Problem of Springs

1. Sep 13, 2010

rubecuber

1. The problem statement, all variables and given/known data
"If the work required to stretch a spring 3 ft beyond its natural length is 8 ft-lb, how much work W is needed to stretch it 9 in. beyond its natural length?"

2. Relevant equations
F=kx.

3. The attempt at a solution
F=kx so 8 = 3k. k= 8/3 then I use an integral of from 0 to .25 (9 inches is .25 of three feet) of (8/3)x. I do it manually and get 1/12. I'm wrong so I do it on a calculator to check for error and I get the same answer. I must be approaching the problem incorrectly, but how?

2. Sep 13, 2010

Mandeep Deka

Re: Springs

"F=kx so 8 = 3k"
How did you do this??

You are being given the amount of work done to extend the spring by 3 ft. How did you use the force relation??

3. Sep 13, 2010

Staff: Mentor

Re: Springs

F = kx is Hooke's law describing the force to stretch a spring. What formula describes the work done when you stretch a spring? (Don't mix up force and work!)

4. Sep 13, 2010

rubecuber

Re: Springs

Okay so I set 8 equal to the integral from 0 to 3 of the spring constant dx and then solve. How am I supposed to find the spring constant? Is there a better way?

5. Sep 13, 2010

Staff: Mentor

Re: Springs

You would integrate the force*dx, thus kx dx (not just k dx).
The spring constant is your only unknown.
Sure. Just look up the formula for the work done to stretch a spring. (Hint: It's the same formula for the energy stored in a stretched spring.) That way you won't have to do any integrating.

6. Sep 13, 2010

rubecuber

Re: Springs

Great, so W= .5kx^2. Then 16=kx^2. Since the initial displacement is 3, 16/9 =k. Now knowing the spring constant to be 16/9 I can find the work to move it .25 of three feet as [(.5)(16/9)(.25^2) which gives me 1/18. But I'm still wrong? Did I make another stupid mistake?

7. Sep 13, 2010

rubecuber

Re: Springs

And by the way, thank you for bearing with me

8. Sep 13, 2010

Staff: Mentor

Re: Springs

9 inches = how many feet?

9. Sep 13, 2010

rubecuber

Re: Springs

9 inches is 3/4 of a foot but I'm solving for three feet so that would makes nine inches a quarter of three feet. Am I really missing something this obvious?

10. Sep 13, 2010

rubecuber

Re: Springs

Wow I'm stupid. I just figured it out. Thanks for putting up with this.