# Homework Help: Energy question, displacement of a spring

1. Jan 28, 2010

### holezch

1. The problem statement, all variables and given/known data

A 4000 elevator breaks off and falls from rest 12 ft above the surface of a relaxed spring. The elevator has a safety system so that the elevator will always experience a frictional force with a constant magnitude of 1000 lb. The spring constant k is equal to 10 000 lb/ft

a) find velocity just before it hits the spring
b) how much does the elevator compress the spring before it stops?

2. Relevant equations

3. The attempt at a solution

I solved part a and for 24 ft/s. For part b, I tried using this equation:

sum of conservative forces work + work of friction = change in K

work of gravity + work of spring + work of friction = change in K

then it would be

-ky^2 + (4000 - 1000)y + 1/2mv^2 = 0

because change in K = 0 - K-initial..
plugging in the rest, it should be solvable. But it didn't work out. The reasoning seems fine? what happened? thanks

2. Jan 28, 2010

### PhanthomJay

Looks OK, did you solve the quadratic equation correctly for y? Or maybe there's no more friction when the elevator hits the spring? Did you convert the elevator weight to its mass unit??

Last edited: Jan 28, 2010
3. Jan 28, 2010

### holezch

thanks a lot, I thought it looked okay as well.. I solved it for y just using the quad. equation.. and I did convert the weight to mass. I'm not sure about the friction thing, I'd imagine that it would still be there, since it's like an elevator shaft and the elevator would still be touching the walls.. If you could, could you please try it and tell me what you get? :S it should be 3 feet but I'm not getting that :S

4. Jan 29, 2010

### PhanthomJay

We both overlooked the fact that the work done by the spring is -1/2ky^2, not -ky^2. Making that correction, y=3 feet.

Note: It is sometimes easier to use the conservation of energy principle rather than the work energy theorem when conservative and nonconservative forces are involved (W_nc = delta PE + delta KE), in that it helps from getting mixed up with the plus and minus signs, and uses the PE of the spring rather than the work done by the spring. PE of a spring is 1/2ky^2, and the work done by the spring is -1/2 ky^2. I am not sure if you just had a typo, or if you assumed that the work done by a spring is -(ky)(y), which it is not, because the spring force is not constant.

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