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Elastic potential

  1. Nov 15, 2013 #1

    please just remind me

    if I have elastic potential U=( K(x-X0)^2 )/2 for the flexible spring moved on X distance from the initial X0 position So I have I have elastic force F=K(x0+vt-x) wich is the derivative of the U. Its not clear for me why vt term which should be equal to x in this equation is present ? From the general assumption (x-X0)^2 should be x^2-2*X*X0+X0^2 where is the vt term here?

  2. jcsd
  3. Nov 15, 2013 #2
    what's vt?

    looks like you differentiated incorrectly.
  4. Nov 15, 2013 #3
    x = V (velosity) * t (time)
    both of that formulas are from textbooks.

    Reasonable U= (Kx^2)/2 so force = kx. wehat fould be force if U=( K(x-X0)^2 )/2 in case when we moved body from the initial position (Xo) on X. The delta X should give us velosity should it?
  5. Nov 15, 2013 #4

    Philip Wood

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    x = vt is for a body moving at constant velocity for a time t. I wouldn't have thought this was the case here. So we have simply

    [itex]U = \frac{1}{2}k(x - x_0)^2[/itex]

    [itex]F = -\frac{dU}{dx} = -k(x - x_0)[/itex] (having used chain rule)

    The minus sign in front of the k indicates that the force is in the -x direction.
  6. Nov 15, 2013 #5

    Some calculus:
    In the F=K(x0+vt-x) dealing with the constant velosity vt should be equal to 0, shouldnt it?

    In the chain rule we have f(x)= x-x0 and g(y)=y^2 so solving it we obtain that (x-xo)' should be 1. I dont understand why its not 0.

  7. Nov 15, 2013 #6

    Philip Wood

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    Gold Member

    Don't understand why you're bringing in vt, especially since you seem to have both x and vt in this equation. Stick with just x, unless you're sure that x really does vary with time as x = vt.

    [itex] \frac{d}{dx}\frac{1}{2}k(x - x_0)^2 = \frac{1}{2}\frac{d}{du}u^2 \frac {du}{dx}[/itex]

    in which [itex]u = (x - x_0)[/itex] so [itex] \frac{du}{dx} = 1[/itex]

    and [itex]\frac{d}{du}u^2 = 2u[/itex]

    So [itex] \frac{d}{dx}\frac{1}{2}k(x - x_0)^2 =k(x - x_0)[/itex].
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