Conservation of energy of an elevator

[knives out]

Homework Statement

The cable of the 1800 kg elevator in Fig. 8-62 snaps when the elevator is at rest at the first floor, where the bottom is a distance d = 3.7 m above a cushioning spring whose spring constant is k = 0.15 MN/m. A safety device clamps the elevator against guide rails so that a constant frictional force of 4.4 kN opposes the motion of the elevator.

Using conservation of energy, find the approximate total distance that the elevator will move before coming to rest.

**I already found the speed of the elevator just before it hits the spring (7.377 m/s), the distance the spring is compressed (0.90 m), and the distance the elevator bounces back up the shaft (2.764 m) in case any of that is useful.

Homework Equations

Change in Mechanical energy + change in thermal energy = work
Change in Potential energy = - Work
W = F * d
Wspring = 1/2k(x^2)

The Attempt at a Solution

I tried finding the distances the elevator bounced back up after every time it compressed the spring, and added them up. This did not get the correct answer. I also tried to manipulate Emec + Eth = W to apply to this problem, but I am not sure of how to do this. Please help

 

[learningphysics]

Work done by friction = final mechanical energy - initial mechanical energy

ie:

work done by friction = (final kinetic energy + final gravitational potential energy + final spring elastic potential energy) - (initial kinetic energy + initial gravitational potential energy + initial spring elastic potential energy)

 

[knives out]

i just got the answer, thank you very much for your help