(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Change in Mechanical energy + change in thermal energy = work

Change in Potential energy = - Work

W = F * d

Wspring = 1/2k(x^2)

3. 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

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# Homework Help: Conservation of energy of an elevator

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