course Phy 231 {Ӷaassignment #014
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14:41:39 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 = F/m F / m * 'dt, gives us a velocity. Specifically a change in velocity, which we recall: a = 'dv / 'dt, so 'dv = a * 'dt I determined this by looking at the units: F = Newtons, m = kg, 'dt = s Newtons / kg = m/s^2 m/s^2 * s = m/s, which suggests a velocity measurement 'ds = v0 * 'dt + 0.5 * a * 'dt^2 vf = sqrt( v0^2 + 2 a 'ds ) Fnet * 'ds = work performed or 'dKE I determined this by again looking at the units: Fnet = Newtons, 'ds = m. Newton meters = Joules, which is a measurement of energy 'dKE = Fnet * 'ds confidence assessment: 3
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14:42:51 **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! My response were slightly different, but I believe I supplied the required information self critique assessment: 3
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14:56:01 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 --> Net work done by system is equal to the net work done by a system against nonconservative forces and the change in PE, which is equal to the net work done by the system against conservative forces. ___ 'dKE + 'dWnonconservative + 'dPE = 0 'dPE = work done by a system against conservative forces confidence assessment: 2
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15:43:29 ** 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 - I think I had the general point, though I didn't word it the same way Conservative forces can be regained if the system is somehow ""released"" whereas nonconservative forces can not be regained. An example of a conservative force would be a weight being lifted, where the conservative force acting against the system is gravity. If the weight is ""released"" it gains that force back in the form of KE. While the weight is being raised and is working against the force, the weight is building up PE, which will then be transferred to KE when released. self critique assessment: 3
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16:07:55 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 by the rubber band on the rail is equal to the amount of work done to pull the rubber band and rail backwards. The work done by the rail against friction until it comes to rest is equal to the amount of work done on the rail as the rubber band accelerates it. confidence assessment: 2
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16:09:23 ** 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 --> OK interesting note on the heating and cooling of the rubber band and how that affects the amount of energy. self critique assessment: 3
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16:14:48 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> distance traveled by the rail should be proportional to the F * 'ds (work) total for the rubberband because the force required to pull the rail backwards (stretch the rubber band) is equal to the amount of force required to bring the rail to rest through friction. Therefore the 'ds acheived in pulling the rail backwards is proportional to the 'ds achieved after release of the rail. confidence assessment: 2
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16:17:21 ** 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 I think I was on the right track, though I failed to mention the equations of motion and how they inter-relate in this instance. it seems you are saying this: work done by friction after release must be equal and opposite to work done by stretching rubber band self critique assessment: 2
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16:17:35 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 --> skip, different class confidence assessment: 3
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16:17:41 ** 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 --> skip, different class self critique assessment: 3
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16:33:20 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 --> I don't see this problem in the 12th edition (I checked ch 4 - 7), but here goes: F = 25 kN (I'm assuming that means kilo-Newtons) a = -1.2 m/s^2 once F = 10 kN, a = 0.8 m/s^2 weight at surface? F = m * a, so when F = 25 kN and a = 1.2 m/s^2, m must equal: m = F / a = 25 kN / 1.2 m/s^2 m = 20.83 k kg (which I believe is a Mega gram, or Mg) weight = m * g = 20.83 Mg * 9.8 m/s^2 = 204.17 kN when F = 10 kN: m = 10 kN / 0.8 m/s^2 = 12.5 k kg weight = m * g = 12.5 Mg * 9.8 m/s^2 = 122.5 kN confidence assessment: 1
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16:35:12 ** 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 --> Interesting... so I should have set the equations up as: m * a + m * g, which would have allowed me to eliminate the m * a terms and arrive at 16000 kN for the m * g (weight) component. self critique assessment: 2
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