- #1

tristanslater

- 14

- 0

## 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.

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

## Homework Equations

Not 100% sure, but probably: 1/2mv^2, PE = mgh, -1/2kx^2

## The Attempt at a Solution

I am pretty lost on this one, but have been attempting to figure out the spring constant based on the information that the elevator is stopped at 2 m. I tried F_net = F_spring - F_g + F_f, F_net at 2 m is 0, so F_spring = F_g - F_f, F_f is given, and F_g is easy to workout (mg), from there I get: k = (mg-F_f) / delta-y, which gets me 2600 as a spring constant. I also calculated the KE_i = 1/2m(v_i)^2.

After that I get kind of stuck. I feel like I am supposed to figure out the change in energy, but I'm not sure how friction and the spring factor into that.

Thanks,

Tristan