#$&* course Phy 121 009. Forces exerted by gravity; nature of force; units of force
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Given Solution: Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points. The equation of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2. If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 142 cm/s^2 / (.16) = 890 cm/s^2 (very approximately). Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points. STUDENT COMMENT If slope is rise / run (190 - 21) / (.20 - .02) = 939, then how is 925 the slope? INSTRUCTOR RESPONSE 925 is the ideal slope, which you are unlikely to achieve by eyeballing the position of the best-fit line. Your selected points will be unlikely to give you the ideal slope, and the same is so for the point selected in the given solution. The last paragraph says 'If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) ...' That paragraph doesn't say this selection of point will give the ideal slope. You are unlikely to get the ideal slope based on a graphical selection of points; however you can come reasonably close. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't know that I would have to put all of those points in, like .02, 21cm. So I guess that's why my slope was 887.5. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My points are randomly scattered around the straight line, but they are pretty close to it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended. STUDENT QUESTION Is this because the best fit line does not completely go through all the data points, even though it looks linear, or am I plotting wrong?????????????????????????? INSTRUCTOR RESPONSE The slope you gave in the first problem is the average slope between two actual data points. The line through these two data points is unlikely to be the best fit to the data. You should have sketched the line you think fits the data best, coming as close as possible on the average to the data points (and likely not actually going through any of the data points), then used two points on this line to calculate the slope. However, whatever straight line you used, it won't go through all of the data points, because they don't all lie on a straight line. So there is some variation in the straight lines people will sketch in their attempt to approximate the best-fit line, and there is variation in the slopes of these estimated lines. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, the straight line almost passes through the origin. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0. Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin. In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The gravitational forces exerted on the system are exerted two objects: gravitational force is exerted on the suspended mass, i.e., the part of the original mass that has been removed from the cart and is now suspended gravitational force is exerted on the cart (including the masses that remain in it). The frictional force and the force exerted by the ramp together counter the force of gravity on the cart and the masses remaining in it. The gravitational force on the cart and the masses in it therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses. The net force on the system is therefore just the gravitational force acting which acts on the suspended mass. The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration. STUDENT QUESTION I am pretty sure I understand this question. Basically the only thing acting in this equation to cause a different net force or any type of acceleration is gravity? INSTRUCTOR RESPONSE Gravity is the source of the unbalanced force, so in a sense it is the cause. However the net force is different when different proportions of the mass are suspended, and to say that gravity is the cause does not explain the differing accelerations. The cause of the different accelerations is the configuration of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result. In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated. In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement. How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration each time was almost the same as before. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions. If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit? STUDENT QUESTION so, normally 1 force unit = 9.8m/s^@ acceleration from gravity? The force unit has nothing to do with the acceleration of gravity. Gravity does exert forces. First we define our unit of force, then we use the observed acceleration of gravity to figure out how much force gravity exerts. 1 force unit accelerates 1 mass unit at 1 m/s^2. how many force units does gravity exert on one mass unit so force must be equal to 9.8 force units to be 1m/s^2 9.8 force units would accelerate 1 mass unit at 9.8 m/s^2 INSTRUCTOR COMMENT The action of the gravitational force of Earth on object near its surface acts as follows: The gravitational force on any freely falling mass accelerates it at 9.8 m/s^2. In particular 1 mass unit would, like anything else, accelerate at 9.8 m/s^2 if in free fall. Thus the gravitational force accelerates 1 mass unit at 9.8 m/s^2. Our force unit is defined as follows: 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2. So, for example, 2 force units acting on 1 mass unit would result in an acceleration of 2 m/s^2. 3 force units acting on 1 mass unit would result in an acceleration of 3 m/s^2. etc. We can say the same thing in slightly different words: To accelerate 1 mass unit at 2 m/s^2 would require 2 force units. To accelerate 1 mass unit at 3 m/s^2 would require 3 force units. etc. Following this pattern, we come to the conclusion that To accelerate 1 mass unit at 9.8 m/s^2 would require 9.8 force units. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration is 9.8m/s^2 on any mass unit There should be 9.8 force units exerted on one mass unit. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would be 9.8 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would still be 9.8 Newtons on 1 kg mass. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. How much force would gravity exert on a mass of 8 kg? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To get this, it would be 9.8 * 8 to = 78.4 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg. STUDENT COMMENT I think I did it right but am I not supposed to multiply out 8 * 9.8 Newtons to get 78.6 Newtons? I understand the units wouldn’t cancel, but am I supposed to leave it without “completing” it? INSTRUCTOR RESPONSE It's fine to complete it, and that's what you are generally expected to do. The given solution didn't do that because I wanted to emphasize the logic of the solution without the distraction of the (obvious) arithmetic. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No it is 20 Newtons since the acceleration isn't 9.8cm/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons. STUDENT QUESTION 9.8 = 5kg * 4m/s^2=9.8newtons = 20 kg ok, so force is equal to mass time acceleration, But I am a little confused as to the 9.8 newtons in question 8...it seems you would incorporate this into the calculation somehow? INSTRUCTOR RESPONSE 9.8 Newtons came out of a previous situation in which the acceleration was 9.8 m/s^2. That isn't the acceleration here, so 9.8 Newtons isn't relevant to this question. 9.8 Newtons is the force exerted by gravity on 1 kg. This is because 1 Newton of force accelerates 1 kg at 1 m/s^2, and gravity accelerates a mass at 9.8 m/s^2. 1 Newton accelerates 1 kg at 1 m/s^2, so 9.8 N are required to accelerate 1 kg at 9.8 m/s^2. However the number 9.8 doesn't enter into the current situation at all. We are asked in this question how much force is required to accelerate 5 kg at 4 m/s^2. We could use the following chain of reasoning: 1 Newton accelerates 1 kg at 1 m/s^2, so to accelerate 1 kg at 4 m/s^2 requires 4 Newtons. 4 Newtons accelerate 1 kg at 4 m/s^2, so to accelerate 5 ig at 4 m/s^2 takes 5 times as much, or 20 Newtons. Our calculation 5 kg * 4 m/s^2 = 20 Newtons summarizes this reasoning process. The result of our reasoning can be generalized: To get the force required to give a given mass a given acceleration, we multiply the mass by the acceleration. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would require 2400 Newtons. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. A net force F_net accelerates a certain mass m at 30 cm/s^2. What would be the acceleration of the same mass if subjected to a net force twice as great? What would be the acceleration if the original net force acted on a mass half as great? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the same mass of a net force twice as great should be 60 cm/s^2. The acceleration of the same mass of a net force half as great should be 15 cm/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. What net force would be required to accelerate a 50 kg mass at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 200 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. What would be the acceleration of a 40 kg mass subjected to a net force of 20 Newtons? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration would be 2m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q015. What is the mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass would be 25 kg. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Questions related to Class Notes 1. An ideal cart of mass 8 kg rolls frictionlessly down an incline. The incline has a length of 50 cm and its 'rise', from low end to high end, is 3 cm. According to the relationship between the accelerating force, the slope and the weight of an object on an incline, as described in the Class Notes: What are the magnitude and direction of the force that accelerates the cart down the incline? What therefore is the acceleration of the cart? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. A net force F_net accelerates a certain mass m at 30 cm/s^2. What would be the acceleration of the same mass if subjected to a net force twice as great? What would be the acceleration if the original net force acted on a mass half as great? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration of the same mass of a net force twice as great should be 60 cm/s^2. The acceleration of the same mass of a net force half as great should be 15 cm/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. What net force would be required to accelerate a 50 kg mass at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 200 Newtons confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q013. What would be the acceleration of a 40 kg mass subjected to a net force of 20 Newtons? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration would be 2m/s^2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q015. What is the mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The mass would be 25 kg. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Questions related to Class Notes 1. An ideal cart of mass 8 kg rolls frictionlessly down an incline. The incline has a length of 50 cm and its 'rise', from low end to high end, is 3 cm. According to the relationship between the accelerating force, the slope and the weight of an object on an incline, as described in the Class Notes: What are the magnitude and direction of the force that accelerates the cart down the incline? What therefore is the acceleration of the cart? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy 121
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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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: `dKE = ½ * 1500 kg * 10.9 m/s ^2= 89,107.5 J `ds = 89,107.5 J / -441 N = 202 m 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dKE = 202 m * 441 N = 89, 082 J confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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 despite which direction you are looking at the object traveling, it is still having the same net force acted upon it and was still traveling throughout the same displacement, resulting in the same change of KE. 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: When the car is traveling up an incline, the force of gravity does negative work on the object, if the motion caused by the force to go up the incline is considered positive. In contrast, when the object travels down the incline, gravity would be negative, if traveling down the incline is considered negative as in the previous example. If you were traveling down the incline, the forward progression is positive, so you would look at gravity as a positive force in that context. 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: If gravity is exerting positive force on the object than it will increase the `dKE. If gravity is exerting a negative force on the object than it will decrease the `dKE. 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): The direction of the motion determines positive or negative. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: Since the object is traveling up the incline, then gravity is exerting force to push the object back down the incline. Since the object is moving upward, then the object is doing positive work against gravity. Since gravity is exerting a force causing the object to move down an incline, then pursuant to Newton’s law the object is exerting some kind of force back, but the object is moving because of the force of gravity, so the object is doing negative work against gravity. 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: 10,000 J / (400 N + 100 N) = 20 m
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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): So is the KE 0, or is the `dKE = 300 N * 20 m = 6,000 J ???? ------------------------------------------------ Self-critique rating: 1
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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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* Question: `q010. When the automobile was 200 meters 'up the incline' from 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because whether you are dropping the object or shoving it down the ramp, it still requires the same amount of work to move the object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): This may change dependent upon forces that could be exerted either by air resistance or friction. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7 kg * 9.8 m/s^2 = 68.6 N 68.6 N * 45 m = 3,087 J confidence 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: The motion of the ball in respect to the force of gravity on the object would make the force of gravity positive. This would then make the force exerted by air resistance negative. Since that force is negative, then it would cause a decrease in KE. 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): NA ------------------------------------------------ Self-critique rating: NA ********************************************* 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: 7 kg * 9.8 N = 68.6 N 68.6 N - 10 N = 58.6 N 58.6 N * 45 m = 2,637 J 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): NA ------------------------------------------------ Self-critique rating: NA " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy 121 October 10/ around 6:00 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: Net force is the sum of all forces acting on an object. Typically a number of forces act on a given object. The word 'force' can be used to refer to any of these forces, but the word 'net force' refers exclusively to the sum of all the forces (for future reference note that the word 'sum' refers to a vector sum; this idea of a vector sum will be clarified later). If you're pushing your car on a level surface you are exerting a force, friction is opposing you, and the net force is the sum of the two (note that one is positive, the other negative so you end up with net force less than the force you are exerting). Your heart rate responds to the force you are exerting and the speed with which the car is moving. The acceleration of the car depends on the net force. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the net force exerted on the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The formula for work is `dW = Fnet * `ds so if we just solve for Fnet, then it’s Fnet = `dw / `ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Knowing the distance `ds and the work `dW we use the basic relationship • `dW = F_net * `ds Solving this equation for F we obtain • F_net = `dW / `ds &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Kinetic Energy change is the same thing as the formula `dW = Fnet * `ds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First answer to the question (work = force * distance): This first answer serves to give you the main idea: • the KE change is equal to the work done by the net force. • the work done by the net force is the product of the force and the distance through which it acts so • the KE change is equal to the product of the force and the distance. First answer modified to consider directions of force and motion (work = force * displacement in direction of force): The previous answer applies only if the net force is in same the direction as the motion. More correctly: • the KE change is equal to the work done by the net force. • the work done by the net force is the product of the force and the displacement (not 'distance') in the direction of the force • theKE change is equal to the product of the force and the displacement in the direction of the force. The key difference here is the use of the word 'displacement' rather than 'distance'. Since a displacement, unlike a distance, can be positive or negative, so the work done by a force can be positive or negative. Another thing to keep in mind for the future is that the displacement is to be in the direction of the force. A negative displacement therefore denotes a displacement in the direction opposite the force. We will later encounter instances where the force is not directed along the line on which the object moves, in which case the work will be defined as the force multiplied by the component of the displacement in the direction of the force. ________________________________________ Sometimes we will want to think in terms of the forces exerted ON objects, sometimes in terms of the forces exerted BY objects. The above statement of the work-KE theorem is in terms of the forces exerted ON an object. The basic idea is simple enough. • If a force is exerted ON an object in its direction of motion, the work is positive and the object tends to speed up. • On the other hand if the object exerts a force in its direction of motion, it tends to slow down. • Positive work done ON an object tends to speed it up (increasing its PE), • positive work done BY an object tends to slow it down (decreasing its PE). The above ideas are expanded below to consider forces exerted ON objects vs. forces exerted BY objects. ________________________________________ Synopsis of work-kinetic energy: First be aware that because of Newton's Second Law, there are typically two equal and opposite net forces, the net force which acts on a system and the net force which is exerted by the system. It is necessary to be careful when we label our forces; it's easy to mix up forces exerted by a system with forces exerted on the system. • The first basic principle is that the work by the net force acting ON the system is equal and opposite to the work done by the net force exerted BY the system. The KE, on the other hand, is purely a property OF the system. • The kinetic energy change OF the system is equal to the work done by the net force acting ON the system. • The kinetic energy change OF the system is therefore equal and opposite to the work done by the net force exerted BY the system. Intuitively, when work is done ON a system things speed up but when the system does work things have to slow down. A more specific statement would be • If positive work is done ON a system, the total kinetic energy of the system increases. • If positive work is done BY a system, the total kinetic energy of the system decreases. (We could also state that if negative work is done ON a system, its total KE decreases, which should be easy to understand. It is also the case that if a system does negative work, its total KE increases; it's easy to see that this is a logical statement but most people fine that somehow it seems a little harder to grasp). Below we use `dW_net_ONfor the work done by the net force acting ON the system, and `dW_net_BYfor the work done by the net force being exerted BY the system. The work-kinetic energy theorem therefore has two basic forms: The first form is • `dW_net_ON = `dKE which states that the work done by the net force acting ON the system is equal to the change in the KE of the system. The second form is • `dW_net_BY + `dKE = 0 which implies that when one of these quantities is positive the other is negative; thus this form tells us that when the system does positive net work its KE decreases. ________________________________________ Summary: • work = force * distance gives us the general idea but needs to be refined • work = force * displacement is a correct definition as long as motion is along a straight line parallel to the force • work = force * displacement in the direction of the force is true for all situations • If the net force does positive work on the system, the system speeds up. Negative work on the system slows it down. More precisely: • `dW_net_ON is the work done by the net force acting ON the system, and is equal to the KE change of the system. This is the work-kinetic energy theorem. One alternative way of stating the work-kinetic energy theorem: Forces exerted on the system are equal and opposite to forces exerted by the system, so • If the net force exerted by the system does positive work the system slows down. Negative work done by the system speeds it up: • `dW_net_BY is the work done by the net force exerted BY the system, and is equal and opposite to the KE change of the system This expanded discussion is in a separate document at the link Expanded Discussion of Work and KE . It is recommended that you bookmark this discussion and refer to it often as you sort out the ideas of work and energy. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qWhy is KE change equal to the product of net force and displacement? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m really not sure why this is so. I read the above summary of the theorem, but still can’t answer this question. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE, so that F `ds = `dKE. Here F is the net force acting on the system, so we could more specifically write this as • F_net_ON = `dKE. STUDENT QUESTION: I do not see how you go from KE = 1/2 m v^2 to F_net 'ds = kEf - Ke0 INSTRUCTOR RESPONSE: If KE = 1/2 m v^2, then KEf = 1/2 m vf^2 stands for the KE at the end of the interval and KE0 = 1/2 m v0^2 stands for the KE at the beginning of the interval. Then F_net `ds = 1/2 m vf^2 - 1/2 m v0^2 becomes F_net `ds = KEf - KE0. STUDENT COMMENT In my answer I simply related it to work, I didn’t realize It was supposed to be derived from a formula. Either way, I have read through the solution and almost fully understand. I am only slightly confused by the initial choice of formula. Was this just because these were the units that were given? INSTRUCTOR RESPONSE The definition of KE can be regarded as coming from the formula. The formula is there, and when we substitute a = F / m we get quantities which we define as work and KE. The question that motivates us to do this is 'what happens when a certain force is exerted over a certain distance?' This question can be contrasted with 'what happens when a certain force is exerted over a certain time interval?'. When we answer this question, we get the quantities we define as impulse and momentum. University Physics Students Note: The formula approach outline above is based on the equations of uniformly accelerated motion. However the concept of work and kinetic energy applies whether acceleration is uniform or not. If a force F(x) is applied over a displacement interval from x_0 to x_f, we define the work to be the definite integral of F(x) with respect to x, over this interval, and it isn't difficult to show that the result is the change in the KE. If F(x) is constant, then the result is equivalent to what we get from the equations of uniform acceleration. Similarly if force F(t) is applied over a time interval, an integral leads us to the general definitions of impulse and momentum. STUDENT QUESTION I do not understand why you related this to that one specific equation INSTRUCTOR RESPONSE I assume you mean the equation vf^2 = v0^2 + 2 a `ds. The original question concerned the effect on velocity of applying a given force on a given mass through a given displacement, starting with a given initial velocity. The given force and mass imply the acceleration. Acceleration, initial velocity and acceleration imply the final velocity. So the equations arise naturally from the question. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sure like the other students where the equation came from. Also, I was confused on the part where we went from KE = 1/2 m v^2 to F_net 'ds = kEf - Ke0. But seeing as Kef is = 1/2 m vf^2 and KE0 is = 1/2m v0^2 ------------------------------------------------ Self-critique Rating: ********************************************* Question: When we push an actual object with a constant force, why do we not expect that the KE change is equal to the product F * `ds of the force we exert and the distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The change in KE is equal to the net force. The net force is not considered if we are pushing it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Change in KE is equal to the work done by the net force, not by the force I exert. i.e., `dKE = F_net * `ds The net force is not generally equal to the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!