course phy 231 Nϧ،Fwӳxؿassignment #011 011. `query 11 Physics I 02-13-2008
......!!!!!!!!...................................
09:53:21 set 3 problems 15-19. Explain the difference between a conservative and a nonconservative force, and give an example of each.
......!!!!!!!!...................................
RESPONSE --> conservative forces areforces that store energy. an example of this is gravity and eletric forces. when lifting an object, you are moving against gravity. nonconsecutive deals with friction and do not store energy. resistance is an example. so pushing a book on the floor is an example of that confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:53:27 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 0
.................................................
......!!!!!!!!...................................
10:06:51 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.
......!!!!!!!!...................................
RESPONSE --> the force changes from potential energy to kinetic energy. when the energy is released from the object and it hits something, it converts to kinetic energy, thereby causing the ground to absorb the potential energy. as the weight is held in place with the rubber band on the incline, the potential energy is being stored. as it is released down the incline and stretches, the potential energy is then converted to kinetic energy. But since theres friction, then there should be no potential energy. confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:06:58 ** `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). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:10:32 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?
......!!!!!!!!...................................
RESPONSE --> potential energy is lost confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:10:40 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:13:27 Give a specific example of such a process.
......!!!!!!!!...................................
RESPONSE --> anexample would be if a ball was dropped in a cup. the ball is at rest at the bottom, so potential energy is lost. confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:13:33 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:22:58 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?
......!!!!!!!!...................................
RESPONSE --> because the total mass would be the same. when the object is pushed in the same direction, the work done is equal to the force times the distance. the force used must be eual to its weight. W = Fd force time distance would equal the work done on the object confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:23:09 ** 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. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:29:25 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?
......!!!!!!!!...................................
RESPONSE --> the work done will be equal to the upward force times vertical distance. the same goes for it being pushed in one direction. again, W = Fd. the work done to raise the cart is greater, because of the stored potential energy. the relationship between the 3 would probably be that they deal with force and distance confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:29:31 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:30:45 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
......!!!!!!!!...................................
RESPONSE --> the formula Fnet = mass * acceleration confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:30:58 ** the graph of acceleration vs. number of washers should be linear **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:35:58 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?
......!!!!!!!!...................................
RESPONSE --> the minimum value would be 0 confidence assessment: 1
.................................................
......!!!!!!!!...................................
10:39:36 univ phy what is Superman's initial velocity, and what does the graph look like (be specific)?
......!!!!!!!!...................................
RESPONSE --> tiem would be on the x axis and the inital velocity would be onthe y axis. supermans inital velocity would be a slight curve.from the top of his 0v, the curve goes downward as it goes straight as it reaches the time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
10:39:56 ``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. **
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment:
.................................................