Spring and work

  • Thread starter superdave
  • Start date
  • #1
150
3
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
But that's not the answer.

Help?
 

Answers and Replies

  • #2
Doc Al
Mentor
45,084
1,394
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:
  • #3
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
Why are you using work, it would be much easier to use one of Hooke's laws;
[tex]F = - kx[/tex]

Can you arrive at the correct answer using this?

[Edit] It seems Doc Al got there before me, damn he's so quick! :mad: [Edit]
 
  • #4
150
3
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
 

Related Threads on Spring and work

  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
15K
Top