Physics Help..
posted by Rick on .
The cable of an elevator of mass M = 1900 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 = 10.6 m above a cushioning spring whose spring constant is k = 8500 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of f = 14499 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.

net force down = m g  14499
= 1900 (9.81)  14499
= 4140 Newtons
Work done by downward force = 4140(10.6+x)Joules
Energy stored in spring = (1/2) k x^2
so
4140(10.6+x) = 4250 x^2
4250 x^2  4140 x  43884 = 0
x^2  .974 x  10.3 = 0
x = [ .974 +/ sqrt (.949+41.2) /2
= [ .974+/6.49 ]/2
= 3.73 
Thank you sooo much!