course Phy 121 sh抹ǚۓassignment #010 wy Physics I Class Notes 02-18-2008 l|[Ns⧩IՍ assignment #011 wy Physics I Class Notes 02-18-2008 iЅºxWJϏٕpՖ assignment #011 011. `query 11 Physics I 02-18-2008
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23:19:36 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 --> The difference between conservative and nonconservative forces is that anything that can store energy is a conservative force and anything that cannot store energy is a nonconservative force. An example of conservative force is when you lift a book because you have lifted the book it now has potential energy. An example of nonconservative force is the act of droping the book because it is not gaining energy and once it lands it has no energy. confidence assessment: 3
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23:21:10 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 --> confidence assessment: 1
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23:25:02 ** `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 --> `dKE is equal to the NET work done ON the system `dKE = -W1 + W2 `dPE = - W2. PE decreases, thereby tending to increase KE self critique assessment: 3
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23:26:27 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 --> 'dPE = 'dKE - W confidence assessment: 1
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23:26:55 ** 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 --> self critique assessment: 3
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23:29:10 ** 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 --> I understand your example and understand the reasoning bhind it. I was not exactly sure so I didn't give an example. self critique assessment: 3
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23:32:08 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 --> For some reason I was unable to open up the class notes link. confidence assessment: 0
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23:32:52 ** 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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23:35:08 ** 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: 1
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23:36:33 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
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RESPONSE --> Our evidence is that forces acting against the cart is the cart's mass and the force of gravity acting upon it. confidence assessment: 1
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23:36:56 ** the graph of acceleration vs. number of washers should be linear **
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RESPONSE --> the graph of acceleration vs. number of washers should be linear self critique assessment: 3
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00:06:55 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 --> First we convert 85km/hr to m/s: 85km/hr * (1000m / 1km) * (1hr / 3600sec) = 23.6m/s Now we substitute our known variables into the equation y = v0(t2) + 1/2(a)(t1^2). After substituting we get the equation 23.6m/s = 0 + 1/2(9.8m/s^2)(t^2). 23.6m/s = 4.9m/s^2 * t^2 23.6m/s / 4.9m/s^2 = t^2 4.82s = t^2 'sqrt4.82 = t = 2.2 seconds confidence assessment: 2
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00:23:36 **** 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 --> I estimate that the car was put into fourth gear at coordinate (28, 44) and it was put into fifth gear at coordinate (40,50). First we need the average velocity of the car in this interval. We find this as follows: (50-44) / (40-28) = 6/12 = .5 m/s^2
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00:25:26 ** 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 --> I was way off. my coordinates were much different than yours. Not sure why. self critique assessment: 3
Deconstruct the given solution and explain in detail what you do and do not understand about every part.. &#
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