#$&* course PHY 121 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: 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. STUDENT QUESTION: I assume the perpendicular force is balanced beause of the normal force downward of gravity and mass and then upward from the road. INSTRUCTOR RESPONSE: Good, but there's a little more to it: The normal force balances the component of the gravitational force which is perpendicular to the road. The component of the gravitational force parallel to the incline is the for that tends to accelerated objects downhill. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This is an application of the work-kinetic energy theorem. In words, this theorem says that the change in KE is equal to the work done by the net force acting ON the system In symbols, this is expressed • `dW_net = `d(KE). KE is kinetic energy, equal to 1/2 m v^2. 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 acting ON the system, 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 • `dW_net = -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. STUDENT QUESTION I set `dKE = Fnet * `ds and had `dKE as negative so I came up with the correct solution but just above you say that work is negative. I don’t understand how work is negative, especially going by the equation because I thought work was opposite `dKE. INSTRUCTOR RESPONSE You're thinking about exactly the right things. The specific statement of the work-KE theorem is that the work done by the net force acting ON the system is equal to the change in the kinetic energy of the system. This is abbreviated • `dW_ON_net = `dKE. Rather that talking about 'the work', it's very important to get into the habit of labeling 'the work' very specifically. You have two basic choices. You can think in terms of the work done on the system or object by a force the work done by the system or object against a force The two are equal and opposite. • Note that the words 'on' and 'by' modify the word 'system', not the word 'force'. • The key phrases are 'on the system' and 'by the system'. In the present case if you were to choose to think in terms of the work done by the net force exerted by object, then this force would be labeled `dW_BY_net and would be equal and opposite to `dW_ON_net, the work exerted by the net force acting on the object. Formally we would have `dW_BY_net = - `dW_ON_net so that `dW_BY_net = - `dKE. This last equation is often written `dW_BY_net + `dKE = 0, and is another equivalent formulation of the work-kinetic energy theorem. STUDENT COMMENT I am getting all the equations mixed up is there any way you can just send the different equations? I understand the 4 from the major quiz. I can do the algebra I just don’t know which equation to plug it in for. INSTRUCTOR RESPONSE Physics is about more than figuring out what to plug into what equation. It's necessary to understand the words and the concepts to know which equation to plug into. In other words, the concepts are what keep us from getting the equations mixed up. However I have observed in your work that you do very well with the algebra. So the equations might well be your most appropriate starting point. You can use the equations to understand the words and the concepts, just as less algebraically adept students might use the words and concepts to understand the equations. The relevant relationships here are • `dW_net = `dKE and • KE = 1/2 m v^2. The relationship • `dW_BY_net = - `dW_ON_net is also invoked in the additional comments at the end, which mention an alternative formulation of the work-kinetic energy theorem. However this relationship is not used in solving this particular problem. STUDENT COMMENT OK, I understand the solution and will use _on and _by descriptors in my answers from now on. INSTRUCTOR RESPONSE Good. Remember that ON and BY are adjectives applied to the word 'system', not to the word 'force'. That is, you have to determine whether the force is acting ON the system, or is exerted BY the system. Your choice of point of view will determine whether you use the equation • `dW_NC_ON = `dPE + `dKE or • `dW_NC_BY + `dPE + `dKE = 0. STUDENT COMMENT Ok. I see why my ‘ds was negative. The F is negative in this system because it is working against the positive motion of the car UP the ramp. For every force, there is an equal and OPPOSITE force. INSTRUCTOR RESPONSE Good. If you assume the positive direction to be up the incline, F_net is negative, as you say, but the specific reason is slightly different than the one you give. It's good to think in terms of equal and opposite forces, but the motion of the car is not a force. In this case it really just comes down to signs: The force used to calculate `dW_net was the net force acting on the car. That force acts down the incline, in the direction opposite motion. Therefore F_net and `ds are of opposite sign, and the net force acting on the car does negative work. This decreases the KE, as your solution indicates. The question of whether `ds is positive or negative depends on which direction you choose for the positive direction. In your solution you apparently thought of upward as the positive direction; you should have explicitly stated this. Relative to this choice F_net is negative. As I mentioned, you should have declared the positive direction in your solution. You could have chosen either upward (which is the direction of the displacement, and is the direction you implicity chose) or downward (which is the direction of the net force) to be the positive direction. Either choice of positive direction would have been perfectly natural. If you had chosen 'down the incline' to be the positive direction, then `ds would have been negative (therefore opposite to the downward direction, so up the incline). In either case, `ds would have been up the incline. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The work on would have to equal to that of work by in order to stop it correct? So, that would be -441 N. Is that right? I’m really confused on this one. ------------------------------------------------ Self-critique rating:2
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Given Solution: 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 on 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because, it is only gravity pushing against it. The car will regain everything because the gravity will be replaced. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The car initially had some KE. The gravitational component parallel to the incline is in the direction opposite to the direction of motion up the incline and therefore does negative work ON the object as it travels up the incline. The gravitational component is the net force on the object, so the work done by this net force on the system 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 ON the system. • 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 by this force 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 remains the same. Thus the Fnet * `ds products for the motion up and the motion down equal and opposite. When the object reaches its original point, the work that was done on it by the net force, as it rolled up the incline, must be equal and opposite to the work done on it while coasting down 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. STUDENT QUESTION I still really don’t understand how it can return back to its original position because of what we saw in class It never returned back to its original position. INSTRUCTOR RESPONSE There are a number of situations in which an object doesn't return to its original position. The one that's relevant to this situation: When you rolled the ball up the single incline, it slowed, came to rest for an instant, and then rolled back down. It did return to its initial position. Of course when it got there is was moving pretty fast so if you didn't stop it, it kept going until something else did. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand this after reading the given solution. The forces must be equal and opposite so this will make it return back to normal. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Negative. \Positive. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As the object travels up the incline the net force is directed opposite its direction of motion, so that Fnet_ON and `ds have opposite signs and as a result `dWnet = Fnet_ON * `ds must be negative. As the object travels down the incline the net force is in the direction of its motion so that Fnet_ON and `ds have identical signs and is a result `dWnet = Fnet_ON * `ds must be positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Postiive work, increase. Negative work, decrease. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The KE change of an object must be equal to the work done ON the system by the net force. Therefore if positive work is done on an object by the net force its KE must increase, and if negative work is done by the net force the KE must decrease. STUDENT QUESTION (instructor comments in parentheses) Ok, after reading a couple of time I get it dw will be equal to KE. If dw is - Ke will be - . KE can't be negative; `dW is not equal to KE, but to `dKE, the change in KE. `dKE can certainly be positive or negative (or zero), depending on the situation. I am still a little unclear about if the dw done on an object is negative then what direction is it moving?? The sign of `dW by the net force does not determine the direction of motion of the object. It determines only the change in its kinetic energy. In the present case, the net force is the component of gravity along the incline. The direction of motion of the object determines whether this force is in the direction of motion or opposite that direction, and so determines whether the displacement is in the direction of motion (implying positive work) or opposite the direction of motion (implying negative work). The direction of motion thus determines, for this situation, whether `dW_net is positive or negative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. While traveling up the incline, does the object do positive or negative work against gravity? Answer the same question for motion down the incline. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Negative. Positive. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT COMMENT: A little shaky on this problem because I feel its easy to get confused on the positive and negative. INSTRUCTOR RESPONSE This is the most common point of confusion at this stage of the course. To sort out positive and negative, you would answer the following questions: Are you thinking about the work done ON the system or BY the system (i.e., are you thinking about the forces acting ON the system or a forces exerted BY the system)? The ON and the BY are equal and opposite. Whichever force you are thinking about, it does positive work when it is in the direction of motion and negative work when it is opposite the direction of motion. STUDENT QUESTION Ok, so i get his really mixed up. The work done BY the object is positve, against gravity which is doing negative work ON the object going up the inlcine. When going down the incline work done BY the object is negative as work done ON the object by gravity is positive. Is this right? INSTRUCTOR RESPONSE Your statement is correct. And, until it 'clicks', this is certainly confusing. It takes most students a few assignments before this becomes clear. You are progressing nicely. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The net force is 500 Newtons. ‘ds = -10,000 Joules / -500 Newtons = 20 meters It has all of its KE when it returns. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 ON the automobile 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. STUDENT QUESTION I’m still not sure how I would find the final KE with all the other forces. I read the given solution multiple times but I am still really confused. INSTRUCTOR RESPONSE Your solution was fine for motion up the incline, and agreed with the given solution to that point. The first line of the given solution that you didn't address in your solution begins 'The automobile will then coast 20 meters back down the incline, ... ' The point you need to understand is that the 400 N gravitational component is still there, but since the car is moving down the incline the 100 N frictional force is now directed up the incline. So the net force on the car is 300 N down the incline. You therefore know the net force acting on the system, and you know the displacement, so you can easily calculate the work `dW_ON done by this net force. This is equal to the change in KE as the car travels down the ramp. The car began with 0 KE at the top (it came to rest for an instant), so you can figure out its final KE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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 from the first few problems in this q_a_ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 14700 N * 6 meters = 88200 Joules confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. This is the work done by the entire gravitational force, not just by its component parallel to the incline. It is multiplied by the 6 meter vertical displacement of the automobile, since it acts along the same line as that displacement. However this is identical to the work done on the automobile by the parallel component of the gravitational force in the original problem. In that problem the parallel component was multiplied by the displacement along the incline (which was much more than 6 meters), since it acts along the same line as that displacement. STUDENT QUESTION I do not see how it’s the same amount of work? 88200 Joules INSTRUCTOR RESPONSE The first few problems in this qa obtained KE = 88200 Joules. They also obtained the result that the work done by the parallel component of the gravitational force acting on the system was 88200 Joules. In this problem we see that in a straight 6 meter drop the work done by gravity is the same, 88200 Joules. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I also do not understand how it is the same. ------------------------------------------------ Self-critique rating:2
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Given Solution: 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. STUDENT QUESTION I am not sure about this concept could you elaborate? I am not really even sure what the question is asking in the first place. INSTRUCTOR RESPONSE Subsequent questions will also reinforce this idea. It was shown in previous problems that an automobile at the top of the ramp is in a position to gain about 88000 J of kinetic energy, by rolling without friction down the incline. It was also shown that an automobile that falls freely from the top of the ramp to the level of the bottom of the ramp will gain the same amount of kinetic energy. In both cases the vertical position of the automobile changed by the same amount. We therefore conjecture that there's something in the change in the vertical position of the automobile that determines how much energy it can gain or lose to gravity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am not sure about this concept. Its just going over my head. ------------------------------------------------ Self-critique rating:
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It wouldn’t have an affect on the potential energy but it would have an effect on the kinetic energy. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ADDITIONAL INSTRUCTOR COMMENT: The PE change is the result of only conservative forces; in this case the PE change depends only on the vertical displacement and the gravitational force. The force of air resistance is nonconservative and has no effect on PE change. KE change depends on net force, which includes both conservative and nonconservative forces. So KE change is affected by both PE change and the work `dW_nc_ON done by nonconservative forces. In this case PE change is negative, which tends to increase KE, while nonconservative forces do negative work on the mass, which tends to reduce KE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 N * 45 m = -450 Joules 3100 J - 450 J = 2650 J would be the ending kinetic energy. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!