course PHY 121
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13:22:47 `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 --> v0 = 0 m/s 'ds = 200 m a = F / m a = 9.8 N / 1500 kg a = (0.007 m/s^2) * 0.03 a = 0.00021 m/s^2 W = m * a W = 1500 kg * 0.00021 m/s^2 W = 0.315 N vf = 'sqrt (2 a 'ds) vf = 'sqrt (2 * 0.00021 m/s^2 * 200 m) vf = 0.29 m/s
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13:30:28 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 --> I did not understand how to solve the question before reading the solutions. I now understand that you must find the weight of the automobile and multiply this by the slope. Then 'dWnet = F * 'ds and this value is in Joules (J). Then vf is found using the formula KEf = 1/2 m vf^2 and KEf is equal to the change in KE ('dWnet) added to the KE0.
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13:45:57 `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 --> v0 = 10.9 m/s vf = 0 m/s I am not sure where to go with this problem.
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13:56:31 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 --> I did not realize that KEinit could be found the same way as KEf, except for using v0 instead of vf. v0 = 10.9 m/s vf = 0 m/s KEinit = 1/2 m v0^2 KEinit = 1/2 (1500 kg) (10.9 m/s)^2 KEinit = 89107.5 J 'dWnet = F * 'ds 'ds = 'dWnet / F 'ds = (-89,107.5 J) / (-441 N) 'ds = 202 m
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13:58:58 `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 --> 0 Joules.
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14:03:10 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 --> I understand that it started out with 88,200 J and it lost 88,200 J upon reaching its displacement up the ramp and then gained 88,200 J on the way back down.
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14:05:49 `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 gravitational force is the same going up the ramp as it is coming down the ramp, only in different directions.
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14:07:01 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 --> OK.
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14:07:58 `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 --> Gravity does negative work on the object as it travels up the incline. Gravity does positive work on the object as it travels down the incline.
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14:08:23 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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14:09:04 `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 --> Positive work will increase KE. Negative work will decrease KE.
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14:09:18 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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14:10:10 `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 --> While traveling up the incline, the object does negative work against gravity. While traveling down the incline, the object does positive work against gravity.
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14:11:41 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 --> While traveling up the incline, the object does positive work against gravity. While traveling down the incline, the object does negative work against gravity. I have revised my answer.
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14:16:29 `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 --> 'dWnet = F * 'ds 'ds = 'dWnet / F 'ds = 10,000 J / (400 N + 100 N) 'ds = 20 m 'dWnet = F * 'ds 'dWnet = (400 N - 100 N) * 20 m 'dWnet = 6,000 J KEf = 6,000 J
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14:18:58 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 --> OK.
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14:22:24 `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 --> 'dWnet = F * 'ds 'dWnet = (1500 kg * 9.8 m/s^2) * 6 m 'dWnet = 88,200 J
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14:22:42 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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14:24:23 `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 --> You could prove this by a similar situation that was performed in these questions.
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14:24:44 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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14:27:49 `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 --> 'dWnet = F * 'ds 'dWnet = (7 kg * 9.8 m/s^2) * 45 m 'dWnet = 3087 J PE = 3087 J KE = 3087 J
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14:28:37 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 --> OK.
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14:29:27 `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 --> It would decrease both the PE and the KE.
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14:30:18 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 --> OK. I understand that the PE is not affected.
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14:32:27 `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 --> 'dWnet = F * 'ds 'dWnet = (68.6 N - 10 N) * 45 m 'dWnet = 2637 J KE = 2637 J
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14:33:11 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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Éμ~^JĘة㰚{ assignment #012 }_M_⣅|G͙ Physics I Class Notes 10-13-2005
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15:35:22 We put energy into the rubber-band-and-rail system, in the form of the work we do stretching the rubber band. In detail, what happens to this energy from the instant we start pulling back to the instant the rail stops?
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RESPONSE --> The energy decreases as we pull back the rubber band, then the energy increases as we release the rubber band, and then the energy decreases until the rail stops.
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15:35:39 The system stores most of the energy as elastic PE of the rubber band. When the system is released the PE decreases and goes into the KE of the system. There are also thermodynamic effects. As the rubber band is stretched the thermal energy that results from the stretching is released into the air. When the rubber band becomes unstretched it cools off, but the thermal energy that was released does not return to the rubber band.
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RESPONSE --> OK.
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15:35:55 What happens to the KE of the rail as it slides across the floor? Where does it go?
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RESPONSE --> The KE goes into work against friction.
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15:36:04 ** The KE was gained at the expense of the rubber band's PE, but as it slides across the floor it is lost to friction. The result of friction is thermal energy, or heat, which is dissipated from the rail and the surface. **
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RESPONSE --> OK.
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15:36:17 From the point of view of the rail, is it doing positive work or negative work as it slides across the floor?
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RESPONSE --> It is doing positive work.
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15:36:23 ** It's doing positive work, but the reason is that it is exerting a force against friction, and this force is in the direction of its motion. **
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RESPONSE --> OK.
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15:36:58 Does the rubber band supply more or less energy to the rail than the energy we put into it, and where does the difference go?
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RESPONSE --> Less energy, as some goes into thermal energy.
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15:37:19 When we pull the rail back, we do work, most (but not all) of which will be recovered after we release the rail. This work will therefore be present in the rail in the form of energy of motion. Before we release the rail, this work is potentially there; the potential will become reality after we release the rail. At the instant the rail reaches its original position the rubber band ceases to accelerate it, and it has its maximum energy of motion. As the rail slides across the floor, this energy is dissipated in the form of work done against friction. This continues until all the energy is dissipated and the rail stops.
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RESPONSE --> OK.
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15:37:48 Does the rubber band exert more or less average force when it accelerates the rail than when it was pulled back?
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RESPONSE --> The rubber band exerts less average force.
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15:37:55 ** It can't exert more average force because if it did it would release more energy than was put into it. **
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RESPONSE --> OK.
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15:38:22 Is a cooler rubber band stiffer or less stiff than a warmer one?
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RESPONSE --> less stiff
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15:38:28 The cooler rubber band is less stiff than the warmer one, contrary to our intuition about how things stiffen when they are cooled.
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RESPONSE --> OK.
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oa̯Vԫztz assignment #013 }_M_⣅|G͙ Physics I Class Notes 10-13-2005
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16:50:43 What would a graph of potential energy vs. stretch look like for the rubber band?
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RESPONSE --> It would be increasing at a decreasing rate.
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16:51:09 * potential energy increases with stretch; and since the force required to stretch the rubber band increases with stretch, the PE increases at an increasing rate with respect to the stretch. **
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RESPONSE --> I was thinking about something different, and do understand.
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16:51:43 What would a graph of kinetic energy vs. distance look like for the rail sliding across the floor? Would the graph be linear or would it curve? If it would curve, in what direction would it curve?
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RESPONSE --> It would be increasing at an increasing rate.
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16:52:24 ** since the rail is moving more and more slowly, and since the work done against friction depends on the distance moved, the rail will lose KE more and more slowly. So the graph of KE vs. distance will decrease, but at a decreasing rate. **
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RESPONSE --> OK. I had my thinking backwards.
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16:53:10 How is it that we can regard the force of gravity as equivalent to two forces, one parallel and one perpendicular to the ramp?
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RESPONSE --> Because there is a force acting against the ramp and with the ramp.
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16:53:35 The forces wParallel and wPerpendicular are to be regarded as completely equivalent to the weight. We should understand that if we pulled on an object with two forces equal to these, and in the indicated directions, they could have an effect identical to that of the downward force of gravity. So the force of gravity could be replaced by these two forces. Given a high enough coefficient of friction, the force acting parallel to the incline could be equal and opposite to the frictional force. The normal force is the elastic force required to exactly balance the weight component perpendicular to the ramp.
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RESPONSE --> OK.
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16:53:41 ""
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RESPONSE --> OK.
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\٠rނ͌s assignment #014 }_M_⣅|G͙ Physics I Class Notes 10-13-2005
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17:20:29 How does the velocity vs. clock time trapezoid give us the two basic equations of motion?
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RESPONSE --> We can find 'ds and a.
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17:20:44 The displacement between any two clock times is equal to the area of the trapezoid between these two clock times. This area is equal to the product of the average altitude of the trapezoid and its width, which gives us the first basic equation of uniformly accelerated motion: `ds = (v0 + vf) / 2 * `dt. Under the same conditions the slope of the graph will be the acceleration a, and the change in velocity between two clock times will be represented by the rise of a slope triangle who slope is a and whose run is the time interval `dt between the two clock times. The velocity change will therefore be `dv = a * `dt, and the final velocity will be equal to the initial velocity plus this change: vf = v0 + a `dt. This is the second basic equation of motion.
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RESPONSE --> OK.
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17:21:03 When a wound-up friction car is released on the level surface, what do we see as a result of the potential energy conversion?
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RESPONSE --> It turns into KE.
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17:21:14 Most of the work done to wind the spring goes into the elastic potential energy of the spring. When it is released, most of this energy is then converted to kinetic energy. In each part of the process there is some friction loss in the mechanism.
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RESPONSE --> OK.
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17:21:50 When a wound-of friction car is released up a ramp, what to we see as a result of the potential energy conversion?
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RESPONSE --> It is turned into KE, but some works against gravity and friction.
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17:22:01 The car climbs the ramp, increasing its gravitational PE. The car also speeds up, increasing its KE. Some of the energy is dissipated in the form of thermal energy as a result of friction.
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RESPONSE --> OK.
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17:22:12 What forces act on an object sliding up or down an incline?
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RESPONSE --> gravity and friction
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17:22:22 A normal force will be exerted by the ramp. Gravity acts on the object; usually the resulting weight is expressed as two components, one parallel and one perpendicular to the incline. There is also a frictional force acting in the direction opposite to the motion of the object.
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RESPONSE --> OK.
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17:22:36 What energy changes take place as an object slides up or down an incline?
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RESPONSE --> PE becomes KE
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17:23:44 The parallel component of the weight is wParallel = w sin(`theta), and to slide the object through displacement `ds at a constant velocity requires a force equal and opposite to this component. As a result work `dW = w sin(`theta) * `ds, if the positive direction is chosen as up the incline. If we simply raise the object vertically through a distance equal to its vertical rise, we will have to exert a force equal in magnitude to its weight. The vertical distance through which we lift the object will be `dy = `ds sin(`theta), so the work done will be `dW = w * `dy = w * (`ds sin(`theta)). It should be clear that this is the same work contribution found before.
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RESPONSE --> OK. I understand that this is the same work contribution.
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17:24:20 If we know the mass and length of a pendulum, how can we determine the force required to displaced pendulum a given small distance from equilibrium?
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RESPONSE --> F/mg = x/L
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17:24:29 When a pendulum is displaced a certain distance from its equilibrium, the forces on the pendulum will be in equilibrium if the tension force directed along the pendulum string has a vertical component equal to the weight and a horizontal component equal to F. Therefore, we will have equilibrium if the ratio of F to the weight is the same as the ratio of the displacement x to the length L. That is, F / mg = x / L. It follows that F = mg * (x / L), which is the weight of the pendulum multiplied by the ratio x / L of displacement to length.
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RESPONSE --> OK.
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yPc߰| assignment #015 }_M_⣅|G͙ Physics I Class Notes 10-13-2005
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17:53:57 Why do we expect the velocity attained by a ball on a ramp to be proportional to the square root of the vertical position change of the ball?
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RESPONSE --> 'dx = const * 'sqrt('dy) v = Konst. * 'sqrt('dy)
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17:54:17 If the object was released from rest and and allowed to fall freely through a downward distance `dy equal to the vertical distance traveled on the ramp, its gravitational potential energy will convert to kinetic energy. In this case, setting the potential energy decrease equal to the kinetic energy increase (i.e., `dKE = -`dPE) gives.5 m v^2 = m g `dy. We solve to obtain v = `sqrt(2 g `dy). This demonstrates that v is proportional to`sqrt(`dy).
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RESPONSE --> OK.
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17:55:21 Why do we expected distance traveled by a ball after being projected horizontally off of a ramp and falling a fixed distance to be proportional to the velocity with which the ball leaves the ramp?
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RESPONSE --> because the velocity that it leaves the ramp at will determine its distance.
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17:55:46 If v^2 is proportional to dy, then for some constant k we have v^2 = k * dy. From experimental results it appears that dx^2 is also proportional to dy, so that dx is emprircally proportional to sqrt(dy). Since `dy is proportional to v^2, it follows that sqrt(`dy) is proportional to v, and we finally conclude that dx is proportional to the velocity v with which the ball is projected horizontally from the ramp.
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RESPONSE --> OK.
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17:56:17 Why is the distance traveled by the projectile in this situation less than that predicted from the velocity v, where v is determined by setting 1/2 m v^2 equal to the potential energy loss on the ramp? Why is the distance still less even if frictional losses are taken into consideration?
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RESPONSE --> Because some of the PE is not converted to KE
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17:56:31 This situation is due mostly to the fact that in addition to attaining a velocity v the ball also rolls down the ramp and therefore gains kinetic energy associated with its rotational motion. As a ball rolls (without slipping) down a smooth ramp, it will turn out that 2/7 of its PE loss converted to rotational KE. This means that the KE associated with its velocity v on the ramp, which is called translational kinetic energy, theoretically, should only be 5/7 of the PE loss. Frictional losses further reduce v.
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RESPONSE --> OK.
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