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Understanding derivation of kinetic energy from impulse?

  1. May 24, 2009 #1
    I have a question that annoys my basic understanding of kinetic energy.

    I know if I have a force-time plot then the area under the curve is equivalent to the impulse imparted on an object (in units Newton-Seconds). I know that this is also equivalent to the change in momentum of the object i.e. Ft = delta mV

    I know that i can get these values from the plot by simply intgrating the Force function wrt to time. Now I want to determine the kinetic energy involved in this event. I know that k.e. = 1/2mV^2 so I know that is simply the integral of the impulse w.r.t. to velocity

    Now that is the part conceptually I don't really grasp. What does it really mean to integrate with respect to velocity? This means I am suddenly on a velocity :uhh: domain, not time, and I don't really understand this?

  2. jcsd
  3. May 24, 2009 #2

    Doc Al

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    Staff: Mentor

    While impulse is ∫Fdt, work is ∫Fdx. Since force can be viewed as the rate of change of momentum, you have F = dp/dt = m dv/dt. Thus:
    W = ∫Fdx = ∫m (dv/dt) * dx = m∫dv*(dx/dt) = m∫v dv = ½mv² (which is kinetic energy).

    Make sense?
  4. May 24, 2009 #3
    After some head scratching yes it does, finaly.

    I think my mistake has always been trying to integrate wrt time. I really had to step back and ask myself what i was after, work that is, which is not a time integral.

    Manipulating dv/dt to dx/dt and then subbing in V now makes it algebraicly understandable.

    Thanks alot.

  5. May 24, 2009 #4
    Similar to questions which annoys my basic understanding of electrostatics?

    Why not just calculate the final velocity of the body and then calculate the K.E.

    Take as a caveat that the momentum of a body might not be directly proportional to K.E possesed by it.
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