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## Homework Statement

In a "worst-case" design scenario, a 2000-kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 2.00 m as it does so. During the motion a safety clamp applies a constant 17000-N frictional force to the elevator.

1) What is the speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring?

2) When the elevator is 1.00 m below point where it first contacts a spring, what is its acceleration?

## Homework Equations

K

_{1}+U

_{1}= K

_{2}+U

_{2}-Work

K=.5*m*v^2

U=mgh; .5*k*x^2

W=FD

## The Attempt at a Solution

Using the formulas I got above, I came up with:

k=8500 N/m??

.5*2000*4^2+2000*9.8*2=.5*2000*v^2+.5*8500*1^2 - 17000.

Doing this gave me v = 8.24 m/s^2 which I don't think is right...

I can't begin part 2 until I understand part 1, so no attempt at the moment.

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