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Hookes law

  1. Nov 9, 2006 #1
    Hello, I'm having some problems with the spring constant "k" together with Hookes law. U=1\2kx^2

    Could someone please explain how you get that integral?
    If you insert it in a diagram and calculate the area as a triangle you would get 1/2kx. Where does the ^2 come from?
  2. jcsd
  3. Nov 9, 2006 #2


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    No, you don't!
    Draw the triangle again.
    How long is its base, and how long is its height?
  4. Nov 9, 2006 #3
    Alright, my bad :rolleyes: I missed that it's (kx * x)/2
    Now, another question.
    The function for F = kx.
    The primitiv function for kx = (kx^2)/2 right? Because then you could calculate it as an integral?
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