course Phy 201 ziӡǨَϬѓĥassignment #014
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14:39:05 `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 --> Parallel weight component = weight * slope = (1500kg * 9.8 m/s/s) * .03 = 441 N a =Fnet/m = 441N / 1500kg = .294m/s/s vf^2 = vo^2 + 2as vf^2 = 0 + 2* .294 * 200 vf = sqrt 117.6 = 10.8 m/s confidence assessment: 1
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14:42:48 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 found the parallel weight component but didn't use that to find the work done by using the formula: `dWnet = 441N * 200 m = 88200 J I think I did the second part of the problem correct, right? self critique assessment: 2
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15:07:53 `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 --> Fnet * `ds = 1/2 m v^2 Fnet * `ds = 1/2 1500kg * 10.9^2 Fnet * `ds = 89100 J ??? confidence assessment: 0
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15:18:06 The automobile starts out with kinetic energy KEinit = 1/2 m v0^2 = 1/2 (1500 kg) ( 10.9 m/s)^2 = 88000 kg m^2/s^2 = 88000 Joules. The gravitational force component parallel to the incline is in this case opposite to the direction of motion so that gravity does negative work on the automobile. Since the change in KE is equal to the work done by the net force, if the gravitational force component parallel to the incline does negative work the KE of the object will decrease. This will continue until the object reaches zero KE. As found previously the gravitational force component along the incline has magnitude 441 Newtons. In this case the forces directed opposite to the direction motion, so if the direction up the incline is taken to be positive this force component must be -441 N. By the assumptions of the problem this is the net force exerted on the object. Acting through displacement `ds this force will therefore do work `dWnet = Fnet * `ds = -441 N * `ds. Since this force must reduce the KE from 88,000 Joules to 0, `dWnet must be -88,000 Joules. Thus -441 N * `ds = -88,000 Joules and `ds = -88,000 J / (-441 N) = 200 meters (approx.). Had the arithmetic been done precisely, using the precise final velocity found in the previous exercise instead of the 10.9 m/s approximation, we would have found that the displacement is exactly 200 meters.
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RESPONSE --> OK I knew the equation `dWnet = Fnet * `ds was involved, but I didn't know where to pull Fnet from. Now I see you just get it from the previous problem so that: -89000 J = -441 N * `ds `ds = 202 m The negative and positive numbers are still a little confusing to me. self critique assessment: 2
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15:35:16 `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 --> It will regain all of its KE because only conservative forces were involved. confidence assessment: 1
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15:35:38 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 --> ok self critique assessment: 3
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15:38:28 `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 --> Gravity is a conservative forces which conserves the energy, meaning all KE can be regained. Friction is a non-conservative force which would not conserve energy causing the final KE to be different than the initial KE. confidence assessment: 2
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15:38:52 The car initially had some KE. The gravitational component parallel to the incline is opposite to the direction of motion and therefore does negative work as the object travels up the incline. The gravitational component is the net force on the object, so the work done by this net force causes a negative change in KE, which eventually decreases the KE to zero so that the object stops for an instant. This happens at the position where the work done by the net force is equal to the negative of the original KE. The gravitational component parallel to the incline immediately causes the object to begin accelerating down the incline, so that now the parallel gravitational component is in the same direction as the motion and does positive work. At any position on the incline, the negative work done by the gravitational component as the object traveled up the incline from that point, and the positive work done as the object returns back down the incline, must be equal and opposite. This is because the displacement up the incline and the displacement down the incline are equal and opposite, while the parallel gravitational force component is the same, which makes the Fnet * `ds product must be equal and opposite. Thus when the object reaches its original point, the work done on it by the net force must be equal and opposite to the work done on it while coasting up the incline. Since the work done on the object while coasting up the incline was the negative of the original KE, the work done while coasting down, being the negative of this quantity, must be equal to the original KE. Thus the KE must return to its original value.
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RESPONSE --> ok self critique assessment: 3
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15:39:58 `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 when the object travels up the incline. Gravity does positive work when the object travels down the incline. confidence assessment: 2
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15:40:11 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 self critique assessment: 3
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15:41:07 `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 done on the object increases KE. Negative work done on the object decreases KE. confidence assessment: 2
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15:41:18 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 self critique assessment:
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15:42:12 `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 --> Positive confidence assessment: 2
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15:43:05 If the object ends up in the same position as it began, the work done on the object by gravity and work done by the object against gravity must be equal and opposite. Thus when the object does positive work against gravity, as when it travels up the incline, gravity is doing negative work against the object, which therefore tends to lose kinetic energy. When the object does negative work against gravity, as when traveling down the incline, gravity is doing positive work against the object, which therefore tends to gain kinetic energy.
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RESPONSE --> OK self critique assessment: 3
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15:55:18 `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 --> Wnet = Fnet * `ds 10,000 J = -500 N * `ds `ds = 20 m 10,000 J / 500 N = 20 J/N 20 J/S * 100 N = 2000 J KE = 10000 J - 2000 J = 8000 J confidence assessment: 0
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16:02:33 The net force on the automobile as it climbs the incline is the sum of the 400 Newton parallel component of the gravitational force, which is exerted down the incline, and the 100 Newton frictional force, which while the automobile is moving up the incline is also exerted down the incline. Thus the net force is 500 Newtons down the incline. This force will be in the direction opposite to the displacement of the automobile up the incline, and will therefore result in negative work being done on the automobile. When the work done by this force is equal to -10,000 Joules (the negative of the original KE) the automobile will stop for an instant before beginning to coast back down the incline. If we take the upward direction to be positive the 500 Newton force must be negative, so we see that -500 Newtons * `ds = -10,000 Joules so `ds = -10,000 Joules / -500 Newtons = 20 Newtons meters / Newtons = 20 meters. After coasting 20 meters up the incline, the automobile will have lost its original 10,000 Joules of kinetic energy and will for an instant be at rest. The automobile will then coast 20 meters back down the incline, this time with a 400 Newton parallel gravitational component in its direction of motion and a 100 Newton frictional force resisting, and therefore in the direction opposite to, its motion. The net force will thus be 300 Newtons down the incline. The work done by the 300 Newton force acting parallel to the 20 m downward displacement will be 300 Newtons * 20 meters = 6,000 Joules. This is 4,000 Joules less than the when the car started. This 4,000 Joules is the work done during the entire 40-meter round trip against a force of 100 Newtons which every instant was opposed to the direction of motion (100 Newtons * 40 meters = 4,000 Newtons). As the car coasted up the hill the frictional force was downhill and while the car coasted down friction was acting in the upward direction. Had there been no force other than the parallel gravitational component, there would have been no friction or other nongravitational force and the KE on return would have been 10,000 Joules.
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RESPONSE --> I messed up caluculating the lost KE when it should have been: Fnet = 300 N Wnet = Fnet * `ds = 300 N *20m = 6000 J self critique assessment: 2
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16:14:51 `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 --> The KE would be the same if the automobile dropped 6 meters rather than rolling down the incline. Fgrav = 1500 kg * 9.8 m/s/s = 14700 N Wnet = 14700 N * 6 m = 88200 J confidence assessment: 2
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16:15:07 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 --> cool self critique assessment: 3
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16:17:41 `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 --> In this example (and others like it) it didn't matter the manner in which the automobile descended. Gravity did the same amount of work on each object since the vertical distances were the same. confidence assessment: 2
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16:18:06 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 self critique assessment: 3
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16:49:09 `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 --> KE = Fnet * `ds = (7 kg * 9.8 m/s/s) * 45 m KE when caught by person= 30871 J So PE at the top of the tower is 30871 J confidence assessment: 2
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16:51:03 The potential energy of the ball at the top of the tower is equal to the work that would be done by gravity on the ball in dropping from the 90-meter altitude at the top to the 45-meter altitude at the middle of the tower. This work is equal to product of the downward force exerted by gravity on the ball, which is 7 kg * 9.8 m/s^2 = 68.6 Newtons, and the 45-meter downward displacement of the ball. Both force and displacement are in the same direction so the work would be work done by gravity if ball dropped 45 meters: 68.6 Newtons * 45 meters = 3100 Joules, approx.. Thus the potential energy of the ball at the top of the tower, relative to the position halfway down the tower, is 3100 Joules. The ball loses 3100 Joules of PE as it drops. If no force acts to oppose the effect of gravity, then the net force is the gravitational force and the 3100 Joule loss of PE will imply a 3100 Joule gain in KE.
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RESPONSE --> I miss-read my notes. I meant to say PE = 3087 J KE = 3087 J self critique assessment: 2
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16:53:17 `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 does affect both `dPE and `dKE confidence assessment: 1
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17:00:18 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, Does this mean PE is only affected by conservative forces while KE is affected by both conservative and nonconservative forces? self critique assessment: 2
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17:02:53 `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 --> Fnet = 68.6 N - 10 N = 58.6 N KE = 58.6 N * 45 m = 2600 J confidence assessment: 2
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17:03:34 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 --> Is the method I used correct? self critique assessment: 2
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