Confused about Work Energy again

[dnovikov]

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So my confusion is in regards to the second question which talks about how much work is done to stop the cart. The answer says that the W= PEfinal- PEinitial since the change in Kinetic energy is 0 and frictional force isn't given. However I was wondering why isn't the frictional force considered to do work on the cart since it hits a surface of friction from point b --> c. Is it because work just deals with the change from initial energy which is all potential to a final energy with no potential.

 

[daFendi]

So my confusion is in regards to the second question which talks about how much work is done to stop the cart. The answer says that the W= PEfinal- PEinitial since the change in Kinetic energy is 0 and frictional force isn't given. However I was wondering why isn't the frictional force considered to do work on the cart since it hits a surface of friction from point b --> c. Is it because work just deals with the change from initial energy which is all potential to a final energy with no potential.

Well the thing is, if the cart stops some distance away from the ramp, that must mean that after it arrives at the horiztonal surface it has some momentum, which must be stopped somehow. the only way is friction, because nothing else seems to be acting on the cart. The reason they do not use the force of friction to figure it out, is because we only know the coefficient of friction on the ramp, not the horizontal surface, thus we must calculate the work done based on the potential energy of the cart between the two points.

At least, thats the way I understood the question.

 

[milski]

So my confusion is in regards to the second question which talks about how much work is done to stop the cart. The answer says that the W= PEfinal- PEinitial since the change in Kinetic energy is 0 and frictional force isn't given. However I was wondering why isn't the frictional force considered to do work on the cart since it hits a surface of friction from point b --> c. Is it because work just deals with the change from initial energy which is all potential to a final energy with no potential.

They are asking about [all] the work done on the cart. It might as well be that all this work was done by the friction force - it really does no matter. At the end of the day, one or more forces did work on that cart and its potential energy changed. The amount of change in that PE is the same as all the work done.

daFendi is right about why 0.2 was not used in finding out the answer - you can find how much work was done by the friction force on the ramp but you don't know if that is all the work done and you have no idea how much other work was done. The only choice is to deal with the total energy change.

 

[dnovikov]

Thanks for the quick responses. I do however have another question. It was said that when for example a weight lifter lifts weight a certain h, there is no work done since there is no x displacement. However if W= change in PE, and since the object is at a different h, shouldn't there be work done on the object?

 

[vashka]

Where did you find this problem? Thanks

 

[dnovikov]

Where did you find this problem? Thanks

Berkeley review physics chapter on work and energy. Its on the first page of the chapter.

 

[milski]

Thanks for the quick responses. I do however have another question. It was said that when for example a weight lifter lifts weight a certain h, there is no work done since there is no x displacement. However if W= change in PE, and since the object is at a different h, shouldn't there be work done on the object?

I'm not sure where the x displacement comes from here? It would be applicable if they were talking about work done by a force parallel to the x direction but I cannot think of such force in a simple lifter case.

There are two foces in that case: gravity and the lifter pushing up. The gravity does work on the weight equal to the change in PE. The lifter does exactly the same amount of work on the weight but with the opposite sign.

 

[indianjatt]

^So are we saying that in the first case a change in PE is equal to work, but in the second case of lifting an object Work is only = to KE and not PE, because of what? Is it because only gravity is acting on the object? Could the ramp be potentially doing negative work to counterbalance the positive work of gravity on the block?

 

[milski]

^So are we saying that in the first case a change in PE is equal to work, but in the second case of lifting an object Work is only = to KE and not PE, because of what? Is it because only gravity is acting on the object? Could the ramp be potentially doing negative work to counterbalance the positive work of gravity on the block?

You have to be specific about which work you are talking about. The total work done on an object (by all forces acting on it) is always going to be ΔΚΕ. That's what's going on with the lifter, or at least what I was talking about.

If the example that started this thread we are talking about the work done by a single force on an object. The PE in this case is related to the gravity force and as such any changes in the PE will be due to work done by the gravity force. (The total work again is 0, but that is not something that we care about in this case).