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Where does 1/2 come from in 1/2kx^2?

  1. Aug 15, 2004 #1
    Where does 1/2 come from in 1/2kx^2??

    Sorry is this sounds like a silly question, but i was just wondering where the 1/2 in PE = 1/2kx^2 comes from. If F=kx, then wouldnt it be kx*x = kx^2 ?
     
  2. jcsd
  3. Aug 15, 2004 #2

    arildno

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    In this case the (force*distance) formula must be represented as an integral, for example:

    [tex]PE=\int_{0}^{x}kx'dx'=\frac{1}{2}kx^{2}[/tex]
     
  4. Aug 15, 2004 #3
    errr....I havnt taken calculas yet, so i have no idea what that means, lol.
     
  5. Aug 15, 2004 #4

    Doc Al

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    force is not constant

    Even without calculus you should be able to see where the half comes in. In the equation Work = Force * Distance, the force is changing so you can't just put F = kx, since that's only true when the spring is fully stretched. Instead, use the average force: The force varies uniformly from 0 to kx, so the average force is kx/2. So Work = (kx/2)*(x) = 1/2 k x^2. Make sense?
     
  6. Aug 15, 2004 #5
    basically, what arildno did was integrate. You dont need much calculus to understand. If you have
    [tex] F = kx [/tex]

    now you find the anit - derivative, do you know what that means?

    [tex] P_e = \frac {1}{2}kx^2 [/tex]

    if you still dont understand, pm me.
     
  7. Aug 15, 2004 #6

    chroot

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    It doesn't take much calculus to perform an integration, Nenand?

    - Warren
     
  8. Aug 15, 2004 #7

    chill, chill, chroot, hmmm,..., Nenand, chroot ???

    he was only saying integrating a constant multiplied by x is one of them elementary integranda...

    marlon
     
  9. Aug 15, 2004 #8

    robphy

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    Plot force F(x) vs displacement x.
    The "area under the curve" is the "work done by that force".

    Since F(x) is linear, you are finding the area of a triangle with base (x) and height (-kx).
    So, (Work done by F)=(1/2)(x)(-kx).

    Since that force is conservative, the potential energy is minus the work done by that force.
    So, U=-(-(1/2)kx^2)=(1/2)kx^2.
     
  10. Aug 15, 2004 #9
    integration is calculus. And I wasnt shure if he had any experience with polynomial functions or calculus.
     
  11. Aug 15, 2004 #10

    chroot

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    I'm aware.
    He said he didn't.

    - Warren
     
  12. Aug 15, 2004 #11
    Oh wow, that makes sense, at least the explanations without the use of "intergration" or "derivative" and that calculus stuff. I have no idea what those words mean, although ive heard them many times. I'm going to take BC calc when school starts. Anyways, thanks a lot for the responses. :)
     
  13. Aug 16, 2004 #12

    russ_watters

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    To get you started, an integral is the area under a curve and the derivative is the slope of the curve. Its more complicated than that, of course...
     
  14. Aug 16, 2004 #13

    NateTG

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    You should already know that the force that a spring exerts is proportional to the displacement (i.e. [tex]F=kx[/tex]), and be familiar with the work-energy equation for a constant force ([tex]W=F d[/tex]).

    Now, we're not entirely sure about what the work that is done by the spring is, but we can approximate it. For example, we could cut the path of the spring into lots of little pieces, and approximate the work done on each of the peices of the path by picking the smallest force that the spring exerts on that segment to calculate the work for that segment. It turns out that it's possible to show that the total of these sengments tends toward [tex]\frac{1}{2}kx^2[/tex] as the segments get smaller and smaller.
     
  15. Aug 16, 2004 #14
    Thanks for the tip. :smile:

    Yes, I already know that...
     
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