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Work done by a Variable Force

  1. Nov 4, 2006 #1
    A single force acts on a 7.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

    I want to use the equation [tex]W = \int F(x)dx = \int madx[/tex]
    To find acceleration, I took the second derivative of the position function:
    [tex]x(t) = 3t-4t^2+t^3[/tex]
    [tex]v=x'(t) = 3-8t+3t^2[/tex]
    [tex]a=x''(t) = -8+6t[/tex]

    When I go to plug values into my work integral, I get [tex]W=\int m(6t-8)dx[/tex]
    I would then take mass out of the integral, but having t and dx in the integral is evil. Do I say that dx=x'(t) and so substitute my velocity function (with dt tacked onto the end) for dx?
    Last edited: Nov 4, 2006
  2. jcsd
  3. Nov 4, 2006 #2


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    "Yes". The change of variable theorem states that,

    [tex]\int_{x(t=0)}^{x(t=3)}F(x)dx = \int_0^3F(x(t))\dot{x}(t)dt[/tex]

    But as you cleverly noted, F(x(t))=ma(t).
  4. Nov 4, 2006 #3


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    That looks like it will take away the evil and give you the correct solution!l
  5. Nov 4, 2006 #4
    Thank you very kindly!
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