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Problem of Springs

  1. Sep 13, 2010 #1
    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?
    Thanks in advance
     
  2. jcsd
  3. Sep 13, 2010 #2
    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??
     
  4. Sep 13, 2010 #3

    Doc Al

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    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!)
     
  5. Sep 13, 2010 #4
    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?
     
  6. Sep 13, 2010 #5

    Doc Al

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    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. :wink:
     
  7. Sep 13, 2010 #6
    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?
     
  8. Sep 13, 2010 #7
    Re: Springs

    And by the way, thank you for bearing with me
     
  9. Sep 13, 2010 #8

    Doc Al

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    Staff: Mentor

    Re: Springs

    9 inches = how many feet?
     
  10. Sep 13, 2010 #9
    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?
     
  11. Sep 13, 2010 #10
    Re: Springs

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