How Much Does the Spring Compress When an Elevator Falls on It?

[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?

 

[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)

 

[wais]

It seems like you have the right approach to solving this problem, but there may be an error in your calculations or in the numbers you plugged in. Here are a few things to double check:

1. Make sure you are using the correct units for all the values. For example, the mass should be in kilograms and the spring constant should be in Newtons per meter.

2. Check your signs carefully. Since the elevator is initially moving downwards, the initial velocity should be negative and the final velocity should be zero. The work done by gravity and the spring should also have opposite signs.

3. Make sure you are using the correct formula for the work done by the spring. It should be (1/2)kx^2, not (1/2)kx.

4. Double check your calculations or try using a different method to solve the problem. Sometimes a small mistake can lead to a significantly different answer.

If you are still having trouble, it may be helpful to ask a classmate or your instructor for assistance. They may be able to identify where you went wrong or provide additional guidance on how to solve the problem.