Elastic Potential Energy problem

[kadavis2010]

Homework Statement

A spring has a force constant of 30000 N/m. How far must it be stretched for it's potential energy to be 47 J? Answer in units of m.

Homework Equations

Us= 1/2k(∆x)^2

The Attempt at a Solution

47= 1/2 (30000)(x)^2
47=15000(x)^2
-14953= x^2
√14953= 122.28

 

[Doc Al]

The Attempt at a Solution

47= 1/2 (30000)(x)^2
47=15000(x)^2

So far, so good.

-14953= x^2

Oops... you subtracted instead of divided. Redo this step.

 

[hage567]

47= 1/2 (30000)(x)^2
47=15000(x)^2
-14953= x^2
√14953= 122.28

You subtracted when you should have divided.

 

[kadavis2010]

So you are saying to do 47/15000?
that would equal .0031... then would i do the square root of it or leave it at that?

ohhh nvm i did that on one of my attempts but i forgot to put .0 in front of .0557=] Thanks a lot. I apperciate it!