parachutist is 55 kg * 9.8 m/s^2 = 540 N, approx.. So the parachutist experiences a downward force of 540 N and an upward force of 620 N. Choosing upward as the positive direction the forces are -540 N and + 620 N, so the net force is -540 + 620 N = 80 N. Your free body diagram should clearly show these two forces, one acting upward and the other downward. The acceleration of the parachutist is a = Fnet / m = +80 N / (55 kg) = 1.4 m/s^2, approx..
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RESPONSE --> ok
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18:43:39 univ phy (4.34 10th edition) fish on balance, reading 60 N when accel is 2.45 m/s^2 up; find true weight, situation when balance reads 30 N, balance reading when cable breaks What is the net force on the fish when the balance reads 60 N? What is the true weight of the fish, under what circumstances will the balance read 30 N, and what will the balance read when the cable breaks?
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RESPONSE --> true wt=48.0N b/c 9.8+2.45=12.25, 60/12.25=4.90, 4.90*9.8=48.0, net force=12N b/c 60-48=12, balance reads 30N when the acceleration is 3.68m/s^2 down b/c 30/4.90=6.12, 9.8-6.12=3.68, when the cable breaks, the balance reads 0 because the balance is in free fall.
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18:48:13 ** Weight is force exerted by gravity. Net force is Fnet = m * a. The forces acting on the fish are the 50 N upward force exerted by the cable and the downward force m g exerted by gravity. So m a = 50 N - m g, which we solve for m to get m = 50 N / (a + g) = 50 N / (2.45 m/s^2 + 9.8 m/s^2) = 50 N / 12.25 m/s^2 = 4 kg. If the balance reads 30 N then Fnet is still m * a and we have m a = 30 N - m g = 30 N - 4 kg * 9.8 m/s^2 = -9.2 N so a = -9.2 N / (4 kg) = -2.3 m/s^2; i.e., the elevator is accelerating downward at 2.3 m/s^2. If the cable breaks then the fish and everything else in the elevator will accelerate downward at 9.8 m/s^2. Net force will be -m g; net force is also Fbalance - m g. So -m g = Fbalance - m g and we conclude that the balance exerts no force. So it reads 0. **
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RESPONSE --> You said 60N in the problem but 50N in the answer. Otherwise, our answers would have been about the same.
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18:57:11 STUDENT QUESTION: I had trouble with the problems involving tension in lines. For example the Fish prob. Prob#9 A person yanks a fish out of water at 4.5 m/s^2 acceleration. His line is rated at 22 Newtons Max, His line breaks, What is the mass of the fish. Here's what I did. Sum of F = Fup + F down -22 N = 4.5 m/s^2 * m(fish) - 9.8 m/s^2 * m(fish) -22N = -5.3 m/s^2 m(fish) m(fish) = 4.2 kg I know its wrong, I just don't know what to do.I had the same problem with the elevator tension on problem 17.
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RESPONSE --> 1.54kg b/c 9.8+4.5=14.3, 22/14.3=1.54
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18:57:44 ** Think in terms of net force. The net force on the fish must be Fnet = m a = m * 4.5 m/s^2. Net force is tension + weight = T - m g, assuming the upward direction is positive. So T - m g = m a and T = m a + m g. Factoring out m we have T = m ( a + g ) so that m = T / (a + g) = 22 N / (4.5 m/s^2 + 9.8 m/s^2) = 22 N / (14.3 m/s^2) = 1.8 kg, approx.. The same principles apply with the elevator. **
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RESPONSE --> ok
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zyߪ^x assignment #014 yXVH Physics I 09-25-2006
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19:20:44 set 3 intro prob sets If you calculate the acceleration on a mass m which starts from rest under the influence of a constant net force Fnet and multiply by a time interval `dt what do you get? How far does the object travel during this time and what velocity does it attain? What do you get when you multiply the net force by the distance traveled? What kinetic energy does the object attain?
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RESPONSE --> a) final velocity because vf=at b) df=.5*at^2 c) vf=at d) F*`ds=PE e) KE=.5*m*v^2
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19:27:40 **STUDENT ANSWER AND INSTRUCTOR COMMENTS: a*'dt = the final velocity if V0=0. to get the change in position you would divide the final velocity(since V0=0) by 2 to get the average velocity and then multiply that by the 'dt to get the units of distance traveled. Multiply that by the 'dt to get the units of distance traveled. It attains a Vf of a*'dt as shown above because V0=0, if V0 was not zero you would have to add that to the a*'dt to get the final velocity. When you multiply Fnet by 'dt you get the same thing you would get if you multiply the mass by the change in velocity(which in this case is the same as the final velocity). This is the change in momentum. The Kinetic Energy Attained is the forcenet multiplied by the change in time. a = Fnet / m. So a `dt = Fnet / m * `dt = vf. The object travels distance `ds = v0 `dt + .5 a `dt^2 = .5 Fnet / m * `dt^2. When we multiply Fnet * `ds you get Fnet * ( .5 Fnet / m * `dt^2) = .5 Fnet^2 `dt^2 / m. The KE attained is .5 m vf^2 = .5 m * ( Fnet / m * `dt)^2 = .5 Fnet^2 / m * `dt^2. Fnet * `ds is equal to the KE attained. The expression for the average velocity would be [ (v0 + a * `dt) + v0 ] / 2 = v0 + 1/2 a `dt so the displacement would be (v0 + 1/2 a `dt) * `dt = v0 `dt + 1/2 a `dt^2. This is equal to (v0 `dt + 1/2 a `dt^2) * Fnet = (v0 `dt + 1/2 a `dt^2) * m a , since Fnet = m a. **
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RESPONSE --> Ok. It helps me to see the derivation of the equations of motion.
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19:30:10 Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system. How does PE come into this relationship?
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RESPONSE --> PE=KE when work is done against conservative forces, but KE
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19:30:45 ** The work done by the system against all forces will decrease the KE by an equal amount. If some of the forces are conservative, then work done against them increases the PE and if PE later decreases this work will be recovered. Work done against non-conservative forces is not stored and cannot be recovered. STUDENT RESPONSE WITH INSTRUCTOR COMMENTARY: The work done by a system against nonconservative forces is the work done to overcome friction in a system- which means energy is dissipated in the form of thermal energy into the 'atmosphere.' Good. Friction is a nonconservative force. However there are other nonconservative forces--e.g., you could be exerting a force on the system using your muscles, and that force could be helping or hindering the system. A rocket engine would also be exerting a nonconservative force, as would just about any engine. These forces would be nonconservative since once the work is done it can't be recovered. STUDENT RESPONSE WITH INSTRUCTOR COMMENTS: The work done by a system against conservative forces is like the work to overcome the mass being pulled by gravity. INSTRUCTOR COMMENT: not bad; more generally work done against conservative force is work that is conserved and can later be recovered in the form of mechanical energy **
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RESPONSE --> ok
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19:42:57 class notes: rubber band and rail How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest?
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RESPONSE --> The work done to stretch the rubber band equals the work done by the rubber band on the rail because the force is transmitted along the length of the rubber band. The work done by the rail against friction from release of the rubber band to the rail coming to rest, by contrast, is a nonconservative force; therefore, energy is lost out of the system to friction during the release of the rubber band.
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19:51:16 ** The work done to stretch the rubber band would in an ideal situation be available when the rubber band is released. Assuming that the only forces acting on the rail are friction and the force exerted by the rubber band, the work done by the rail against friction, up through the instant the rail stops, will equal the work done by the rubber band on the rail. Note that in reality there is some heating and cooling of the rubber band, so some of the energy gets lost and the rubber band ends up doing less work on the rail than the work required to stretch it. **
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RESPONSE --> I see. The work done by the rail is the same as the work done on the rail because the friction is taken into account both times.
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19:53:00 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> In order to maintain conservation of energy in the system, force times distance must yield an equal energy product for the rail and the rubber band alike.
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19:57:45 ** The F `ds total of the rail when it is accelerated by the rubber band is equal Fave `ds, which is equal to to m * aAve * `ds. Here aAve is the average acceleration of the rail by the rubber band. 2 aAve `ds = vf^2 - v0^2 by the fourth equation of motion. So the F `ds total is proportional to the change in v^2. The rail is then stopped by the frictional force f; since f `ds is equal to m * a * `ds, where a is the acceleration of the sliding rail, it follows that f `ds is also proportional to the change in v^2. Change in v^2 under the influence of the rubber band (rest to max vel) is equal and opposite to the change in v^2 while sliding against friction (max vel back to rest), so work f `ds done by friction must be equal and opposite to F `ds. This ignores the small work done by friction while the rubber band is accelerating the rail. **
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RESPONSE --> I should have further specified what the components of each force were (i.e., m*aAve*`ds and m*-aAve*`ds) to fully explain why the proportionality exists.
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20:02:48 gen phy A person of mass 66 kg crouches then jumps to a height of .8 meters. From the crouches position to the point where the person leaves the ground the distance is 20 cm. What average force is exerted over this 20-cm distance?
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RESPONSE --> 390J b/c .8-.2=.6, 66*9.8*.6=390
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06:37:24 ** the normal force is the force between and perpendicular to the two surfaces in contact, which would be 646.8N if the jumper was in equilibrium. However during the jump this is not the case, and the normal force must be part of a net force that accelerates the jumper upward. In a nutshell the net force must do enough work to raise the person's weight 1 meter while acting through only a .2 meter displacement, and must therefore be 5 times the person's weight. The person still has to support his weight so the normal force must be 6 times the person's weight. The detailed reasoning is as follows: To solve this problem you have to see that the average net force on the jumper while moving through the `dy = 20 cm vertical displacement is equal to the sum of the (upward) average normal force and the (downward) gravitational force: Fnet = Fnormal - m g. This net force does work sufficient to increase the jumper's potential energy as he or she rises 1 meter (from the .20 m crouch to the .8 m height). So Fnet * `dy = PE increase, giving us ( Fnormal - m g ) * `dy = PE increase. PE increase is 66 kg * 9.8 m/s^2 * 1 meter = 650 Joules approx. m g = 66 kg * 9.8 m/s^2 = 650 Newtons, approx.. As noted before `dy = 20 cm = .2 meters. So (Fnormal - 650 N) * .2 meters = 650 Joules Fnormal - 650 N = 650 J / (.2 m) Fnormal = 650 J / (.2 m) + 650 N = 3250 N + 650 N = 3900 N. An average force of 3900 N is required to make this jump from the given crouch. This is equivalent to the force exerted by a 250-lb weightlifter doing a 'squat' exercise with about 600 pounds on his shoulders. It is extremely unlikely that anyone could exert this much force without the additional weight. A 20-cm crouch is only about 8 inches and vertical jumps are typically done with considerably more crouch than this. With a 40-cm crouch such a jump would require only half this total force and is probably feasible. **
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RESPONSE --> I see. You get 1 meter total height in the PE calculation because the jumper is already .2m above the ground before jumping .8m. You then use this PE to find out how much normal force is required to overcome gravity and propel .2m upward.
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02:56:01 univ phy text prob 4.42 (4.40 in 10th edition) Mercury lander near surface upward thrust 25 kN slows at rate 1.2 m/s^2; 10 kN speeds up at .8 m/s^2; what is weight at surface?
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RESPONSE --> This is hard to interpret. I need to find the spacecraft weight. I know what the deceleration is on this unknown mass if an upward thrust of 25kN is applied (-a given F). I don't know what this means: 10 kN speeds up at .8 m/s^2. Does that mean a 10kN upward thrust on the same unknown mass fails to overcome gravity and the spacecraft falls at .8m/s^2? If so, I know the thrust equal to the gravitational pull must be between 10kN and 25kN. So, a=0 when F=1/3 the distance between the two values, or when F=15kN
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02:57:17 ** If a landing craft slows then its acceleration is in the direction opposite to its motion, therefore upward. If it speeds up while landing that its acceleration is in the direction of its motion, therefore downward. If the upward motion is taken as the positive direction, then the acceleration under a thrust of 25 kN is + 1.2 m/s^2, and the acceleration when under thrust of 10 kN is - .8 m/s^2. In either case m * a = net force. Net force is thrust force + gravitational force. 1 st case, net force is 25 kN so m * 1.2 m/s/s + m * g = 25 kN. 1 st case, net force is 10 kN so m * (-.8 m/s/s ) + m * g = 10 kN. Solve these equations simultaneously to get the weight m * g (multiply 1 st eqn by 2 and 2d by 3 and add equations to eliminate the first term on the left-hand side of each equation; solve for m * g). The solution is m * g = 16,000 kN. Another solution: In both cases F / a = m so if upward is positive and weight is wt we have (25 kN - wt) / (1.2 m/s^2) = m and (10 kN - wt) / (-.8 m/s^2) = m so (25 kN - wt) / (1.2 m/s^2) = (10 kN - wt) / (-.8 m/s^2). Solving for wt we get 16 kN. **
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RESPONSE --> ok, accurate within 1kN.
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dߟЈˬǂH Student Name: assignment #014
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06:46:07 `q001. An automobile of mass 1500 kg coasts from rest through a displacement of 200 meters down a 3% incline. How much work is done on the automobile by its weight component parallel to the incline? If no other forces act in the direction of motion (this assumes frictionless motion, which is of course not realistic but we assume it anyway because this ideal situation often gives us valuable insights which can then be modified to situations involving friction), what will be the final velocity of the automobile?
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RESPONSE --> 88200J b/c 1500*9.8=14700, 14700*.03=441, 441*200=88200. vf=sqrt.((88200*2)/1500)=10.8m/s
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06:46:11 The weight of the automobile is 1500 kg * 9.8 m/s^2 = 14,700 Newtons. The weight component parallel to the incline is therefore very close to the small-slope approximation weight * slope = 14700 Newtons * .03 = 441 Newtons (small-slope approximation). If no other forces act parallel to the incline then the net force will be just before 441 Newtons and the work done by the net force will be `dWnet = 441 Newtons * 200 meters = 88200 Joules. [ Note that this work was done by a component of the gravitational force, and that it is the work done on the automobile by gravity. ] The net work on a system is equal to its change in KE. Since the automobile started from rest, the final KE will equal the change in KE and will therefore be 88200 Joules, and the final velocity is found from 1/2 m vf^2 = KEf to be vf = +_`sqrt(2 * KEf / m) = +-`sqrt(2 * 88200 Joules / (1500 kg) ) = +-`sqrt(2 * 88200 kg m^2/s^2 / (1500 kg) ) = +- 10.9 m/s (approx.). Since the displacement down the ramp is regarded as positive and the automotive will end up with a velocity in this direction, we choose the +10.9 m/s alternative.
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RESPONSE --> ok
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06:53:06 `q002. If the automobile in the preceding problem is given an initial velocity of 10.9 m/s up the ramp, then if the only force acting in the direction of motion is the force of gravity down the incline, how much work must be done by the gravitational force in order to stop the automobile? How can this result be used, without invoking the equations of motion, to determine how far the automobile travels up the incline before stopping?
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RESPONSE --> -89100J b/c .5*1500*-(10.9^2)=89100. 202m b/c 1500*9.8*.03=441, 89100/441=202.
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06:53:14 The automobile starts out with kinetic energy KEinit = 1/2 m v0^2 = 1/2 (1500 kg) ( 10.9 m/s)^2 = 88000 kg m^2/s^2 = 88000 Joules. The gravitational force component parallel to the incline is in this case opposite to the direction of motion so that gravity does negative work on the automobile. Since the change in KE is equal to the work done by the net force, if the gravitational force component parallel to the incline does negative work the KE of the object will decrease. This will continue until the object reaches zero KE. As found previously the gravitational force component along the incline has magnitude 441 Newtons. In this case the forces directed opposite to the direction motion, so if the direction up the incline is taken to be positive this force component must be -441 N. By the assumptions of the problem this is the net force exerted on the object. Acting through displacement `ds this force will therefore do work `dWnet = Fnet * `ds = -441 N * `ds. Since this force must reduce the KE from 88,000 Joules to 0, `dWnet must be -88,000 Joules. Thus -441 N * `ds = -88,000 Joules and `ds = -88,000 J / (-441 N) = 200 meters (approx.). Had the arithmetic been done precisely, using the precise final velocity found in the previous exercise instead of the 10.9 m/s approximation, we would have found that the displacement is exactly 200 meters.
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RESPONSE --> ok
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06:55:32 `q003. If the automobile in the previous example rolls from its maximum displacement back to its original position, without the intervention of any forces in the direction of motion other than the parallel component of the gravitational force, how much of its original 88200 Joules of KE will it have when it again returns to this position?
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RESPONSE --> 88200J b/c 1500*9.8=14700, 14700*.03=441, 441*200=88200
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06:55:39 In the first exercise in the present series of problems related to this ramp, we found that when the automobile coasts 200 meters down the incline its KE increases by an amount equal to the work done on it by the gravitational force, or by 88200 Joules. Thus the automobile will regain all of its 88200 Joules of kinetic energy. To summarize the situation here, if the automobile is given a kinetic energy of 88200 Joules at the bottom of the ramp then if it coasts up the ramp it will coast until gravity has done -88200 Joules of work against it, leaving it with 0 KE. Coasting back down the ramp, gravity works in the direction of motion and therefore does 88200 Joules of work on it, thereby increasing its KE back to its original 88200 Joules.
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RESPONSE --> ok
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06:57:30 `q004. Explain why, in the absence of friction or other forces other than the gravitational component parallel to incline, whenever an object is given a kinetic energy in the form of a velocity up the incline, and is then allowed to coast to its maximum displacement up the incline before coasting back down, that object will return to its original position with the same KE it previously had at this position.
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RESPONSE --> Because the total energy is this case is always the proportion of kinetic to potential energy. Whenever one is minimum, the other is maximum.
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06:58:52 The car initially had some KE. The gravitational component parallel to the incline is opposite to the direction of motion and therefore does negative work as the object travels up the incline. The gravitational component is the net force on the object, so the work done by this net force causes a negative change in KE, which eventually decreases the KE to zero so that the object stops for an instant. This happens at the position where the work done by the net force is equal to the negative of the original KE. The gravitational component parallel to the incline immediately causes the object to begin accelerating down the incline, so that now the parallel gravitational component is in the same direction as the motion and does positive work. At any position on the incline, the negative work done by the gravitational component as the object traveled up the incline from that point, and the positive work done as the object returns back down the incline, must be equal and opposite. This is because the displacement up the incline and the displacement down the incline are equal and opposite, while the parallel gravitational force component is the same, which makes the Fnet * `ds product must be equal and opposite. Thus when the object reaches its original point, the work done on it by the net force must be equal and opposite to the work done on it while coasting up the incline. Since the work done on the object while coasting up the incline was the negative of the original KE, the work done while coasting down, being the negative of this quantity, must be equal to the original KE. Thus the KE must return to its original value.
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RESPONSE --> This is a more detailed explanation. I might have also mentioned the fact that the component of velocity up the incline is opposite and equal to the component of gravity down the incline.
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06:59:37 `q005. As the object travels up the incline, does gravity do positive or negative work on it? Answer the same question for the case when the object travels down the incline.
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RESPONSE --> Negative up the incline because the motion is opposite to the direction of gravity. Positive down the incline because the motion is in the direction of gravity.
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06:59:44 As the object travels up the incline the net force is directed opposite its direction of motion, so that Fnet and `ds have opposite signs and as a result `dWnet = Fnet * `ds must be negative. As the object travels down the incline the net force is in the direction of its motion so that Fnet and `ds have identical signs and is a result `dWnet = Fnet * `ds must be positive.
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RESPONSE --> ok
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07:01:26 `q006. If positive work is done on the object by gravity, will it increase or decrease kinetic energy of the object? Answer the same question if negative work is done on the object by gravity.
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RESPONSE --> a) Increase because kinetic energy is motion energy, and positive work increases the object's motion. b) Decrease because kinetic energy is motion energy and negative work decreases the object's motion.
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07:01:31 The KE change of an object must be equal to the work done on by the net force. Therefore if positive work is done on an object by the net force its KE must increase, and if negative work is done by the net force the KE must decrease.
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RESPONSE --> ok
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07:03:07 `q007. While traveling up the incline, does the object do positive or negative work against gravity? Answer the same question for motion down the incline.
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RESPONSE --> a) Positive work because the object moves forward to oppose gravity. b) Negative work because the object moves backward with the pull of gravity.
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07:04:19 If the object ends up in the same position as it began, the work done on the object by gravity and work done by the object against gravity must be equal and opposite. Thus when the object does positive work against gravity, as when it travels up the incline, gravity is doing negative work against the object, which therefore tends to lose kinetic energy. When the object does negative work against gravity, as when traveling down the incline, gravity is doing positive work against the object, which therefore tends to gain kinetic energy.
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RESPONSE --> ok
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07:12:38 `q008. Suppose that the gravitational force component exerted parallel to a certain incline on an automobile is 400 Newtons and that the frictional force on the incline is 100 Newtons. The automobile is given an initial KE of 10,000 Joules up the incline. How far does the automobile coast up the incline before starting to coast back down, and how much KE does it have when it returns to its starting point?
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RESPONSE --> a) 20m b/c 400+100=500, 10000/500=20 b)10000J b/c 500*20=10000
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07:14:03 The net force on the automobile as it climbs the incline is the sum of the 400 Newton parallel component of the gravitational force, which is exerted down the incline, and the 100 Newton frictional force, which while the automobile is moving up the incline is also exerted down the incline. Thus the net force is 500 Newtons down the incline. This force will be in the direction opposite to the displacement of the automobile up the incline, and will therefore result in negative work being done on the automobile. When the work done by this force is equal to -10,000 Joules (the negative of the original KE) the automobile will stop for an instant before beginning to coast back down the incline. If we take the upward direction to be positive the 500 Newton force must be negative, so we see that -500 Newtons * `ds = -10,000 Joules so `ds = -10,000 Joules / -500 Newtons = 20 Newtons meters / Newtons = 20 meters. After coasting 20 meters up the incline, the automobile will have lost its original 10,000 Joules of kinetic energy and will for an instant be at rest. The automobile will then coast 20 meters back down the incline, this time with a 400 Newton parallel gravitational component in its direction of motion and a 100 Newton frictional force resisting, and therefore in the direction opposite to, its motion. The net force will thus be 300 Newtons down the incline. The work done by the 300 Newton force acting parallel to the 20 m downward displacement will be 300 Newtons * 20 meters = 6,000 Joules. This is 4,000 Joules less than the when the car started. This 4,000 Joules is the work done during the entire 40-meter round trip against a force of 100 Newtons which every instant was opposed to the direction of motion (100 Newtons * 40 meters = 4,000 Newtons). As the car coasted up the hill the frictional force was downhill and while the car coasted down friction was acting in the upward direction. Had there been no force other than the parallel gravitational component, there would have been no friction or other nongravitational force and the KE on return would have been 10,000 Joules.
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RESPONSE --> I should have calculated force down the incline as 400-100, otherwise ok.
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07:15:52 `q009. The 1500 kg automobile in the original problem of this section coasted 200 meters, starting from rest, down a 3% incline. Thus its vertical displacement was approximately 3% of 200 meters, or 6 meters, in the downward direction. Recall that the parallel component of the gravitational force did 88200 Joules of work on the automobile. This, in the absence of other forces in the direction of motion, constituted the work done by the net force and therefore gave the automobile a final KE of 88200 Joules. How much KE would be automobile gain if it was dropped 6 meters, falling freely through this displacement?
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RESPONSE --> 88200J b/c 1500*9.8*6=88200
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07:15:56 Gravity exerts a force of gravitational force = 1500 kg * 9.8 m/s^2 = 14700 N on the automobile. This force acting parallel to the 6 meter displacement would do `dW by gravitational force = 14700 N * 6 meters = 88200 Joules of work on the automobile. Note that this is identical to the work done on the automobile by the parallel component of the gravitational force in the original problem.
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RESPONSE --> ok
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07:17:22 `q010. When the automobile was 200 meters above the lower end of the 3% incline, it was in a position to gain 88200 Joules of energy. An automobile positioned in such a way that it may fall freely through a distance of 6 meters, is also in a position to gain 88200 Joules of energy. The 6 meters is the difference in vertical position between start and finish for both cases. How might we therefore be justified in a conjecture that the work done on an object by gravity between two points depends only on the difference in the vertical position between those two points?
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RESPONSE --> It appears that an object can follow any descending slope from one height to another because the horizontal component of the path taken does not factor into the energy calculation.
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07:17:33 This is only one example, but there is no reason to expect that the conclusion would be different for any other small incline. Whether the same would be true for a non-constant incline or for more complex situations would require some more proof. However, it can in fact be proved that gravitational forces do in fact have the property that only change in altitude (or a change in distance from the source, which in this example amounts to the same thing) affects the work done by the gravitational force between points.
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RESPONSE --> ok
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07:20:49 `q011. If an object has a way to move from a higher altitude to a lower altitude then it has the potential to increase its kinetic energy as gravity does positive work on it. We therefore say that such an object has a certain amount of potential energy at the higher altitude, relative to the lower altitude. As the object descends, this potential energy decreases. If no nongravitational force opposes the motion, there will be a kinetic energy increase, and the amount of this increase will be the same as the potential energy decrease. The potential energy at the higher position relative to the lower position will be equal to the work that would be done by gravity as the object dropped directly from the higher altitude to the lower. A person holds a 7-kg bowling ball at the top of a 90-meter tower, and drops the ball to a friend halfway down the tower. What is the potential energy of the ball at the top of the tower relative to the person to whom it will be dropped? How much kinetic energy will a bowling ball have when it reaches the person at the lower position, assuming that no force acts to oppose the effect of gravity?
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RESPONSE --> a) 6200J b/c 7*9.8*90=6200 b) 6200J b/c KE at the bottom of a fall equals PE at the top of the fall neglecting other forces.
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07:21:53 The potential energy of the ball at the top of the tower is equal to the work that would be done by gravity on the ball in dropping from the 90-meter altitude at the top to the 45-meter altitude at the middle of the tower. This work is equal to product of the downward force exerted by gravity on the ball, which is 7 kg * 9.8 m/s^2 = 68.6 Newtons, and the 45-meter downward displacement of the ball. Both force and displacement are in the same direction so the work would be work done by gravity if ball dropped 45 meters: 68.6 Newtons * 45 meters = 3100 Joules, approx.. Thus the potential energy of the ball at the top of the tower, relative to the position halfway down the tower, is 3100 Joules. The ball loses 3100 Joules of PE as it drops. If no force acts to oppose the effect of gravity, then the net force is the gravitational force and the 3100 Joule loss of PE will imply a 3100 Joule gain in KE.
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RESPONSE --> I overlooked the fact that the friend was only halfway down the 90m tower. Otherwise ok.
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19:02:54 `q012. If a force such as air resistance acts to oppose the gravitational force, does this have an effect on the change in potential energy between the two points? Would this force therefore have an effect on the kinetic energy gained by the ball during its descent?
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RESPONSE --> a) Yes, air resistance would decrease acceleration and thus total PE b) Yes, final v would be lower due to air resistance, thus so would total KE
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19:04:28 The change in potential energy is determined by the work done on the object by gravity, and is not affected by the presence or absence of any other force. However the change in the kinetic energy of the ball depends on the net force exerted, which does in fact depend on whether nongravitational forces in the direction of motion are present. We can think of the situation as follows: The object loses gravitational potential energy, which in the absence of nongravitational forces will result in a gain in kinetic energy equal in magnitude to the loss of potential energy. If however nongravitational forces oppose the motion, they do negative work on the object, reducing the gain in kinetic energy by an amount equal to this negative work.
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RESPONSE --> I see. Less acceleration with air resistance, but PE remains the same because g is a constant.
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19:06:45 `q014. If an average force of 10 Newtons, resulting from air resistance, acts on the bowling ball dropped from the tower, what will be the kinetic energy of the bowling ball when it reaches the halfway point?
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RESPONSE --> 2700J b/c 7*9.8=69, 69-10=59, 59*45=2700.
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19:07:18 The ball still loses 3100 Joules of potential energy, which in the absence of nongravitational forces would increase its KE by 3100 Joules. However the 10 Newton resisting force does negative work -10 N * 45 m = -450 Joules on the object. The object therefore ends up with KE 3100 J - 450 J = 2650 J instead of the 3100 J it would have in the absence of a resisting force.
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RESPONSE --> ok
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Ο痄|ђ assignment #014 yXVH Physics I 09-27-2006 ܗs{Ḟӣ assignment #015 yXVH Physics I 09-27-2006
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18:57:52 Set 4 probs 1-7 If we know the net force acting on an object and the time during which the force acts, we can find the change in what important quantity?
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RESPONSE --> Change in momentum
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18:57:55 ** You can find the change in the momentum. Fnet * `ds is change in KE; Fnet * `dt is change in momentum. **
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RESPONSE --> ok
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18:58:05 What is the definition of the momentum of an object?
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RESPONSE --> mass times velocity
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18:58:08 ** momentum = mass * velocity. Change in momentum is mass * change in velocity (assuming constant mass). UNIVERSITY PHYSICS NOTE: If mass is not constant then change in momentum is change in m v, which by the product rule changes at rate dp = m dv + v dm. If mass is constant `dm = 0 and dp = m dv so `dp = m * `dv. **
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RESPONSE --> ok
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18:58:41 How do you find the change in the momentum of an object during a given time interval if you know the average force acting on the object during that time interval?
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RESPONSE --> Average force times time interval
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18:58:47 ** Since impulse = ave force * `dt = change in momentum, we multiply ave force * `dt to get change in momentum. **
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RESPONSE --> ok
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19:03:25 How is the impulse-momentum theorem obtained from the equations of uniformly accelerated motion and Newton's Second Law?
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RESPONSE --> Force is mass times acceleration. Acceleration is rate of velocity change. Acceleration times time interval gives velocity. Combined together, they give mass times velocity, which is change in momentum.
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19:03:55 ** First from F=ma we understand that a=F/m. Now if we take the equation of uniformly accelerated motion vf= v0 + a'dt and subtract v0 we get vf-v0 = a'dt. Since vf-v0 = 'dv, this becomes 'dv = a'dt. Now substituting a=F/m , we get 'dv = (F/m)'dt Multiplying both sides by m, m'dv = F'dt **
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RESPONSE --> ok
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19:08:40 If you know the (constant) mass and the initial and final velocities of an object, as well as the time required to change from the initial to final velocity, there are two strategies we can use to find the average force exerted on the object. What are these strategies?
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RESPONSE --> 1) Change in velocity over time interval gives acceleration. Acceleration times mass gives average force. 2) Mass times change in velocity gives momentum. Momentum divided by time gives average force.
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19:08:54 ** The impulse-momentum theorem for constant masses is m `dv = Fave `dt. Thus Fave = m `dv / `dt. We could alternatively find the average acceleration aAve = (vf - v0) / `dt, which we then multiply by the constant mass to get Fave. **
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RESPONSE --> ok
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19:23:32 Class notes #14. How do we combine Newton's Second Law with an equation of motion to obtain the definition of energy?
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RESPONSE --> F=ma, vf^2=vi^2+2ax, vf^2=vi^2+2(F/m)x, Fx=.5mv^2-.5mv^2, or change in energy equals final KE - initial KE.
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19:23:45 ** a = F / m. vf^2 = v0^2 + 2 a `ds. So vf^2 = v0^2 + 2 (Fnet / m) `ds. Multiply by m/2 to get 1/2 mvf^2 = 1/2 m v0^2 + Fnet `ds so Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2--i.e., work = change in KE. **
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RESPONSE --> ok
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19:26:40 What is kinetic energy and how does it arise naturally in the process described in the previous question?
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RESPONSE --> Kinetic energy is the energy of motion. It arises whenever a mass is set into motion by a force, as in the previous question.
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19:27:20 ** KE is the quantity 1/2 m v^2, whose change was seen in the previous question to be equal to the work done by the net force. **
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RESPONSE --> Work done by net force would have been good for me to mention.
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19:29:38 What forces act on an object as it is sliding up an incline?
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RESPONSE --> Gravity and friction are resisting the motion up the incline. The normal force is pushing back against the object at an angle perpendicular to the surface. The remaining force overcomes gravity and friction to move the object up the incline.
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19:29:53 ** Gravitational force can be broken into two components, one parallel and one perpendicular to the ramp. The normal force exerted by the ramp is an elastic force, and unless the ramp breaks the normal force is equal and opposite to the perpendicular component of the gravitational force. Frictional force arises from the normal force between the two surfaces, and act in the direction opposed to motion. The gravitational force is conservative; all other forces in the direction of motion are nonconservative. COMMON ERROR: The Normal Force is in the upward direction and balances the gravitational force. COMMENT: The normal force is directed only perpendicular to the incline and is in the upward direction only if the incline is horizontal. The normal force cannot balance the gravitational force if the incline isn't horizontal. Friction provides a component parallel to the incline and opposite to the direction of motion. **
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RESPONSE --> ok
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19:34:49 For an object sliding a known distance along an incline how do we calculate the work done on the object by gravity? How do we calculate the work done by the object against gravity?
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RESPONSE --> work done on the object by gravity: mass times component of gravity on the incline times (+)distance. work done by the object against gravity: mass times component of gravity on the incline times (-)distance.
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19:42:35 ** The gravitational force is m * g directly downward, where g is the acceleration of gravity. m * g is the weight of the object. If we know change in vertical position then we can simply multiply weight m * g with the vertical displacement `dy, being careful to keep track of which is positive and/or negative. Alternatively it is instructive to consider the forces in the actual direction of motion along the incline. For small inclines the component of the gravitational force which is parallel to the incline is approximately equal to the product of the weight and the slope of the incline, as seen in experiments. The precise value of the component parallel to the incline, valid for small as well as large displacements, is m g * sin(theta), where theta is the angle of the incline with horizontal. This force acts down the incline. If the displacement along the incline is `ds, measured with respect to the downward direction, then the work done by gravity is the product of force and displacement, m g sin(theta) * `ds. If `ds is down the incline the gravitational component along the incline is in the same direction as the displacement and the work done by gravity on the system is positive and, in the absence of other forces in this direction, the KE of the object will increase. This behavior is consistent with our experience of objects moving freely down inclines. If the displacement is upward along the incline then `ds is in the opposite direction to the gravitational force and the work done by gravity is negative. In the absence of other forces in the direction of the incline this will result in a loss of KE, consistent with our experience of objects coasting up inclines. The work done against gravity is the negative of the work done by gravity, positive for an object moving up an incline (we have to use energy to get up the incline) and negative for an object moving down the incline (the object tends to pick up energy rather than expending it) **
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RESPONSE --> I should have specified that work done by gravity is positive only if down is defined as the positive direction and the object does not overcome the downward pull of gravity. Vice versa for work done against gravity.
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19:47:09 For an object sliding a known distance along an incline how do we calculate the work done by the object against friction? How does the work done by the net force differ from that done by gravity?
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RESPONSE --> Mass times acceleration up the incline must be subtracted from the friction force to give the actual force up the ramp. Friction differs from gravity because it is a nonconservative force, meaning it dissipates energy out of the system.
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19:52:55 ** The work done against friction is the product of the distance moved and the frictional force. Since the force exerted by friction is always opposed to the direction of motion, the force exerted by the system against friction is always in the direction of motion so the work done against friction is positive. The net force on the system is sum of the gravitational component parallel to the incline and the frictional force. The work done by the net force is therefore equal to the work done by gravity plus the work done by the frictional force (in the case of an object moving up an incline, both gravity and friction do negative work so that the object must do positive work to overcome both forces; in the case of an object moving down an incline gravity does positive work on the system while friction, as always, does negative work on the system; in the latter case depending on whether the work done by gravity on the system is greater or less than the frictional work done against the system the net work done on the system may be positive or negative) **
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RESPONSE --> I should have also mentioned F(friction)*`ds=Work(friction); that Work(friction) is always positive unlike gravity; and that an object must overcome both gravity and friction to progress upward.
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20:10:23 Explain why the restoring force on a simple pendulum is in nearly the same proportion to the weight of the pendulum as its displacement from equilibrium to its length, and explain which assumption is made that makes this relationship valid only for displacements which are small compared to pendulum length.
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RESPONSE --> The longer a pendulum is, the farther distance it must travel to return to equilibrium from a given degree of displacement. Likewise, the heavier a pendulum is, the greater the gravitational force acting to pull the pendulum back toward equilibrium. This relationship is valid only for displacements which are small compared to pendulum length because as the displacement approaches pendulum length, the center of mass of the pendulum is displaced closer to the point of attachment and the weight on the end of the pendulum is less of a factor.
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06:04:34 ** In terms of similar triangles: The reason the approximation only works for small displacements is because the sides used on one triangle are not the same as the sides used on the other. From the triangle we see that the restoring force and the weight are at right angles, while the length and horizontal displacement of the pendulum from equilibrium are the hypotenuse and the horizontal leg of a triangle and hence are not at right angles. For small angles the two long sides of the triangle are approximately equal so the discrepancy doesn't make much difference. For larger angles where the two long sides are significantly different in length, the approximation no longer works so well. In terms of components of the vectors: The tension force is in the direction of the string. The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement (just draw the picture). The vertical component of the tension force must be equal to the weight of the pendulum, since the pendulum is in equilibrium. If the displacement is small compared to the length the vertical component of the tension force will be very nearly equal to the tension force. So the previous statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the total tension as the length of the pendulum to the horizontal displacement' can be replaced by the statement that 'The component of the tension force in the horizontal direction is therefore seen geometrically to be in the same proportion to the weight of the pendulum as the length of the pendulum to the horizontal displacement. **
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RESPONSE --> Drawing the right triangle diagram helps me see the comparison between the horizontal tension component:total tension ratio and the pendulum length:horizontal displacement ratio. I also see how the components become increasingly less equal further away from equilibrium.
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06:13:35 prin and gen phy: 6.4: work to push 160 kg crate 10.3 m, horiz, no accel, mu = .50.
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RESPONSE --> 8090J b/c 160*9.8=1570, 1570*.5=785, 785*10.3=8090.
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06:13:48 The net force on the crate must be zero, since it is not accelerating. The gravitational force on the crate is 160 kg * 9.8 m/s^2 = 1570 N, approx. The only other vertical force is the normal force, which must therefore be equal and opposite to the gravitational force. As it slides across the floor the crate experiences a frictional force, opposite its direction of motion, which is equal to mu * normal force, or .50 * 1570 N = 780 N, approx.. The only other horizontal force is exerted by the movers, and since the net force on the crate is zero the movers must be exerting a force of 780 N in the direction of motion. The work the movers do in 10.3 m is therefore work = Fnet * `ds = 780 N * 10.3 m = 8000 N m = 8000 J, approx..
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RESPONSE --> ok
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06:16:38 gen phy prob 6.9: force and work accelerating helicopter mass M at .10 g upward thru dist h.
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RESPONSE --> F=ma=m(9.8+.10)=m*9.9. W=m*9.9*h.
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06:18:43 To accelerate the helicopter at .10 g it must experience net force Fnet = mass * acceleration = M * .10 g = .10 M g. The forces acting on the helicopter are its upward thrust T and the downward pull - M g of gravity, so the net force is T - M g. Thus we have T - M g = .10 M g, and the upward thrust is T = .10 M g + M g = 1.10 M g. To exert this force through an upward displacement h would therefore require work = force * displacement = 1.10 M g * h = 1.10 M g h.
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RESPONSE --> I misread .10g as .10m/s^2, otherwise ok.
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06:26:46 **** Univ: 6.58 (6.50 10th edition). chin-up .40 m, 70 J/kg of muscle mass, % of body mass in pullup muscles of can do just 1. Same info for son whose arms half as long.
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RESPONSE --> I see the distance to perform the chinup is .40m and the muscular energy output is 70J/kg, but the wording past there makes no sense to me.
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06:32:33 ** For each kg of mass the weight is 1 kg * 9.8 m/s^2 = 9.8 N. Work done to lift each kg of mass .4 m would then be 9.8 N * .4 m = 3.92 J. The chin-up muscles generate 3.92 J per kg, which is 3.92 / 70 of the work one kg of muscle mass would produce. So the proportion of body mass in the pullup muscles is 3.92 / 70 = .056, or 5.6%. For the son each kg is lifted only half as far so the son only has to do half the work per kg, or 1.96 J per kg. For the son the proportion of muscle mass is therefore only 1.96 / 70 = 2.8%. The son's advantage is the fact that he is lifting his weight half as high, requiring only half the work per kg. **
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RESPONSE --> I see now that I was supposed get the weight per kg of mass and multiply that by .4m to get the work per kg of mass. By finding what % that is of 70, I get the % if body mass in the pullup. I also see that someone half the size would only perform half the work and the proportion of muscle mass would only be 2.8%.
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06:36:23 Univ. 6.72 (6.62 10th edition). net force 5 N/m^2 * x^2 at 31 deg to x axis; obj moves along x axis, mass .250 kg, vel at x=1.00 m is 4.00 m/s so what is velocity at x = 1.50 m?
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RESPONSE --> Not for gen phy students, I'm guessing. I haven't seen this type of problem before.
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06:36:36 ** Force is variable so you have to integrate force with respect to position. Position is measured along the x axis, so you integrate F(x) = - k / x^2 with respect to x from x1 to x2. An antiderivative of - k / x^2 is k / x so the integral is k / x2 - k / x1. If x2 > x1, then k / x2 < k / x1 and the work is negative. Also, if x2 > x1, then motion is in the positive x direction while F = - k / x^2 is in the negative direction. Force and displacement in opposite directions imply negative work by the force. For slow motion acceleration is negligible so the net force is practically zero. Thus the force exerted by your hand is equal and opposite to the force F = - k / x^2. The work you do is opposite to the work done by the force so will be - (k / x2 - k / x1) = k/x1 - k/x2, which is positive if x2 > x1. This is consistent with the fact that the force you exert is in the opposite direction to the force, therefore in the positive direction, as is the displacement. Note that the work done by the force is equal and opposite to the work done against the force. **
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RESPONSE --> n/a
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sܣ~Jքާ|h Student Name: assignment #015
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08:20:47 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
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RESPONSE --> 15m/s b/c 10/2=5, 5*3=15
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18:42:51 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
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RESPONSE --> ok, Change in mv=30kg m/s Fnet * `dt=30kg m/s. These are equal.
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18:42:56 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
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RESPONSE --> ok
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18:46:55 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
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RESPONSE --> 3000kg m/s b/c 2000*1.5=3000
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18:47:21 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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RESPONSE --> ok
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18:51:01 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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RESPONSE --> 2.4m/s b/c 2000/1200=1.6, 1.6*1.5=2.4
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18:51:09 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
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RESPONSE --> ok
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18:52:39 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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RESPONSE --> 4000N b/c 25-20=5, 5/2=2.5, 2.5*1600=4000
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18:52:42 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE --> ok
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18:52:45 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE -->
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j霴wZ assignment #016 yXVH Physics I 09-28-2006
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08:40:50 Class notes #15 When a projectile rolls off a ramp with its velocity in the horizontal direction, why do we expect that its horizontal range `dx will be proportional to the square root of its vertical displacement `dy rolling down the ramp?
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RESPONSE --> `dy represents how long g acts on the projectile. From the equation on motion vf^2=v0^2+2ax we see that the square of the velocity is proportional to the displacement, so the longer the interval of acceleration, the greater the velocity, and proportionately, the greater the square root of the displacement.
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10:06:48 ** Since the initial vertical velocity is zero the time of fall for a given setup will always be the same. Therefore the horizontal range is proportional to the horizontal velocity of the projectile. The horizontal velocity is attained as a result of vertical displacement `dy, with gravitational PE being converted to KE. PE loss is proportional to `dy, so the KE of the ball as it leaves the ramp will be proportional to `dy. Since KE = .5 m v^2, v is proportional to sqrt( KE ), therefore to sqrt(y). **
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RESPONSE --> I see that horizontal range is proportional to horizontal velocity, horizontal velocity comes from vertical displacement PE converted to KE, where v is proportional to the sqrt of KE and sqrt of vertical displacement.
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10:07:41 In the preceding situation why do we expect that the kinetic energy of the ball will be proportional to `dy?
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RESPONSE --> Because KE is proportional to PE and PE depends upon the vertical displacement over which gravity acts.
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10:12:33 ** This question should have specified just the KE in the vertical direction. The kinetic energy of the ball in the vertical direction will be proportional to `dy. The reason: The vertical velocity attained by the ball is vf = `sqrt(v0^2 + 2 a `ds). Since the initial vertical velocity is 0, for distance of fall `dy we have vf = `sqrt( 2 a `dy ), showing that the vertical velocity is proportional to the square root of the distance fallen. Since KE is .5 m v^2, the KE will be proportional to the square of the velocity, hence to the square of the square root of `dy. Thus KE is proportional to `dy. **
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RESPONSE --> I see how the equations of motion factor in: vf = `sqrt( 2 a `dy ) and KE=.5 m v^2 creating this proportionality.
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10:16:13 Why do we expect that the KE of the ball will in fact be less than the PE change of the ball?
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RESPONSE --> While PE and KE may be equal in ideal conditions, in reality friction may take energy out of the system, reducing the final velocity.
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10:16:35 ** STUDENT RESPONSE: Because actually some of the energy will be dissapated in the rotation of the ball as it drops? INSTRUCTOR COMMENT: Good, but note that rotation doesn't dissipate KE, it merely accounts for some of the KE. Rotational KE is recoverable--for example if you place a spinning ball on an incline the spin can carry the ball a ways up the incline, doing work in the process. The PE loss is converted to KE, some into rotational KE which doesn't contribute to the range of the ball and some of which simply makes the ball spin. ANOTHER STUDENT RESPONSE: And also the loss of energy due to friction and conversion to thermal energy. INSTRUCTOR COMMENT: Good. There would be a slight amount of air friction and this would dissipate energy as you describe here, as would friction with the ramp (which would indeed result in dissipation in the form of thermal energy). **
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RESPONSE --> ok
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10:20:43 prin phy and gen phy 6.18 work to stop 1250 kg auto from 105 km/hr
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RESPONSE --> 1.91E9 J b/c 105*1000=105000, 105000/60=1750, .5*1250*1750^2=1.91E9
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10:21:50 The work required to stop the automobile, by the work-energy theorem, is equal and opposite to its change in kinetic energy: `dW = - `dKE. The initial KE of the automobile is .5 m v^2, and before calculating this we convert 105 km/hr to m/s: 105 km/hr = 105 km / hr * 1000 m / km * 1 hr / 3600 s = 29.1 m/s. Our initial KE is therefore KE = .5 m v^2 = .5 * 1250 kg * (29.1 m/s)^2 = 530,000 kg m^2 / s^2 = 530,000 J. The car comes to rest so its final KE is 0. The change in KE is therefore -530,000 J. It follows that the work required to stop the car is `dW = - `dKE = - (-530,000 J) = 530,000 J.
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RESPONSE --> I divided by 60 instead of 3600, otherwise ok.
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10:22:53 prin and gen phy 6.26. spring const 440 N/m; stretch required to store 25 J of PE.
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RESPONSE --> .057m b/c 25/440=.057
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10:35:03 The force exerted by a spring at equilibrium is 0, and the force at position x is - k x, so the average force exerted between equilibrium and position x is (0 + (-kx) ) / 2 = -1/2 k x. The work done by the spring as it is stretched from equilibrium to position x, a displacment of x, is therefore `dW = F * `ds = -1/2 k x * x = -1/2 k x^2. The only force exerted by the spring is the conservative elastic force, so the PE change of the spring is therefore `dPE = -`dW = - (-1/2 kx^2) = 1/2 k x^2. That is, the spring stores PE = 1/2 k x^2. In this situation k = 440 N / m and the desired PE is 25 J. Solving PE = 1/2 k x^2 for x (multiply both sides by 2 and divide both sides by k, then take the square root of both sides) we obtain x = +-sqrt(2 PE / k) = +-sqrt( 2 * 25 J / (440 N/m) ) = +- sqrt( 50 kg m^2 / s^2 / ( (440 kg m/s^2) / m) )= +- sqrt(50 / 440) sqrt(kg m^2 / s^2 * (s^2 / kg) ) = +- .34 sqrt(m^2) = +-.34 m. The spring will store 25 J of energy at either the +.34 m or the -.34 m position.
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RESPONSE --> All clear. I used the spring force equation instead of the spring work equation.
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10:42:08 gen phy text problem 6.19 88 g arrow 78 cm ave force 110 N, speed? What did you get for the speed of the arrow?
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RESPONSE --> 57m/s b/c 110/.088=1250. sqrt(2*1250*.78)=57
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10:43:31 ** 110 N acting through 78 cm = .78 m does work `dW = 110 N * .78 m = 86 Joules appxo.. If all this energy goes into the KE of the arrow then we have a mass of .088 kg with 86 Joules of KE. We can solve .5 m v^2 = KE for v, obtaining | v | = sqrt( 2 * KE / m) = sqrt(2 * 86 Joules / (.088 kg) ) = sqrt( 2000 kg m^2 / s^2 * 1 / kg) = sqrt(2000 m^2 / s^2) = 44 m/s, approx.. **
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RESPONSE --> I see: F*d, solve KE equation for v.
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10:44:08 query univ phy 6.84 (6.74 10th edition) bow full draw .75 m, force from 0 to 200 N to 70 N approx., mostly concave down. What will be the speed of the .0250 kg arrow as it leaves the bow?
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RESPONSE --> n/a for gen phy?
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10:47:10 ** The work done to pull the bow back is represented by the area beneath the force vs. displacement curve. The curve could be approximated by a piecewise straight line from about 0 to 200 N then back to 70 N. The area beneath this graph would be about 90 N m or 90 Joules. The curve itself probably encloses a bit more area than the straight line, so let's estimate 100 Joules (that's probably a little high, but it's a nice round number). If all the energy put into the pullback goes into the arrow then we have a .0250 kg mass with kinetic energy 100 Joules. Solving KE = .5 m v^2 for v we get v = sqrt(2 KE / m) = sqrt( 2 * 100 Joules / ( .025 kg) ) = sqrt(8000 m^2 / s^2) = 280 m/s, approx. **
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RESPONSE --> n/a but this was simple enough to understand based on drawing and estimating the area under the curve, then solving with KE equation.
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10:56:22 Univ. 6.90 (6.78 10th edition) requires 10-25 watts / kg to fly; 70 g hummingbird 10 flaps/sec, work/wingbeat. Human mass 70 kg, 1.4 kW short period, sustain 500 watts. Fly by flapping?
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RESPONSE --> At least 10 watts/kg needed to fly. Hummingbird is .07kg, needs .7 watts. Human is 70kg, needs 700 watts. How to calculate hummingbird energy output given 10 flaps/second? How much energy per flap? With human, we are told 1.4 kW short period, sustain 500 watts. This is human's maximum energy output? If so, it appears energy output is sufficient for short period flight but not sustained flight.
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11:02:54 ** A 70 gram = .070 kg hummingbird would require between .070 kg * 10 watts / kg = .7 watts = .7 Joules / second in order to fly. At 10 flaps / second that would be .07 Joules per wingbeat. A similar calculation for the 25 watt level shows that .175 Joules would be required per wingbeat. A 70 kg human being would similarly require 700 watts at 10 watts / kg, which would be feasible for short periods (possibly for several minutes) but not for a sustained flight. At the 25 watt/kg level flight would not be feasible. **
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RESPONSE --> For hummingbird, I see .7 watts=.7 J/s, thus .07 J/wingbeat, all else clear.
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ҲVצ[YȎ assignment #015 yXVH Physics I 09-28-2006 t̐KyВ assignment #015 yXVH Physics I 09-28-2006 iѡŲîMƿUG¾ assignment #016 yXVH Physics I 09-28-2006 xʢⓔӽ assignment #016 yXVH Physics I 09-28-2006 ~VrҋԶ{e»ߓ} assignment #017 yXVH Physics I 09-28-2006
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11:29:03 prin phy and gen phy 6.33: jane at 5.3 m/s; how high can she swing up that vine?
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RESPONSE --> 1.4m b/c .5*5.3^2=14, 14/9.8=1.4
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11:29:21 Jane is going to convert her KE to gravitational PE. We assume that nonconservative forces are negligible, so that `dKE + `dPE = 0 and `dPE = -`dKE. Jane's KE is .5 M v^2, where M is her mass. Assuming she swings on the vine until she comes to rest at her maximum height, the change in her KE is therefore `dKE = KEf - KE0 = 0 - .5 M v0^2 = - .5 M v0^2, where v0 is her initial velocity. Her change in gravitational PE is M g `dy, where `dy is the change in her vertical position. So we have `dKE = - `dPE, or -5 M v0^2 = - ( M g `dy), which we solve for `dy (multiply both sides by -1, divide both sides by M g) to obtain `dy = v0^2 / (2 g) = (5.3 m/s)^2 / (2 * 9.8 m/s^2) = 1.43 m.
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RESPONSE --> ok
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11:35:11 prin phy and gen phy 6.39: 950 N/m spring compressed .150 m, released with .30 kg ball. Upward speed, max altitude of ball
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RESPONSE --> .5*950*.15^2=10.7 sqrt(10.7/(.5*.3))=8.45m/s=upward speed, (-8.45^2)/(-2*9.8)=3.64m=max altitude.
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11:50:46 We will assume here that the gravitational PE of the system is zero at the point where the spring is compressed. In this situation we must consider changes in both elastic and gravitational PE, and in KE. The PE stored in the spring will be .5 k x^2 = .5 ( 950 N/m ) ( .150 m)^2 = 107 J. When released, conservation of energy (with only elastic and gravitational forces acting there are no nonconservative forces at work here) we have `dPE + `dKE = 0, so that `dKE = -`dPE. Since the ball is moving in the vertical direction, between the release of the spring and the return of the spring to its equilibrium position, the ball t has a change in gravitational PE as well as elastic PE. The change in elastic PE is -107 J, and the change in gravitational PE is m g `dy = .30 kg * 9.8 m/s^2 * .150 m = +4.4 J. The net change in PE is therefore -107 J + 4.4 J = -103 J. Thus between release and the equilibrium position of the spring, `dPE = -103 J The KE change of the ball must therefore be `dKE = - `dPE = - (-103 J) = +103 J. The ball gains in the form of KE the 103 J of PE lost by the system. The initial KE of the ball is 0, so its final KE is 103 J. We therefore have .5 m vv^2 = KEf so that vf=sqrt(2 KEf / m) = sqrt(2 * 103 J / .30 kg) = 26 m/s. To find the max altitude to which the ball rises, we return to the state of the compressed spring, with its 107 J of elastic PE. Between release from rest and max altitude, which also occurs when the ball is at rest, there is no change in velocity and so no change in KE. No nonconservative forces act, so we have `dPE + `dKE = 0, with `dKE = 0. This means that `dPE = 0. There is no change in PE. The initial PE is 107 J and the final PE must also therefore be 107 J. There is, however, a change in the form of the PE. It converts from elastic PE to gravitational PE. Therefore at maximum altitude the gravitational PE must be 107 J. Since PEgrav = m g y, and since the compressed position of the spring was taken to be the 0 point of gravitational PE, we therefore have y = PEgrav / (m g) = 107 J / (.30 kg * 9.8 m/s^2) = 36.3 meters. The ball will rise to an altitude of 36.3 meters above the compressed position of the spring.
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RESPONSE --> Actually, .5*950*.15^2=10.7 J not 107 J, but my PE calculation still should have taken into account the 4.4J of PE lost before using the PE to calculate initial velocity with the KE equation. For max altitude, I see I should have used y = PEgrav / (m g).
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11:52:23 gen phy problem A high jumper needs to be moving fast enough at the jump to lift her center of mass 2.1 m and cross the bar at a speed of .7 m/s. What minimum velocity does she require?
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RESPONSE --> 6.4 m/s b/c sqrt((9.8*2.1)/.5)=6.4
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11:52:37 FORMAL SOLUTION: Formally we have `dPE + `dKE = 0. `dPE = M * g * `dy = M * 20.6 m^2 / sec^2, where M is the mass of the jumper and `dy is the 2.1 m change in altitude. `dKE = .5 M vf^2 - .5 M v0^2, where vf is the .7 m/s final velocity and v0 is the unknown initial velocity. So we have M g `dy + .5 M vf^2 - .5 M v0^2 = 0. Dividing through by M we have g `dy + .5 vf^2 - .5 v0^2 = 0. Solving for v0 we obtain v0 = sqrt( 2 g `dy + vf^2) = sqrt( 2* 9.8 m/s^2 * 2.1 m + (.7 m/s)^2 ) = sqrt( 41.2 m^2/s^2 + .49 m^2 / s^2) = sqrt( 41.7 m^2 / s^2) = 6.5 m/s, approx.. LESS FORMAL, MORE INTUITIVE, EQUIVALENT SOLUTION: The high jumper must have enough KE at the beginning to increase his PE through the 2.1 m height and to still have the KE of his .7 m/s speed. The PE change is M * g * 2.1 m = M * 20.6 m^2 / sec^2, where M is the mass of the jumper The KE at the top is .5 M v^2 = .5 M (.7 m/s)^2 = M * .245 m^2 / s^2, where M is the mass of the jumper. Since the 20.6 M m^2 / s^2 increase in PE must come at the expense of the initial KE, and since after the PE increase there is still M * .245 m^2 / s^2 in KE, the initial KE must have been 20.6 M m^2 / s^2 + .245 M m^s / s^2 =20.8 M m^s / s^2, approx. If initial KE is 20.8 M m^s / s^2, then .5 M v0^2 = 20.8 M m^s / s^2. We divide both sices of this equation by the jumper's mass M to get .5 v0^2 = 20.8 m^2 / s^2, so that v0^2 = 41.6 m^2 / s^2 and v0 = `sqrt(41.6 m^2 / s^2) = 6.5 m/s, appprox.
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RESPONSE --> ok
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12:25:22 query Univ. 7.42 (7.38 in 10th edition). 2 kg block, 400 N/m spring, .220 m compression. Along surface then up 37 deg incline all frictionless. How fast on level, how far up incline?
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RESPONSE --> .5*400*.22^2=9.68, sqrt(9.68/(.5*2))=3.11m/s on level, 9.8sin 37deg=5.90, 9.68/(2*5.90)=.82m up incline.
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12:30:36 ** The spring exerts a force of 400 N / m * .220 m = 84 N at the .220 m compression. The average force exerted by the spring between equilibrium and this point is therefore (0 N + 84 N) / 2 = 42 N, so the work done in the compression is `dW = Fave * `ds = 42 N * .220 m = 5.0 Joules, approx. If all this energy is transferred to the block, starting from rest, the block's KE will therefore be 5.0 Joules. Solving KE = .5 m v^2 for v we obtain v = sqrt(2 KE / m) = sqrt(2 * 5.0 Joules / (2 kg) ) = 2.2 m/s, approx.. No energy is lost to friction so the block will maintain this speed along the level surface. As it begins to climb the incline it will gain gravitational PE at the expense of KE until the PE is 5.0 J and the KE is zero, at which point it will begin to slide back down the incline. After traveling through displacement `ds along the incline the height of the mass will be `ds sin(37 deg) = .6 `ds, approx., and its gravitational PE will be PE = m g h = m g * .6 `ds = .6 m g `ds. Setting this expression equal to KE we obtain the equation .6 m g `ds = KE, which we solve for `ds to obtain `ds = KE / (.6 m g) = 5.0 Joules / (.6 * 2 kg * 9.8 m/s^2) = .43 meters, approx. **
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RESPONSE --> I should have found average force exerted by the spring, then multiplied by distance to get work done in the compression, which could have given me v from the KE equation and h from the PE equation.
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12:44:24 query univ phy 7.50 62 kg skier, from rest, 65 m high. Frict does -10.5 kJ. What is the skier's speed at the bottom of the slope? After moving horizontally over 82 m patch, air res 160 N, coeff frict .2, how fast is she going? Penetrating 2.5 m into the snowdrift, to a stop, what is the ave force exerted on her by the snowdrift?
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RESPONSE --> 62*9.8*65=39500, 39500-10500=29000, sqrt(29000/(.5*62))=30.6m/s=speed at bottom of slope. 160*82=13100, 608*.2=123, 123*82=10100, 29000-13100-10100=5800, sqrt(5800/(.5*62))=13.7m/s=speed after 82m patch. (-13.7^2)/(2*2.5)=37.5. 62*37.5=2330N=ave force exerted on her by the snowdrift
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12:52:06 ** The gravitational PE of the skier decreases by 60 kg * 9.8 m/s^2 * 65 m = 38 kJ, approx. (this means 38 kiloJoules, or 38,000 Joules). The PE loss partially dissipated against friction, with the rest converted to KE, resulting in KE = 38 kJ / 10.5 kJ = 27.5 kJ. Formally we have `dKE + `dPE + `dWnoncons = 0, where `dWnoncons is the work done by the skier against friction. Since friction does -10.5 kJ of work on the skier, the skier does 10.5 kJ of work against friction and we have `dKE = -`dPE - `dWnoncons = - (-38 kJ) - 10.5 kJ = 27.5 kJ. The speed of the skier at this point will be v = sqrt( 2 KE / m) = sqrt( 2 * 27,500 J / (65 kg) ) = 30 m/s, approx. Over the 82 m patch the force exerted against friction will be .2 * 60 kg * 9.8 m/s^2 = 118 N, approx., so the force exerted against nonconservative forces will be 118 N + 160 N = 280 N approx.. The work done will therefore be `dWnoncons = 280 N * 82 m = 23 kJ, approx., and the skier's KE will be KE = 27.5 kJ - 23 kJ = 4.5 kJ, approx. This implies a speed of v = sqrt( 2 KE / m) = 12 m/s, approx. To stop from this speed in 2.5 m requires that the remaining 4.5 kJ of KE be dissipated in the 2.5 m distance. Thus we have `dW = Fave * `ds, so that Fave = `dW / `ds = 4500 J / (2.5 m) = 1800 N. This is a significant force, about 3 times the weight of the skier, but distributed over a large area of her body will cause a good jolt, but will not be likely to cause injury.**
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RESPONSE --> Ok until last step, when I could have just divided work by distance.
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XWj臭L assignment #016 yXVH Physics I 09-28-2006 vC~] assignment #018 yXVH Physics I 09-28-2006
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13:27:24 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
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RESPONSE --> Vertical initial velocity and displacement is used to find the time interval via df=d0+v0t+.5at^2. Time is used to find horizontal range in the same equation using the initial horizontal velocity.
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13:27:51 ** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
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RESPONSE --> I see how to alternatively use vf^2 = v0^2 + 2a'ds
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13:28:55 Query class notes #17 Why do we expect that in a collision of two objects the momentum change of each object must be equal and opposite to that of the other?
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RESPONSE --> For every action there is an equal and opposite reaction and in a frictionless system, we assume energy is conserved.
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13:29:32 **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding object exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
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RESPONSE --> I see that momentum has nothing directly to do with energy.
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13:32:07 What are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
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RESPONSE --> 1) mass of object 1 before 2) mass of object 2 before 3) velocity of object 1 before 4) velocity of object 2 before 5) mass of combined object after 6) velocity of combined object after These are related by conservation of total momentum before and after collision.
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13:32:21 ** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
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RESPONSE --> ok
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13:39:41 prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
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RESPONSE --> 2(.5*7650*95000^2)=3.68E10J=energy before collision. Since the objects come to rest, all this energy is converted to thermal energy.
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13:40:47 There is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J. This KE is practially all converted to thermal energy.
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RESPONSE --> I didn't convert hours to seconds, otherwise ok.
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13:47:39 Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
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RESPONSE --> 23m/s b/c 9.8*28=275, sqrt(275/.5)=23.
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15:10:29 **GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
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RESPONSE --> I see how to setup using conservation of energy, I also forgot to include friction.
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15:30:24 Univ. 7.74 (7.62 in 10th edition). 2 kg pckg, 53.1 deg incline, coeff kin frict .20, 4 m from spring with const 120 N/m. Speed just before hitting spring? How far compressed? How close to init pos on rebound?
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RESPONSE --> 4sin53.1deg=3.20, 3.20*2*9.8=62.7, 2*9.8*.2=3.92, 62.7-3.92=58.8, sqrt(58.8/(.5*2))=7.67m/s=Speed just before hitting spring, sqrt(58.8/(.5*120))=.990m= How far compressed, 58.8-3.92=54.9. 54.9/(9.8*2)=2.80m=How close to init pos on rebound
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15:36:56 ** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package. The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline. The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx.. The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx.. Friction acts in the direction opposite motion, up the incline in this case. If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is 13.4 N * 4 m = 54 Joules approx. Just before hitting the spring we therefore have .5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s. If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx.. However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed. As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have `dKE + `dPE + `dWnoncons = 0 so -54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation 60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8 meaning 1.07 m or -.8 m (see previous note on units). We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE. If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain `dPE + `dKE + `dWnoncons = 0 so -.5 k x^2 + Fparallel * xMax + 0 + 33 N * xMax = 0 or -.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0 We obtain 18 N * xMax = 72 N m, approx., so that xMax = 72 N m / (18 N) = 4 meters, approx.. This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
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RESPONSE --> First two answers are approximately the same. For the third answer, I see how I could have used the conservation of energy and quadratic equation.
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z˾yƿ`| assignment #018 yXVH Physics I 09-28-2006 ϲKNyK assignment #019 yXVH Physics I 09-28-2006
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16:25:34 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
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RESPONSE --> x component=magnitude*cos(angle) y component=magnitude*sin(angle)
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16:27:10 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
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RESPONSE --> ok
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16:29:58 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
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RESPONSE --> Components represent the horizontal and vertical force that combine together to give the actual force. Therefore, the same force is applied whether using the combined force or the two component forces.
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16:30:03 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
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RESPONSE --> ok
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̽ݢ assignment #019 yXVH Physics I 09-28-2006
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16:32:03 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
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RESPONSE --> already done
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16:32:04 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
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RESPONSE -->
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16:32:06 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
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RESPONSE --> already done
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16:32:09 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
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RESPONSE --> already done
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16:32:16 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
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RESPONSE --> already done
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16:49:22 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
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RESPONSE --> Given a point, I can determine the horizontal and vertical components on a grid, then use the arctan(y/x) to determine the angle, then use the pythagorean theorem to determine the magnitude.
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16:49:31 ** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. **
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RESPONSE --> ok
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16:52:43 Explain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity.
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RESPONSE --> The pythagorean theorem gives us the magnitudes of the components. Using the direction given, we orient to the grid to determine which way the vectors point.
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16:54:33 ** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). **
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RESPONSE --> I confused techniques. I remember now to use sine and cosine functions to obtain components given the magnitude and angle of a vector.
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17:24:49 Univ. 8.58 (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?
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RESPONSE --> sqrt(2/(.5*9.8))=.639, .639*9.8=6.26, .626*.04=.025kg*m/s=impulse, .025/.0002=125N=Fave
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19:25:19 ** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.002 s) = -240 Newtons, approx. **
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RESPONSE --> I should have used vf^2=2ax for v, have taken difference between initial and final velocities to find change in momentum, then force calculation would be correct.
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19:26:02 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I understand better how to work problems involving momentum now
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hjۿоm assignment #019 yXVH Physics I 09-28-2006 ֏rŊxC̪ك assignment # yXVH Physics I 09-28-2006 XHсhЛXLxڭ{p assignment #020 yXVH Physics I 09-28-2006
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20:37:11 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> If you have the components of two vectors, you have four vectors to begin with. You take the sum of the y components and the sum of the x components to get a combined x component and a combined y component. You can also use the arctan(y/x) rule to get the direction of the new vector. You can repeat this process again to find the resultant created by these two new vectors.
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20:37:59 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
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RESPONSE --> ok
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20:39:07 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> x component=magnitude*cos(angle). y component=magnitude*sin(angle).
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20:39:13 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
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RESPONSE --> ok
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20:41:04 prin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?
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RESPONSE --> -25/65=-.38, -.38*20=-7.6m/s change.
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20:41:11 If the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity. By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec. The change in momentum is m * `dv, so we have m `dv = impulse and `dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.
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RESPONSE --> ok
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20:43:44 gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block
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RESPONSE --> (.023*230)=(.023*170)+(2*v), v=.69m/s
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20:43:50 **STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.**
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RESPONSE --> ok
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20:54:46 **** Univ. 8.70 (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet?
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RESPONSE --> .75/.25=3, .5*3*15^2=338, sqrt(338/(.5*.992))=26.1m/s block velocity, sqrt(338/(.5*.008))=291m/s bullet velocity.
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20:56:05 ** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **
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RESPONSE --> Didn't convert cm to m, otherwise ok.
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ƾkyp˰eMݥ}㌁ assignment #020 yXVH Physics I 09-28-2006 `R漑Ň assignment #021 yXVH Physics I 09-28-2006
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21:11:02 Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.
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RESPONSE --> vf^2=2ax gives velocity. vf=at gives time. x=vt gives final horizontal position. Using final horizontal and vertical positions and velocities as x and y components, the hypotenuse gives the combined magnitude and arctan(y/x) gives the direction.
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21:11:15 ** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **
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RESPONSE --> ok
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21:19:33 Give at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.
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RESPONSE --> 1) aircraft taking off has a velocity up and away from the runway. y component is vertical from the ground and x component is horizontal away from the runway. 2) debris from an explosion projects toward the ground following a parabola. y component is acceleration toward the pull of gravity and x component is solely caused by the explosion. 3) plotting a trip through wilderness northeasterly, the x component may be eastward component of the trip and the y component may be northward.
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21:19:42 ** GOOD STUDENT RESPONSE: Examples might include: A force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant.
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RESPONSE --> ok
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ߦZɢvS` assignment #022 yXVH Physics I 09-28-2006
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21:23:56 Query gen phy 7.19 95 kg fullback 4 m/s east stopped in .75 s by tackler due west
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RESPONSE --> 95*4=380, 380/.75=507N=stopping force
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21:26:00 ** We'll take East to be the positive direction. The origional magnitude and direction of the momentum of the fullback is p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East. The magnitude and direction of the impulse exerted on the fullback will therefore be impulse = change in momentum or impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s. Impulse is negative so the direction is in the negative x direction, i.e., West. Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N The direction is in the negative x direction, i.e., West. The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N. The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0). The iimpulse on the tackler is to the East. **
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RESPONSE --> I should also have used +,- to indicate east/west as opposite vectors
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ױɢaם Student Name: assignment #015
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\YvlIxdwږr Student Name: assignment #016
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11:11:54 `q001. Note that this assignment contains 4 questions. . How long does it take for an object dropped from rest to fall 2 meters under the influence of gravity?
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RESPONSE --> .6s b/c sqrt((2*2)/9.8)=.6
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11:12:00 The object has initial velocity 0, acceleration 9.8 meters/second^2, and displacement 2 meters. We can easily solve this problem using the equations of motion. Using the equation `ds = v0 `dt + .5 a `dt^2 we note first that since initial velocity is zero the term v0 `dt will be zero and can be dropped from the equation, leaving `ds = .5 a `dt^2. This equation is then easily solved for `dt, obtaining `dt = `sqrt(2 *`ds / a ) = `sqrt(2 * 2 meters / (9.8 m/s^2) ) = .64 second.
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RESPONSE --> ok
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11:13:27 `q002. While an object dropped from rest falls 2 meters under the influence of gravity, another object moves along a level surface at 12 meters/second. How far does the second object move during the time required for the first object to fall?
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RESPONSE --> 7.7m b/c 12*.64=7.7m
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11:13:32 As we have seen in the preceding problem, the first object requires .64 second to fall. The second object will during this time move a distance of 12 meters/second * .64 second = 8 meters, approximately.
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RESPONSE --> ok
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11:17:27 `q003. An object rolls off the edge of a tabletop and falls to the floor. At the instant it leaves the edge of the table is moving at 6 meters/second, and the distance from the tabletop to the floor is 1.5 meters. Since if we neglect air resistance there is zero net force in the horizontal direction, the horizontal velocity of the object will remain unchanged. Since the gravitational force acts in the vertical direction, motion in the vertical direction will undergo the acceleration of gravity. Since at the instant the object leaves the tabletop its motion is entirely in the horizontal direction, the vertical motion will also be characterized by an initial velocity of zero. How far will the object therefore travel in the horizontal direction before it strikes the floor?
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RESPONSE --> 3.3m b/c sqrt((2*1.5)/9.8)=.55, 6*.55=3.3
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11:17:32 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 1.5 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 1.5 m / (9.8 m/s^2) ) = .54 sec, approx., so the object falls for about .54 seconds. The horizontal motion will therefore last .54 seconds. Since the initial 6 meter/second velocity is in the horizontal direction, and since the horizontal velocity is unchanging, the object will travel `ds = 6 m/s * .54 s = 3.2 m, approximately.
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RESPONSE --> ok
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11:24:21 `q004. An object whose initial velocity is in the horizontal direction descends through a distance of 4 meters before it strikes the ground. It travels 32 meters in the horizontal direction during this time. What was its initial horizontal velocity? What are its final horizontal and vertical velocities?
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RESPONSE --> sqrt((2*4)/9.8)=.90, .90*9.8=8.8m/s=final vertical velocity. 32/.9=36m/s=final and initial horizontal velocity because no friction slows down the horizontal velocity.
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11:24:33 We analyze the vertical motion first. The vertical motion is characterized by initial velocity zero, acceleration 9.8 meters/second^2 and displacement 4 meters. Since the initial vertical velocity is zero the equation `ds = v0 `dt + .5 a `dt^2 becomes `ds = .5 a `dt^2, which is easily solved for `dt to obtain `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * 4 m / (9.8 m/s^2) ) = .9 sec, approx., so the object falls for about .9 seconds. The horizontal displacement during this .9 second fall is 32 meters, so the average horizontal velocity is 32 meters/(.9 second) = 35 meters/second, approximately. The final vertical velocity is easily calculated. The vertical velocity changes at a rate of 9.8 meters/second^2 for approximately .9 seconds, so the change in vertical velocity is `dv = 9.8 m/s^2 * .9 sec = 8.8 m/s. Since the initial vertical velocity was zero, final vertical velocity must be 8.8 meters/second in the downward direction. The final horizontal velocity is 35 meters/second, since the horizontal velocity remains unchanging until impact.
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RESPONSE --> ok
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ƧzǿۀYJy Student Name: assignment #016
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zxՆ{Uʣ Student Name: assignment #017
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13:04:02 `q001. Note that this assignment contains 5 questions. . A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.
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RESPONSE --> 80N b/c 2/.3=6.7, 10+2=12, 12*6.7=80
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13:07:21 By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.
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RESPONSE --> I used .3 instead of .03 and I shouldn't have included the 2kg mass in the calculation, otherwise ok.
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13:16:07 `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.
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RESPONSE --> We already determined that -667N is the average force exerted on the second object by the first. (10*5)+(2*0)=(2*v)+(10*3), after-collision velocity of the 2 kg mass=10m/s.
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13:16:13 Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second. This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object. A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s. Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.
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RESPONSE --> ok
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13:19:28 `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?
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RESPONSE --> Equal, because the ratio of mass to velocity remains the same on both sides of the equation.
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13:22:38 The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic) is involved.
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RESPONSE --> I see that extra energy arises in the products when I use the KE equation.
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13:23:12 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
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RESPONSE --> While KE differs, it is momentum that actually remains the same, as determined through the impulse-momentum equation.
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13:23:20 The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision. The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second. The total momentum after collision is therefore equal to the total momentum before collision.
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RESPONSE --> ok
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13:23:54 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
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RESPONSE --> Whatever change in v occurs during the collision is equally distributed between the two masses.
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13:24:01 Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.
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RESPONSE --> ok
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yېۯf Student Name: assignment #018
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15:44:17 `q001. Note that this assignment contains 7 questions. . The Pythagorean Theorem: the hypotenuse of a right triangle has a length c such that c^2 = a^2 + b^2, where a and b are the lengths of the legs of the triangle. Sketch a right triangle on a set of coordinate axes by first locating the point (7, 13). Then sketch a straight line from the origin of the coordinate system to this point to form the hypotenuse of the triangle. Continue by sketching a line straight down from (7, 13) to the x axis to form one leg of the triangle, then form the other leg by continuing straight along the x axis back to the origin. How long are these two legs? How long therefore is the hypotenuse?
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RESPONSE --> The vertical leg is 13, the horizontal leg is 7, hypotenuse is sqrt((13^2)+(7^2))=14.8
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15:44:20 The leg from (7, 13) to the x axis drops from the point (7, 13) to the point (7,0) and so has length 13. The second leg runs from (7,0) to the origin, a distance of 7 units. The legs therefore have lengths a = 13 and b = 7, so that the hypotenuse c satisfies c^2 = a^2 + b^2 and we have c = `sqrt(a^2 + b^2) = `sqrt( 13^2 + 7^2 ) = `sqrt( 216) = 14.7, approximately.
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RESPONSE --> ok
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15:46:08 `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle?
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RESPONSE --> sqrt(15^2-12^2)=9
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15:46:11 Let c stand for the length of the hypotenuse and a for the length of the known side, with b standing for the length of the unknown side. Then since c^2 = a^2 + b^2, b^2 = c^2 - a^2 and b = `sqrt(c^2 - a^2) = `sqrt( 15^2 - 12^2) = `sqrt(225 - 144) = `sqrt(81) = 9.
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RESPONSE --> ok
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15:58:12 `q003. If a line of length L is directed from the origin of an x-y coordinate system at an angle `theta with the positive x axis, then the x and y coordinates of the point where the line ends will be y = L * sin(`theta) and x = L * cos(`theta). Sketch a line of length 10, directed from the origin at an angle of 37 degrees with the positive x axis. Without doing any calculations estimate the x and y coordinates of this point. Give your results and explain how you obtained your estimates.
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RESPONSE --> x=8, y=6, I measured and drew a 37degree line from the origin out 10 units then looked down and over to see where this point intercepted the x and y axes.
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15:58:20 The line will run closer to the x axis then to the y axis, since the line is directed at an angle below 45 degrees. It won't run a whole lot closer but it will run significantly closer, which will make the x coordinate greater than the y coordinate. Since the line itself has length 10, it will run less than 10 units along either the x or the y axis. It turns out that the x coordinate is very close to 8 and the y coordinate is very close to 6. Your estimates should have been reasonably close to these values.
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RESPONSE --> ok
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15:58:44 `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis?
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RESPONSE --> x=7.99, y=6.01
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15:58:55 sin(37 deg) should give you a result very close but not exactly equal to .6. cos(37 deg) should give you a result very close but not exactly equal to .8. Since the x coordinate is L cos(`theta), then for L = 10 and `theta = 37 deg we have x coordinate 10 * cos(37 deg) = 10 * .8 = 8 (your result should be slightly different than this approximate value). Since the y coordinate is L sin(`theta), then for L = 10 and `theta = 37 deg we have y coordinate 10 * sin(37 deg) = 10 * .6 = 6 (your result should be slightly different than this approximate value).
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RESPONSE --> ok
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15:59:26 `q005. Show that the x and y coordinates you obtained in the preceding problem in fact give the legs of a triangle whose hypotenuse is 10.
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RESPONSE --> I got x=7.99, y=6.01. Actual should be x=8, y=6
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15:59:34 If c stands for the hypotenuse of the triangle, then if a = 8 and b = 6 are its legs we have c = `sqrt(a^2 + b^2) = `sqrt(8^2 + 6^2) = `sqrt(100) = 10. The same will hold for the more precise values you obtained in the preceding problem.
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RESPONSE --> ok
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16:12:02 `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin. The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin(`theta). What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units?
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RESPONSE --> 30sin120deg=y=26.0, 30cos120deg=x=-15
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16:12:07 The x component of this vector is vector along the x axis, from the origin to x = 30 cos(120 degrees) = -15. The y component is a vector along the y axis, from the origin to y = 30 sin(120 degrees) = 26, approx.. Note that the x component is to the left and the y component upward. This is appropriate since the 120 degree angle, has measured counterclockwise from the positive x axis, takes the vector into the second quadrant, which directs it to the left and upward.
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RESPONSE --> ok
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16:21:17 `q007. The angle of a vector as measured counterclockwise from the positive x axis is easily determined if the components of the vector are known. The angle is simply arctan( y component / x component ) provided the x component is positive. If the x component is negative the angle is arctan( y component / x component ) + 180 deg. If the x component is positive and the y component negative, arctan( y component / x component ) will be a negative angle, and in this case we generally add 360 degrees in order to obtain an angle between 0 and 360 degrees. The arctan, or inverse tangent tan^-1, is usually on a calculator button marked tan^-1. Find the angle and length of each of the following vectors as measured counterclockwise from the positive x axis: A vector with x component 8.7 and y component 5. A vector with x component -2.5 and y component 4.3. A vector with x component 10 and y component -17.3.
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RESPONSE --> A vector with x component 8.7 and y component 5: 29.9 A vector with x component -2.5 and y component 4.3: 120.2 A vector with x component 10 and y component -17.3: 300
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16:24:16 A vector with x component 8.7 and y component 5 makes an angle of arctan (5/8.7) = 30 degrees with the x axis. Since the x component is positive, this angle need not be modified. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(8.7^2 + 5^2) = 10. A vector with x component -2.5 and y component 4.3 makes an angle of arctan (4.3 / -2.5) + 180 deg = -60 deg + 180 deg = 120 degrees with the x axis. Since the x component is negative, we had to add the 180 degrees. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt((-2.5)^2 + 4.3^2) = 5. A vector with x component 10 and y component -17.3 makes an angle of arctan (-17.3 / 10) = -60 degrees with the x axis. Since the angle is negative, we add 360 deg to get 300 deg. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(10^2 + (-17.3)^2) = 20.
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RESPONSE --> I overlooked that I needed to find the lengths as well, which I understand how to do via the pythagorean theorem
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UcƑݓT}xڸw Student Name: assignment #018
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pGeҗWE´` Student Name: assignment #020
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19:34:59 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> 2.8m/s^2 b/c 2*9.8=19.6, 2+5=7, 19.6/7=2.8
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19:35:17 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> ok
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19:40:32 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> 2.1m/s^2 b/c 5*9.8*.1=4.9, 19.6-4.9=14.7, 14.7/7=2.1
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19:40:36 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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RESPONSE --> ok
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19:44:18 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
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RESPONSE --> 3.56m/s^2 b/c 5*9.8*sin12deg=10.2, 14.7+10.2=24.9, 24.9/7=3.56
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19:44:23 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
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RESPONSE --> ok
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]ryIJFHeҗ Student Name: assignment #019
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19:49:38 `q001. Note that this assignment contains 5 questions. . If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
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RESPONSE --> sqrt(3^2+4^2)=5mi. from starting point. arctan(4/3)=53.1deg relative to the eastward position.
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19:49:45 If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
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RESPONSE --> ok
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20:01:10 `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
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RESPONSE --> 15000sin265deg=14900N=y component. 15000cos265deg=1310N=x component.
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20:01:37 The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline.
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RESPONSE --> ok except I omitted the negative signs
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20:05:47 `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
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RESPONSE --> sqrt(300^2+400^2)=500N=how much total force we exert on the object, (arctan(-400/300))+360=307degrees.
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20:05:51 Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
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RESPONSE --> ok
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20:15:20 `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction?
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RESPONSE --> 200sin30=100N=your y component. 200cos30=173N=your x component. 300sin150=150N=my y component. 300cos150=-260N=my x component. 100+150=250N=our y component. 150-260=-110N=our x component. sqrt(250^2+110^2)=273N=our total force. (arctan(250/-110))+180=177degrees.
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20:19:12 My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.
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RESPONSE --> I just plugged in a mistaken value, otherwise it would be correct
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20:30:35 `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision?
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RESPONSE --> x1=120cos60=60, y1=120sin60=104, x2=80cos330=69, y2=80sin330=-40 x3=60+69=129, y3=104-40=64, sqrt(129^2+64^2)=144kg*m/s=total momentum. arctan(64/129)=26degrees.
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20:30:40 The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
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RESPONSE --> ok
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r}ąʟxpЙm Student Name: assignment #021
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21:33:42 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> y component=3, sqrt(2*9.8*3)=7.67, 7.67/9.8=.783, 12*.783=9.40=x component, arctan(3/9.40)=17.7degrees. sqrt(12^2+7.67^2)=14.2m/s=total velocity.
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21:35:55 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> I should have made 7.7 negative, otherwise ok.
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06:21:21 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> y component=20sin30=10, x component 20cos30=17.3. total time: 0=-4.9t^2+10t+20. positive quadratic root: .724s.
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06:25:51 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> I mistakenly used velocity in place of distance in the equation of motion. I also see how to avoid using the quadratic equation now.
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06:28:39 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> 46.7m b/c x component 20cos30=17.3, 17.3*2.7=46.7
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06:28:43 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> ok
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FdYտ_体~Rx assignment #023 yXVH Physics I 09-29-2006
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08:34:49 Query gen phy 7.27 bumper cars 450 kg at 4.5 m/s, 550 kg at 3.7 m/s, collision from back, elastic
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RESPONSE --> (450*4.5)+(550*3.7)=4060=450v1'+550v2'. 3.7-4.5=-(v2'-v1'), v1'=v2'-.8. Substituting and solving for v2'=4.42m/s. Substituting and solving for v1'=3.62.
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08:34:54 ** For an elastic collision we have m1 v1 + m2 v2 = m1 v1' + m2 v2' and v2 - v1 = -( v2' - v1'). We substitute m1, v1, m2 and v2 to obtain 450 kg * 4.5 m/s + 550 kg * 3.7 m/s = 450 kg * v1 ' + 550 kg * v2 ', or 4060 kg m/s = 450 kg * v1 ' + 550 kg * v2 ' . Dividing by 10 and by kg we have 406 m/s = 45 v1 ' + 55 v2 '. We also obtain 3.7 m/s - 4.5 m/s = -(v2 ' - v1 ' ) or v1 ' = v2 ' - .8 m/s. Substituting this into 406 m/s = 45 v1 ' + 55 v2 ' we obtain 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' . We easily solve for v2 ' obtaining v2 ' = 4.42 m/s. This gives us v1 ' = 4.42 m/s - .8 m/s = 3.62 m/s. Checking to be sure that momentum is conserved we see that the after-collision momentum is pAfter = 450 kg * 3.62 m/s + 550 kg * 4.42 m/s = 4060 m/s. The momentum change of the first car is m1 v1 ' - m1 v1 = 450 kg * 3.62 m/s - 450 kg * 4.5 m/s = - 396 kg m/s. The momentum change of the second car is m2 v2 ' - m2 v2 = 550 kg * 4.42 m/s - 550 kg * 3.7 m/s = + 396 kg m/s. Momentum changes are equal and opposite. NOTE ON SOLVING 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' FOR v2 ': Starting with 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' use the Distributive Law to get 406 m/s = 45 v2 ' - 36 m/s + 55 v2 ' then collect the v2 ' terms to get 406 m/s = -36 m/s + 100 v2 '. Add 36 m/s to both sides: 442 m/s = 100 v2 ' so that v2 ' = 442 m/s / 100 = 4.42 m/s. *
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RESPONSE --> ok
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08:49:36 Univ. 3.48. (not in 11th edition) ball 60 deg wall 18 m away strikes 8 m higher than thrown. What are the Init speed of the ball and the magnitude and angle of the velocity at impact?
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RESPONSE --> 0=v0^2-(2*9.8*8), y component v0=12.5m/s, 0=12.5-9.8t, t=1.28, 18=v0*1.28, x component v0=14.1m/s. Angle given as 60 degrees.
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퉹wЈܿnߏ Student Name: assignment #022
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07:35:07 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
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RESPONSE --> 5cos210deg=-4.33, -4.33/4=-1.08, 10/1.08=9.26s=How long it will take before the velocity in the x direction decreases to 0, 5sin210deg=-2.5, -2.5/4=-.625, .625*9.26=5.79m/s=the y velocity of the object at this instant
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07:35:18 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
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RESPONSE --> ok
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07:55:53 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
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RESPONSE --> vf^2=10^2-(2*1.08*30), x component v=5.93. I think this is a typo: assume that there is the limit to the extent of the field in the y direction. Is that supposed to be NO limit to the extent of the field in the y direction? If so, 5.93=10-(1.08*t), t=3.77, 3.77*-.625=-2.36m/s=y component, sqrt(5.93^2+2.36^2)=6.38m/s=the magnitude of the velocity of the mass as it exits this region, arctan(-2.36/5.93)+360=338degrees.
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07:55:59 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
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RESPONSE --> ok
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