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.

Question

This is what I did but the answer was wrong!:
Wtotal = (mg- f) 18.4m
Wtotal = 1/2 kx^2

Answer 1

Your approach looks OK.
Check for any errors in copying the problem or any math errors.

Answer 2

Wtotal = [(1020)(9.8)-6678] 18.4
= 61051.2J
61051.2J = 1/2 (6500)x^2
x = 4.33m

STill wrong :( Help

Answer 3

The displacement of the safety spring has to be considered in the total potential energy of the elevator car.

Answer 4

Um are you saying
Wtotal = [(1020)(9.8)-6678] (18.4+x) ?..
In that case, I did the quadratic and got x=17.33m, but still wrong answer :( argg

Answer 5

For the roots of the quadratic I got
+4.875
-3.854