course Phy 121 h頛יGassignment #014
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14:33:24 `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 --> 'dW=Fn * 'ds =45 * 200 =9000J vf=... 9000=750v^2 vf=10.9 m/s confidence assessment: 1
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14:35:00 `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 --> wouldn't it have to be the equal and opposite of the work done by the automobile? -88200 J confidence assessment: 3
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14:37:05 `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 --> i understand the KE will return, is it because it is conservative energy? confidence assessment: 2
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14:38:22 `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 --> as the object travels up in a positive direction, gravity does negative work on it when it comes down, gravity does positive work on it confidence assessment: 3
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14:39:19 `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 --> if gravity is doing positive work on an object, the KE must increase if gravity is doing negative work, the KE must decrease confidence assessment: 3
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14:40:15 `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 --> going up the incline, the object does negative work against gravity, going down it does postive work against gravity confidence assessment: 2
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14:43:08 `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 --> don't know confidence assessment: 0
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14:47:28 `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 --> if gravity did 88200 J of work then the diff in KE must be 88200 J over a 'ds of 6m confidence assessment: 1
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14:48:45 `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 --> that seems to be justified here even though in the first case the car drifted 200 m horizontally confidence assessment: 3
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14:51:35 `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 --> the PE at the top would be completely opposite the person at the bottom, the person at the bottom should have 0 PE, the ball at the top should have the most PE possible for that ball the amount of KE of the ball when it reaches the bottom will be equal to the amount of PE it had when it was at the top confidence assessment: 3
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14:52:50 `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 --> air resistance would decrease KE which should, in turn, keep the PE higher in proportion confidence assessment: 3
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14:55:10 `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 --> 10 N * 45 m = 450 J, that is the loss of KE due to air resistance, so the KE would be 2650 confidence assessment: 3
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҈Nʐeɇjs assignment #014 014. `query 14 Physics I 07-06-2007
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15:26:32 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 --> 1. you get 'ds of that mass 2. you can get the vf if you know the mass and the force by using the change in KE formula, .5mv^2 3. you get the 'dW in Joules confidence assessment: 2
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15:29:25 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 against noncons is lost completely the work done against conservs can be regained at some point, returning the KE to its original position, say for example, the conserv force of gravity pulling a car that has drifted up a hill back down PE is a lot of times opposite the KE, except when you factor in things like air resistance confidence assessment: 2
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15:31:54 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 'dW to stretch the rubber band increases the PE, the work done by the rubber band on the rail increases the KE of the rail, the rail works against friction confidence assessment: 2
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15:32:54 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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RESPONSE --> does it have to do with how KE and PE are changed proportionally? confidence assessment: 1
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You then start the process over with the next question, clicking on the Next Question/Solution button at top left, etc. &#
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13:51:25 `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 --> confidence assessment:
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13:51:29 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 --> self critique assessment:
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13:51:31 `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 --> confidence assessment:
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13:51:33 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 --> self critique assessment:
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13:51:35 `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 --> confidence assessment:
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13:51:38 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 --> self critique assessment:
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13:51:41 `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 --> confidence assessment:
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13:51:43 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 --> self critique assessment:
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13:51:46 `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 --> confidence assessment:
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13:51:48 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 --> self critique assessment:
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13:51:50 `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 --> confidence assessment:
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13:51:53 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 --> self critique assessment:
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13:51:56 `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 --> confidence assessment:
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13:51:58 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 --> self critique assessment:
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13:52:01 `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 --> confidence assessment:
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13:52:03 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 --> self critique assessment:
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13:52:06 `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 --> confidence assessment:
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13:52:09 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 --> self critique assessment:
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13:52:11 `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 --> confidence assessment:
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13:52:14 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 --> self critique assessment:
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13:52:17 `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 --> confidence assessment:
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13:52:19 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 --> self critique assessment:
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13:52:21 `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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13:52:24 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 --> self critique assessment:
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13:52:26 `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 --> confidence assessment:
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13:52:29 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 --> self critique assessment:
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C|bȋ assignment #014 014. `query 14 Physics I 07-07-2007
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13:53:29 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 --> confidence assessment:
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13:53:33 **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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13:53:35 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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13:53:38 ** 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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13:53:40 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 --> confidence assessment:
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13:53:43 ** 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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13:53:45 Why should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?
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13:53:48 ** 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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13:53:50 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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13:53:52 ** 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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13:54:00 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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13:54:03 ** 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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