course Phy 121 ???\??????????assignment #013
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16:03:38 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 --> F = 265N a = 2.30m/s/s F = ma 265 = (m)(2.30) m = 115.217 confidence assessment: 3
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16:03:46 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 self critique assessment: 3
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16:04:41 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 --> m = 700 g `dv= 125 `ds = .800 a = 125/.800 a = 156.25 F = (700) (156.25) F = 109375 confidence assessment: 3
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16:06:15 ** 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 --> I did the change in velocity ovr the change in distance as acceleration for some reason. Obviously it is change in velocity over change in time. self critique assessment: 2
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16:06:21 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 --> confidence assessment:
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16:06:24 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 --> self critique assessment:
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16:06:28 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 --> confidence assessment:
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16:06:32 Describe the free body diagram you drew.
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RESPONSE --> confidence assessment:
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16:06:36 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 --> self critique assessment:
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16:06:42 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 --> confidence assessment:
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16:06:46 ** 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 --> self critique assessment:
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16:07:08 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 --> confidence assessment:
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16:07:59 ** 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 --> I didn't have this question, but I guess the way you solved it is something I haven't learned that from being in principles. self critique assessment: 2
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Y???????}?y?i?assignment #014 014. `query 14 Physics I 06-24-2007
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17:23:56 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 --> If you have acceleration and multiply by a time interval, you would get the change in velocity When you multiply the net force * distance traveled, you get the work done. The kinetic energy will be .5mv^2. The object would travel the average velocity divided by the time interval. confidence assessment: 3
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17:25:05 **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 --> That's a really good explanation. self critique assessment: 3
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17:27:33 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 --> The work done by a systema against nonconservative forces would mean that the forces do not conserve energy. The work done by a system against conservative forces consever energy. The KE of the system would increase as a nonconservative force and the KE would decrease against a conservative force. The opposite would be true for PE. confidence assessment: 2
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17:29:20 ** 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 --> I see how it makes sense that KE will be decreased no matter what force is acted upon it. self critique assessment: 2
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17:32:19 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 rubberband is a force acting on the rubberband. In the second scenario the rubberband is doing word on the rail. confidence assessment: 1
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17:34:18 ** 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 --> All I am taking out of this is that the rubberband and the rail perform equal work on eachother. Is there something else I should be getting? self critique assessment: 2
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17:34:51 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> F*ds by the rubberband will give the work of the rubberband which should equal the force of the rail. confidence assessment: 3
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17:35:11 ** 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 --> ok. self critique assessment: 3
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17:35:15 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 --> confidence assessment:
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17:35:17 ** 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 --> self critique assessment:
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17:35:40 ** 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 --> self critique assessment:
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