[SOLVED] Conservation of Energy With Loss Due to Frictionby seichan Tags: conservation, energy, friction, loss, solved 

#1
Mar708, 07:44 PM

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 mgdf. We can set this eqaul to the amount of work done by the spring and solve for x. In other words: sqrt((2(mgdf))/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. 



#2
Mar708, 07:48 PM

Mentor
P: 40,877

When calculating the gravitational PE, don't forget to include the distance that the spring compresses.




#3
Mar708, 07:54 PM

P: 32

1/2kx^2=mg(d+x)f =mgd+mgxf 1/2kx^2mgx=mgdf Also, can't I look at the top of the spring as the xaxis? That was the asumption I was going with. 



#4
Mar708, 08:01 PM

Mentor
P: 40,877

[SOLVED] Conservation of Energy With Loss Due to FrictionYou'll end up with a quadratic equation. Solve it! 



#5
Mar708, 08:05 PM

P: 32

To clarify, it should be:
1/2kx^2=mg(d+x)fx 1/2kx^2=mgd+(mgf)x 1/2kx^2(mgf)xmgd=0 Right? Thank you for your help =) 



#6
Mar708, 08:09 PM

Mentor
P: 40,877

The friction force acts throughout the motion, not just over the distance x. Redo the work done against friction.




#7
Mar708, 08:16 PM

P: 32

Thank you so much =) It's correct



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