# Finding velocity when a mass suspended from a spring passes through equilibrium

1. Nov 11, 2009

### MRM

A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?

PEelastic=1/2kx^2
KE=1/2mv^2

I honestly have no idea what I'm doing. I used the equation above and got: 1/2(100)(10)^2 which equals 5000, but I'm honestly not sure what to do after I get the PEelastic (or if I'm even supposed to solve for that). I think if I could figure out KE, I could solve for velocity, but I'm not sure how to solve for kinetic energy...any help in the right direction would be great...

2. Oct 17, 2010

### Funkmaster W

You are going in the right direction by finding PEelastic. By doing this you also found KE since KE=PEelastic so from there you just plug it into the equation.