Posted by shelby on .
The cable of an elevator of mass = 5320 kg snaps when the elevator is a rest at one of the floors of a skyscraper. At this point the elevator is a distance = 31.6 m above a cushioning spring whose spring constant is = 11700 N/m. A safety device clamps the elevator against the guide rails so that a constant frictional force of = 18158 N opposes the motion of the elevator. Find the maximum distance by which the cushioning spring will be compressed.

physics 
Damon,
Kinetic energy that the elevator has when it hits the spring is:
KE = m*g*hwork done by friction so far
the work done by friction so far =18158*h
where
m = 5320 kg
g = 9.81 m/s^2
h  31.6 m
Now we compress the spring a distance x
Work done by gravity in falling additional distance x = m g x
So at the bottom we have
gained m g x additional
so we have a total available of KE + m g x
That amount goes into compressing the spring and doing further work against friction
(1/2) k x^2 + 18158 x
so in the end
m*g*h18158*h + m g x = (1/2) k x^2 + 18158 x
or
m g (h+x)  18158 (h+x) = (1/2) k x^2
find the two solutions to the quadratic. Probably only one will make sense.