# Elevator and spring problem

1. Feb 23, 2005

### whereisccguys

Problem: An elevator cable breaks when a 754 kg elevator is 20.8 m above the top of a huge spring (k = 6.93×104 N/m) at the bottom of the shaft. Calculate the amount the spring compresses (note that here work is done by both the spring and gravity).

i calculated the speed of the elevator just before it hits the spring which is
Vinitial = 20.19 m/s

the equation i used was (1/2)mV^2-(1/2)mVinitial^2= -(1/2)kx^2 + mgx
x is the distance the spring was compressed
and at distance x the current velocity = 0 because it is not moving at all anymore
therefore the equation becomes -(1/2)mVinitial^2= -(1/2)kx^2 + mgx
i plugged in all the numbers and solved by the quadratic equation.. but still came up with the wrong answer...
can someone help?

2. Feb 23, 2005

### vincentchan

use conservation of energy... before the cable breaks, the elevator is steady, with a potential energy mgh... after the elevator hit the spring and compressed it, the velocity is also zero, potential energy become 1/2kx^2

mgh=1/2kx^2
x=sqrt(2mgh/k)