course Phy 201 assignment #011011. `query 11
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20:31:33 set 3 problems 15-19. Explain the difference between a conservative and a nonconservative force, and give an example of each.
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RESPONSE --> A conservative force is any force whose work is determined by the final displacement of the object acted upon. An example is the gravitational force between the Earth and me (or any other mass). Another example is potential energy. A nonconservative force depends on other factors, such as velocity. These forces are dissipative and no potential energy can be found for them. An example is friction. confidence assessment: 3
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20:56:36 ** A conservative force conserves energy--you can get your energy back. For example: Push something up a hill, climb back down the hill and turn your back and it will probably return your energy to you--all at once--after regaining it as it rolls back down. So you do work against gravity, and gravity can return the energy as it pulls the thing back to you. 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. **
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RESPONSE --> ok self critique assessment: 3
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20:57:40 If 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.
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RESPONSE --> W1 work done BY the system -W1 work done ON the system W2 work done ON the system -W2 work done BY the system -W2 work done BY the system of conservative forces is equal to the change in PE of the system. If an object falls, gravity does positive work ON it so the work done by the object against gravity is negative, which means the PE is negative. Since the change in KE is equal to the work done ON the system by the net force, the KE change is equal and opposite to the work done BY the system against the net force. EX: A marble is released from a rubber band up an incline. The marble goes so far and then begins to come back down. Friction is included so some of the KE is lost but most is returned back as it rolls back down the incline. The work done against gravity when the marble is released up the incline is regained in the form of KE when the marble rolls back down the incline; work done by the marble against the rubber band as it is stretched will be regained as the marble is released. As the marble is fighting against gravity the system does positive work again conservative forces, which increases the potential energy of the system. As the marble falls back down the incline to its original position the PE is reclaimed in the form of KE. confidence assessment: 2
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20:57:49 ** `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). **
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RESPONSE --> ok self critique assessment: 3
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20:59:38 If the KE of an object changes by `dKE while the total nonconservative force does work W on the object, by how much does the PE of the object change?
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RESPONSE --> `dKE + `dPE + `dWbyNoncons = 0 Refering to the object at the system, if W is the work done ON the object by nonconservative forces then work -W is done BY the object against nonconservative forces; thus, `dWnoncons = -W. We therefore have `dKE + `dPE - W = 0 so that `dPE = -`dKE + W. ** confidence assessment: 3
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20:59:44 ** 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 is the work done ON the object by nonconservative forces then work -W is done BY the object against nonconservative forces, and therefore `dWnoncons = -W. We therefore have `dKE + `dPE - W = 0 so that `dPE = -`dKE + W. **
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RESPONSE --> ok self critique assessment: 3
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21:04:09 Give a specific example of such a process.
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RESPONSE --> Ex: Lifting an object weighing 20N. When adding my force and the friction of the object we can obtain the total nonconservative force. The nonconservative force is + 400 J of work on the object. The KE changes by +200J; thus the 400J of work done by my force and friction is used to raise the KE by 200J leaving 200J left over. The 200J is added to the PE of the object. confidence assessment: 3
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21:04:19 ** For example if 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 then 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. This 100 J goes into the PE of the object. **
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RESPONSE --> ok self critique assessment: 3
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21:06:54 Class 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?
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RESPONSE --> The product of the number of washers and the distance through which they fall is proportional to the work done by gravity. This is because all the washers will fall the same amount of distance at the same time and the amount of force from gravity on each washer is the same amount. Therefore you simply multiply the number of washers and the distance they fall to obtain the work done on them by gravity. This is 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 because when the cart ascends we think of work being done on it to raise it. When the cart moves up an incline at constant velocity, the work done on it serves only to raise it. Since a heavier cart moved through a greater vertical distance would be associated with more work, we think of the work done on the cart as the product of its weight and the distance. confidence assessment: 3
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21:06:59 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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21:09:32 How 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?
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RESPONSE --> The force exerted by gravity on the hanging weights tends to move the system up the incline. The force exerted by friction is opposed to the motion of the system. In order for the cart to move with 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. 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. The work friction + work against gravity to raise cart = work by gravity on the hanging weights. ** confidence assessment:
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21:09:42 ** 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. The work friction + work against gravity to raise cart = work by gravity on the hanging weights. **
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RESPONSE --> ok self critique assessment: 3
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21:10:47 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
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RESPONSE --> The evidence is that a graph plotting acceleration vs. the number of washers should be linear. confidence assessment: 2
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21:10:54 ** the graph of acceleration vs. number of washers should be linear **
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RESPONSE --> ok self critique assessment: 3
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21:13:16 prin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr?
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RESPONSE --> 85 km/hr ( 1000 m/km) ( 1 hr / 3600 sec) = 23.6 m/s. With velocity changing at 9.8 m/s every second, it will take between 2 and 3 seconds to reach 23.6 m/s. v0 = 0 `dv = 23.6 m/s a = `dv / `dt, so `dt = `dv / a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec confidence assessment: 3
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21:13:22 We 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 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..
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RESPONSE --> ok self critique assessment: 3
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21:15:52 **** 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?
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RESPONSE --> vAve = (36.5 m/s + 45 m/s) / 2 = 40.75 m/s 'dt = (27.5s - 16s) = 11.5 s '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. `dv = 45 m/s - 36.5 m/s = 8.5 m/s a = `dv / `dt = (8.5 m/s) / (11.5 s) = .77 m/s^2 confidence assessment: 2
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21:16:00 ** 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. **
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RESPONSE --> ok self critique assessment: 3
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21:18:14 **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning?
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RESPONSE --> The slope of the graph is usually defined as the rise over the run. The rise is the vertical coordinate and the run is the horizontal coordinate. In this case, the slope represents the acceleration of the car or `dv/`dt. confidence assessment: 3
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21:18:41 ** 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. **
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RESPONSE --> got a little confused by I understand self critique assessment: 2
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21:20:32 Gen phy what is the meaning of the area under the curve, and why should it have this meaning?
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RESPONSE --> the meaning of the area under the curve is distance travelled. VAve is the average height of the graph and `dt by its width. The area is vAve * `dt which is `ds. confidence assessment: 2
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21:20:44 ** 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 we can annalyze it as one for estimation puposes. **
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RESPONSE --> ok self critique assessment: 3
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21:25:15 Gen phy what is the area of a rectangle on the graph and what does it represent?
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RESPONSE --> area of a rectangle equals length times width or acceleration times time. This implies that an acceleration vs. time graph is equal to the velocity. confidence assessment: 2
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21:25:33 ** The area of a rectange on the graph represents a distance. **
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RESPONSE --> oh ok...so simple! self critique assessment: 2
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21:25:45 univ 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?
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RESPONSE --> does not apply confidence assessment: 0
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21:25:50 univ phy what is Superman's initial velocity, and what does the graph look like (be specific)?
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RESPONSE --> ok confidence assessment: 3
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21:26:16 ``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. **
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RESPONSE --> ok confidence assessment: 0
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21:26:23 ``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
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RESPONSE --> ok confidence assessment: 3
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21:26:34 ** 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. **
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RESPONSE --> ok self critique assessment: 0
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