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Deriving x(t) from work-energy theorem

  1. Feb 17, 2013 #1
    In K&K's text on mechanics, after they present the derivation of the work energy theorem:

    [tex]\frac{1}{2}mv^2-\frac{1}{2}mv_0^2=\int_x_0^x F(x) dx[/tex]

    It is mentioned that since [tex]v=\frac{dx}{dt}[/tex], we could solve for [tex]\frac{dx}{dt}[/tex] and integrate again to find x(t)

    I tried that with [tex]v_0=0[/tex] just to make things a little easier and ended up with something like:

    [tex]x=\int \sqrt{2m(F(x)-F(x_0))}dt[/tex]

    which looks horrible and wrong

    Is this heading down the wrong path? How would you do it when [tex]v_0[/tex] is not zero?
     
  2. jcsd
  3. Feb 17, 2013 #2
    Your derivation is not right; it must be an integral of force rather than force itself under the radical.

    As an example, consider the case when F = -kx (mass attached to a spring).
     
  4. Feb 17, 2013 #3

    haruspex

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    As voko says, plus, you have m where it should be 1/m.
     
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