Friction Work Problem

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Ehwic

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So this question is from the BR:

The question is referring to a figure, where there are three points: Q, Y and Z. Q to Y is a ramp of unknown degrees and Y to Z is some unknown horizontal distance.

The cart is brought to rest because of the frictional force acting from Y to Z (a horizontal unknown distance). The work done by friction to stop the cart under normal conditions is W. Under rainy conditions, how much work is done by friction to bring the cart to rest, assuming the cart reaches point Y (start of the horizontal distance) with its normal speed and the frictional force is reduced by half?)

A) W/2
B) W
C) squrt (2)W
D) 2W

I chose D, correct answer is B.

Can someone explain why B is correct? the Way I thought of it was that since W= frictional force x d, and since force of friction decreases by two, I thought the work would increase by 2 because stopping distance would be twice as long. Why is this wrong considering conservation of energy?
 
hmmm., I could be wrong but I think you answered your own question. if you divide one variable on the right side of the equation by 2, and then multiply the other variable by 2... don't you come up with the same original value? am I missing something?
 
The amount of work done by friction doesn't change. The potential energy of the cart at the top of the incline = mgh. That is the total energy in the system. To bring the cart to rest, W = mgh. Even if the force of friction is reduced, the total work done by friction to bring the cart to rest is still W, it will just take longer for the cart to be brought to rest.
 
The amount of work done by friction doesn't change. The potential energy of the cart at the top of the incline = mgh. That is the total energy in the system. To bring the cart to rest, W = mgh. Even if the force of friction is reduced, the total work done by friction to bring the cart to rest is still W, it will just take longer for the cart to be brought to rest.

Videodrome is one of the weirdest movies I've ever seen. :laugh:
 
Thanks for the responses,

Also, when you raise the height of an object by 2x, why would the velocity increase only by sqrt(2)? shouldnt the equation be 2mgh=.5mv^2? solving for v should get you 2, not sqrt(2).
 
Oh wow yeah I just answered my own question. I don't know what I was thinking. haha.
 
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