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Application of integral calculus: Work (spring)

  1. Sep 7, 2008 #1
    hi there! i'm having some troubles regarding this question:

    "if a force of 5 pounds produces a stretch of 1/10 of the natural length, L , of the spring, how much work is done in stretching the spring to double its natural length?"

    i tried answering this but i'm not sure if my answer is correct.
    will the answer be in terms of L? thanks a lot!
     
  2. jcsd
  3. Sep 7, 2008 #2

    Doc Al

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    Staff: Mentor

    Yes.
     
  4. Sep 7, 2008 #3
    i answered W= 75 L is it correct?
     
  5. Sep 7, 2008 #4

    Doc Al

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    No. How did you get that answer?
     
  6. Sep 7, 2008 #5
    my solution is this:

    F(x)=k(x)
    F(x)= 5 pounds
    x=1/10 of L = L/10

    5 = k(L/10)
    k= 50/L

    then F(x)= 50/L x

    my lower limit is L and my upper limit is 2L.

    then, i did the usual integration in the equation W= integral of F(x) dx (with the upper limit and the lower limit)
     
  7. Sep 7, 2008 #6

    Doc Al

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    So far, so good.
    Careful. In F = kx, x is the amount of stretch, not the total length. (Fix the limits of your integration.)
     
  8. Sep 7, 2008 #7
    yes. x is the amount of stretch. if the length is L, and the amount of stretch is 1/10, wouldn't be x= 1/10 of L?

    did i miss something there or is it just my limits of integration that made my answer wrong?
     
  9. Sep 7, 2008 #8

    Doc Al

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    That part, where you calculated the spring constant, was just fine.
    It's your limits of integration. When you stretch the spring from L (unstretched) to 2L, how does x vary?
     
  10. Sep 7, 2008 #9
    I finally got your point. Thanks a lot! ^^,
     
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