course PHY 121 7/16/09 12:08 am 011. `query 11*********************************************
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Given Solution: `a** A conservative force conserves energy--you can get your energy back. For example: Push something massive up a hill, then climb back down the hill. The object, by virtue of its position, has the potential to return most of your energy to you, after regaining it as it rolls back down. You will have done work against gravity as you move along a path up the hill, and gravity can return the energy as it follows its path back down the hill. In this sense gravity conserves energy, and we call it a conservative force. However, there is some friction involved--you do extra work against friction, which doesn't come back to you. And some of the energy returned by gravity also gets lost to friction as the object rolls back down the hill. This energy isn't conserved--it's nonconservative. ** Another more rigorous definition of a conservative force is that a force is conservative if the work done to get from one point to another independent of the path taken between those two points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qIf a system does work W1 against a nonconservative force while conservative forces do work W2 on the system, what are the change in the KE and PE of the system? Explain your reasoning from a commonsense point of view, and include a simple example involving a rubber band, a weight, an incline and friction. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dKE = Wnet dPE = -W2 Wnet = -W1 + W2 Or dKE = -W1 + W2 Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** `dKE is equal to the NET work done ON the system. The KE of a system changes by an amount equal to the net work done on a system. If work W1 is done BY the system against a nonconservative force then work -W1 is done ON the system by that force. `dPE is the work done BY the system AGAINST conservative forces, and so is the negative of the work done ON the system BY nonconservative forces. In this case then `dPE = - W2. PE decreases, thereby tending to increase KE. If work -W1 is done ON the system by a nonconservative force and W2 is done ON the system by a conservative force, the NET work done ON the system is -W1 + W2. The KE of the system therefore changes by `dKE = -W1 + W2. If the nonconservative force is friction and the conservative force is gravity, then since the system must do positive work against friction, W1 must be positive and hence the -W1 contribution to `dKE tends to decrease the KE. e.g., if the system does 50 J of work against friction, then there is 50 J less KE increase than if there was no friction. If the work done by the nonconservative force on the system is positive, e.g., gravity acting on an object which is falling downward (force and displacement in the same direction implies positive work), the tendency will be to increase the KE of the system and W2 would be positive. If W2 is 150 J and W1 is 50 J, this means that gravity tends to increase the KE by 150 J but friction dissipates 50 J of that energy, so the change in KE will be only 100 J. If the object was rising, displacement and gravitational force would be in opposite directions, and the work done by gravity would be negative. In this case W2 might be, say, -150 J. Then `dKE would be -150 J - 50 J = -200 J. The object would lose 200 J of KE (which would only be possible if it had at least 200 J of KE to lose--think of an object with considerable velocity sliding up a hill). ** STUDENT COMMENT I find this really confusing. Could this be laid out in another way? INSTRUCTOR RESPONSE If you find this confusing at this point, you will have a lot of company. This is a challenge for most students, and these ideas will occupy us for a number of assignments. There is light at the end of the tunnel: It takes awhile, but once you understand this, the basic ideas become pretty simple and even obvious, and once understood they are usually (but not always) easy to apply This could be laid out differently, but would probably be equally confusing to any given student. Different students will require clarification of different aspects of the situation. If you tell me what you do and do not understand about the given solution, then I can clarify in a way that will make sense to you. I also expect that in the process of answering subsequent questions, these ideas will become increasingly clear. In any case feel free to insert your own interpretations, questions, etc. into a copy of this document (mark insertions with &&&& so I can locate them), and submit a copy. STUDENT QUESTION If the system goes against the force will this always make it negative? INSTRUCTOR COMMENT If a force and the displacement are in opposite directions, then the work done by that force is negative. If the system moves in a direction opposite the force exerted BY the system, the work done BY the system is negative. Note, however, that if this is the case then any equal and opposite force exerted ON the system will be in the direction of motion, so the force will do positive work ON the system. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qIf the KE of an object changes by `dKE while the total nonconservative force does work W_nc on the object, by how much does the PE of the object change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W_nc = `dKE + `dPE `dPE = -`dKE + W_nc Confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We have `dKE + `dPE + `dWbyNoncons = 0: The total of KE change of the system, PE change of the system and work done by the system against nonconservative forces is zero. Regarding the object at the system, if W_nc is the work done ON the object by nonconservative forces then work -W_nc is done BY the object against nonconservative forces, and therefore `dWnoncons = -W_nc. We therefore have `dKE + `dPE - W_nc = 0 so that `dPE = -`dKE + W_nc. ** Equivalently, the work-energy theorem can be stated `dW_ON_nc = `dKE + `dPE In this example the work done on the system by nonconservative forces is labeled W_nc, without the subscript ON and without the `d in front. However it means the same thing, so the above becomes STUDENT COMMENT Im still confused on how to understand when the energy is done on the object and when the energy is done against the object. INSTRUCTOR RESPONSE In an application, that can be the difficult question. However in this case it is stated that W_nc is the work done by nonconservative forces ON the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qGive a specific example of such a process. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you move a 10 N box a distance of 8 meters W = F/d W = 10 Newtons / 8 meters W = 1.25 Joules And the KE change by 1 Joule 1.25 J 1 J = .25 J which would be the dPE Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For example suppose I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J. The 300 J of work done by my force and friction is used to increase the KE by 200 J, leaving 100 J to be accounted for. More formally, `dW_noncons_ON = +300 J and `dKE = +200 J. Since `dW_noncons_ON = `dKE + `dPE, So +300 J = +200 J + `dPE, and it follows that `dPE = +100 J. This 100 J goes into the PE of the object. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qClass notes #10. Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall? Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The more clips there are then the greater the mass and the greater the pull of gravity on them. In terms of the cart, the slope will be related to the number of washers. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Informally: The more clips, the more gravitational force, and the more the clips descend the more work is done by that force. The amount of work depends on how many clips, and on how far they descend. The number of clips required is proportional to the slope (as long as the slope is small). More formally, the force exerted by gravity is the same on each clip, so the total gravitational force on the hanging clips is proportional to the number of clips. The work done is the product of the force and the displacement in the direction of the force, so the work done is proportional to product of the number of washers and the vertical displacement. To pull the cart up a slope at constant velocity the number of washers required is proportional to the slope (for small slopes), and the vertical distance through which the cart is raised by a given distance of descent is proportional to the slope, to the work done is proportional to the vertical distance thru which the cart is raised. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qHow does the work done against friction of the cart-incline-pulley-washer system compare with the work done by gravity on the washers and the work done to raise the cart? Which is greatest? What is the relationship among the three? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The gravity on the washers is greater. The work done against the friction and the work done against gravity to raise the cart will equal the work of the gravity on the weights. Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The force exerted by gravity on the hanging weights tends to move the system up the incline. The force exerted by gravity on the cart has a component perpendicular to the incline and a component down the incline, and the force exerted by friction is opposed to the motion of the system. In order for the cart to move with constant velocity up the incline the net force must be zero (constant velocity implies zero accel implies zero net force) so the force exerted by gravity in the positive direction must be equal and opposite to the sum of the other two forces. So the force exerted by gravity on the hanging weights is greater than either of the opposing forces. So the force exerted by friction is less than that exerted by gravity on the washers, and since these forces act through the same distance the work done against friction is less than the work done by gravity on the washers. The work done against gravity to raise the cart is also less than the work done by gravity on the washers. Work done against friction + work against gravity to raise cart = work by gravity on the hanging weights. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ********************************************* Question: `qWhat is our evidence that the acceleration of the cart is proportional to the net force on the cart? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The acceleration vs. number of washers graph would be liner. Confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** the graph of acceleration vs. number of washers should be linear ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `qprin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 85 km/h (1000 m/ km) (1 h / 36000 sec = 23.6 m/s dt = dv /a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec Confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe know that the acceleration of gravity is 9.8 m/s^2, and this is the rate at which the velocity of the car changes. The units of 85 km/hr are not compatible with the units m/s^2, so we convert this velocity to m/s, obtaining velocity 85 km/hr ( 1000 m/km) ( 1 hr / 3600 sec) = 23.6 m/s. Common sense tells us that with velocity changing at 9.8 m/s every second, it will take between 2 and 3 seconds to reach 23.6 m/s. More precisely, the car's initial vertical velocity is zero, so using the downward direction as positive, its change in velocity is `dv = 23.6 m/s. Its acceleration is a = `dv / `dt, so `dt = `dv / a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: `q**** prin phy and gen phy problem 2.52 car 0-50 m/s in 50 s by graph How far did the car travel while in 4 th gear and how did you get the result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve = (36.5 m/d + 45 m/s ) / 2 = 40.75 m/s dt = (27.5s 16s) = 11.5s ds = vAve * dt ds = 40.75 m/s * 11.5 s dst = 468.63 m Confidence rating: 3
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Given Solution: `a** In 4th gear the car's velocity goes from about 36.5 m/s to 45 m/s, between clock times 16 s and 27.5 s. Its average velocity on that interval will therefore be vAve = (36.5 m/s + 45 m/s) / 2 = 40.75 m/s and the time interval is 'dt = (27.5s - 16s) = 11.5 s. We therefore have 'ds = vAve * `dt = 40.75 m/s * 11.5 s = 468.63 m. The area under the curve is the distance traveled, since vAve is represented by the average height of the graph and `dt by its width. It follows that the area is vAve*'dt, which is the displacement `ds. The slope of the graph is the acceleration of the car. This is because slope is rise/run, in this case that is 'dv/'dt, which is the ave rate of change of velocity or acceleration. We already know `dt, and we have `dv = 45 m/s - 36.5 m/s = 8.5 m/s. The acceleration is therefore a = `dv / `dt = (8.5 m/s) / (11.5 s) = .77 m/s^2, approx. ** Self-critique (if necessary): ok Question: `q **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The graph is of velocity vs. clock time, so the rise will be change in velocity and the run will be change in clock time. So the slope = rise/run represents change in vel / change in clock time, which is acceleration. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `qGen phy what is the meaning of the area under the curve, and why does it have this meaning? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt. 'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid it's close enough that we for the purpose of estimation can analyze it as such. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `qGen phy what is the area of a rectangle on the graph and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The area of a rectangle on the graph represents a distance. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `quniv phy problem 2.90 from 10th edition (University Physics students should solve this problem now). Superman stands on the top of a skyscraper 180 m high. A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds. If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: `quniv phy what is Superman's initial velocity, and what does the graph look like (be specific)? ``a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. ** ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2. A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec. For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex. A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0). Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0). Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): STUDENT QUESTIONS I am not clear about the relationship between KE, PE and Work Will nonconservative forces always be - ? And my notes say PE can never be neg. therefore PE will always be conserved? INSTRUCTOR RESPONSE Any force can do positive or negative work, depending on how the direction of the force compares to the direction of the displacement. To determine the sign of a force or a displacement, you first to choose your positive direction. The choice is yours. Having made that choice: Forces are positive or negative according to whether they act in or opposite to the direction you have chosen to be positive. Displacements are positive or negative according to whether they act in or opposite to the direction you have chosen to be positive. The work done by a force is positive or negative according to the signs of the force and the displacement. You multiply the force by the displacement to get the work, and if you include the signs of all quantities, the positives and negatives take care of themselves (i.e., iif force and displacement are in the same direction the work is positive; if they are in the opposite direction the work is negative). KE = 1/2 m v^2. v^2 can't be negative, nor can m, so KE can't be negative. PE can be positive or negative. `dKE represents change in KE on an interval, and can be positive or negative `dPE represents change in PE on an interval, and can be positive or negative. Here is a summary of the work-energy theorem: The work-kinetic energy theorem tells us that `dW_net_ON = `dKE, where `dW_Net_ON is the work done by the net force acting on the system. `dW_net_ON = `dW_nc_ON + `dW_cons_ON (work by the net force acting on the system is broken into work done by the conservative and nonconservative forces acting on the system) `dW_cons_ON = -`dPE (change in PE is the result of conservative forces acting on the system, and is equal and opposite to the work done by those forces) These ideas are combined to give us `dW_net_ON = `dKE `dW_net_ON = `dW_nc_ON + `dW_cons_ON so `dW_nc_ON + `dW_cons_ON = `dKE and `dW_nc_ON - `dPE = `dKE, which can be rearranged to give `dW_nc_ON = `dKE + `dPE. "