# Spring and work

1. Mar 16, 2006

### superdave

Okay,

I am messing something up here. I think it's my logic.

Given k=2.5 * 10^3 N/m

Find how much mass would have to be suspended from the vertical spring to move it 6.0 cm

I get W=1/2kx^2=.5(2500)*(.06m)^2=4.5J
I think that's right.

Then I do
W=F*d=m*a*d or m=W/a*d=4.5/(9.8m/s^2 / .06 m)=7.65 kg

Help?

2. Mar 16, 2006

### Staff: Mentor

Look up Hooke's law and the meaning of k.

Note: I assume the question is asking for the mass that would--if gently added to the spring, not dropped from a height--stretch the spring by the given amount.

Last edited: Mar 16, 2006
3. Mar 16, 2006

### Hootenanny

Staff Emeritus
Why are you using work, it would be much easier to use one of Hooke's laws;
$$F = - kx$$

Can you arrive at the correct answer using this?

 It seems Doc Al got there before me, damn he's so quick! 

4. Mar 16, 2006

### superdave

Oh, so

F=kx
m*9.8=2500 * .06=150/9.8= 15 kg

I got it. Man, I sure went abot that the completely wrong way.

thanks