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Work done by a force on a body

  1. Feb 11, 2006 #1
    Under the action of a force, a 2kg body moves such that its position X as a function of time t is given by X = t^3/3 where X is in metres and t in seconds. What is the work done by the force in the first 2 seconds? Here the symbol ^ denotes power. Please help me in solving this problem.
  2. jcsd
  3. Feb 11, 2006 #2


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    What have you attempted with this problem? You need to show your work, because it is easier to help that way.
  4. Feb 12, 2006 #3

    Meir Achuz

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    1. Find F=m(d^2x/dt^2).
    2. Write F in terms of x.
    3. Integrate Fdx.
  5. Feb 12, 2006 #4
    I solved it the following way :-
    Now X = t^3/3
    Differentiating the above equation w.r.t time we get,
    dX/dt = v = t^2 ( v denotes velocity)
    Differentiating the above equation w.r.t time again we get,
    d^2X/dt^2 = a = 2t (a denotes acceleration)
    a(t = 0) = 0
    a(t = 2) = 4 m/s^2
    Average a = (0+4)/2 = 2 m/s^2
    X( At t = 0) = 0
    X(at t = 2) = 8/3 = 2.67 m
    Displacement in the first 2 seconds = 2.67 m
    Work done = Force x Displacement
    = mass x acceleration x Displacement
    = 2 x 2 x 2.67
    = 10.68 Joules

    But the answer given in my book is 16 Joules. I solved a similar kind of problem but in that problem X = 2t^2. So i got the acceleration as a constant on differentiation.Could you please explain it to me if it is wrong?
  6. Feb 12, 2006 #5
    I did as you said. I got the answer as 10.98 Joules. Is it right?
  7. Feb 12, 2006 #6
    Your work seems fine until you averaged it. I would have said,

    [tex] w= \int F * dx = \int (ma)dx = \int^2_0 (2)(2t)(t^2 dt) = 16 Joules [/tex]

    Because, [tex] dx = t^2 dt [/tex]
  8. Feb 12, 2006 #7
    This is probably not the best way, but this is how I did it.
    [tex] x = \frac{1}{3}t^3 [/tex]
    [tex] W = \int F.dx [/tex]
    Where, [tex]F = ma = m\frac{d^2x}{dt^2} [/tex], so, [tex]\frac{d^2x}{dt^2} = 2t [/tex]
    [tex] W = \int 2mt^3.dt [tex]
    Then solve that.
    edit: dam, beaten to it.
    Last edited by a moderator: Feb 12, 2006
  9. Feb 12, 2006 #8
    Didn't I just post that? No, you were about 33 mins behind me.
    Last edited: Feb 12, 2006
  10. Feb 12, 2006 #9
    yer, just u beat me to it, looking at the times,. you beat me by quite a lot.
  11. Feb 14, 2006 #10

    Thank you very much Sir.
  12. Feb 14, 2006 #11
    The easiest method to do this is work done = change in Kinetic energy
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