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Difference between hooke's law and the work done in a spring?

  1. Oct 26, 2006 #1

    dnt

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    hooke's law says the spring force = -kx

    but the work done in stretching a spring = 1/2 kx^2

    isnt work = Fd?

    so using W=Fd and F=-kx (hooke's law) shouldnt the work come out to:

    W = -kx^2? (arent d and x the same? both are distance stretched)

    where does the 1/2 come from in the equation 1/2kx^2?

    can someone help clarify the difference and when to use each equation. thanks.
     
  2. jcsd
  3. Oct 26, 2006 #2

    OlderDan

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    W = Fd is only correct when the force is constant and in the direction of a displacement of distance d. The force is not constant for a spring. Have you taken calculus?
     
  4. Oct 26, 2006 #3

    Hootenanny

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    That is correct, provided the force is constant. However, generally work can be expressed thus;

    [tex]W=\int^{x_{1}}_{x_{0}}\; F \cdot dx[/tex]

    Edit: Too slow again :frown:
     
  5. Oct 26, 2006 #4

    radou

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    ...which, after plugging in F = - k x, answers your question. :smile:
     
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