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Understanding elastic potential energy

  1. Mar 1, 2009 #1
    where does the formula 1/2 Fx come from - i need to find out where the half comes from for some coursework of mine :S cheers,:cool:
     
  2. jcsd
  3. Mar 1, 2009 #2
    What have you tried so far?
     
  4. Mar 1, 2009 #3
    well its for an experiment i did - How does the speed of a catapulted trolley vary with distance pulled back? - as part of the task it says - word done or enery stored is 1/2Fx. THink why the 1/2 is in there - average? - yh tried a few sites didnt get v far...
     
  5. Mar 1, 2009 #4
    Work done = average force times distance moved in direction of force.
     
  6. Mar 1, 2009 #5
    It comes from the Taylor expansion of the potential energy function. We don't know what U(x) is, so we take the Taylor expansion of it around 0. You can add a constant to any potential energy function and it won't change the physics, so we can let the first term equal zero. If we define the equilibium position to be zero, then U'(0) must be 0 since it is in equilibrium. If it weren't zero, then it wouldn't be the equilibrium position. That leaves us with the third term, 1/2 kx^2, where k = U''(0). Terms beyond the third are ignored, since we assume that three terms of the Taylor expansion will be a good approximation for small displacements (which is was 1/2 kx^2 is valid for). That's where the 1/2 comes from.
     
  7. Mar 2, 2009 #6
    F = -kx

    Work = F*x
    Work for variable force = integral (F dx)
    W = integral (-kx dx) = -k integral(x dx) = -1/2k x^2

    W = -1/2k x^2
    U = -W = 1/2 k x^2
     
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