# [SOLVED] Conservation of Energy With Loss Due to Friction

by seichan
Tags: conservation, energy, friction, loss, solved
 P: 32 1. The problem statement, all variables and given/known data The cable of an elevator of mass m= 2690 kg snaps when the elevator is a rest at one of the floors of a skyscraper. At this point the elevator is a distance d= 75.0 m above a cushioning spring whose spring constant is k= 9700 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f= 9358 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed. 2. Relevant equations U(x)=mgh w(x)=1/2kx^2 3. The attempt at a solution Alright, I thought this problew was pretty straight forward. I first calculated the potential energy for the elevator when it is at rest, in which I found to be U(x)=mgd. This would be equal to the Kinetic friction at the bottom, however, there is energy loss to friction. So, the energy when the elevator hits the spring is U(x)-Force of friction, or mgd-f. We can set this eqaul to the amount of work done by the spring and solve for x. In other words: sqrt((2(mgd-f))/k)=x. The displacement is negative, so we want the negative value of this square root. I have put this answer into my web assignment a few times and cannot figure out where I am going wrong (I've tried using a positive value for the displacement as well). If you could tell me where my reasoning is wrong, that would be great.
 Mentor P: 40,279 When calculating the gravitational PE, don't forget to include the distance that the spring compresses.
P: 32
 Quote by Doc Al When calculating the gravitational PE, don't forget to include the distance that the spring compresses.
Thank you very much- but how do I isolate x in that case?

1/2kx^2=mg(d+x)-f
=mgd+mgx-f
1/2kx^2-mgx=mgd-f

Also, can't I look at the top of the spring as the x-axis? That was the asumption I was going with.

Mentor
P: 40,279

## [SOLVED] Conservation of Energy With Loss Due to Friction

 Quote by seichan Thank you very much- but how do I isolate x in that case? 1/2kx^2=mg(d+x)-f =mgd+mgx-f 1/2kx^2-mgx=mgd-f
That "f" should be the work done by friction, not just the friction force.

You'll end up with a quadratic equation. Solve it!

 Also, can't I look at the top of the spring as the x-axis? That was the asumption I was going with.
Not sure what you mean. You can certainly measure distances from the top of the spring, but in that case your final position will have a gravitational PE term. (If that's what you mean.)
 P: 32 To clarify, it should be: 1/2kx^2=mg(d+x)-fx 1/2kx^2=mgd+(mg-f)x 1/2kx^2-(mg-f)x-mgd=0 Right? Thank you for your help =)
 Mentor P: 40,279 The friction force acts throughout the motion, not just over the distance x. Redo the work done against friction.
 P: 32 Thank you so much =) It's correct

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