Posted by Elle on .
The cable of an elevator of mass M = 1020 kg snaps when the elevator is at rest at one of the floors of a skyscraper. At this point the elevator is a distance d = 18.4 m above a cushioning spring whose spring constant is k = 6500 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 6678 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.
This is what I did but the answer was wrong!:
Wtotal = (mg- f) 18.4m
Wtotal = 1/2 kx^2