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Hookes Law F=-KX

  1. May 6, 2012 #1
    I am a GCSE Physics Student I was hoping somebody would be able to help me when I have been studying Hookes law, we have only looked at a basic equation extension = extented length - original length. We have also look at how the increase in mass is directly proportional to the extension until plastic deformation occurs. I was then looking through my revsion notes and the equation F=-KX came up. I looked this up and found that
    athematically, Hooke's law states that

    F=-kx,

    where

    x is the displacement of the spring's end from its equilibrium position (a distance, in SI units: meters);
    F is the restoring force exerted by the spring on that end (in SI units: N or kg·m/s2); and
    k is a constant called the rate or spring constant (in SI units: N/m or kg/s2).

    this makes no sense to me
    Could somebody please explain this to me
    thanks
     
    Last edited: May 6, 2012
  2. jcsd
  3. May 6, 2012 #2
    Hey, I'm in gr 12 so I don't know how much help I am, but I can try. Think about a spring for one second:

    A spring can be stretched or compressed. When a spring is stretched, or compressed, it has POTENTIAL ELASIC ENERGY. When the spring is at eqilbrium postion, that means it has not been stretched or compressed therefore it has no potential elastic energy. K is just a constant ( a given number) that the spring has. Finally, x is the distance the springs been stretched or compressed FROM EQILBRIUM POSTION.

    If you have anymore questions just ask, I hope this helped.
     
  4. May 6, 2012 #3

    Doc Al

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    Staff: Mentor

    Why doesn't that make sense?
     
  5. May 6, 2012 #4
    Oh yeah, Ee (which is potential elastic energry) = 1/2*k*x^2 this is formula you may use.
     
  6. May 6, 2012 #5
    Anythin we could add Doc?
     
  7. May 6, 2012 #6
    thanks hellohi that was really helpful :smile:
     
  8. May 6, 2012 #7
    Np man, I'm glad it helped:)!!!
     
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