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Kinetic Energy Time Derivative

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data

    So the first part asks to prove the time derivative of kinetic energy is dT/dt=F dot product v which I did not problem. but then the second part of the problem asks to prove that if the mass is changing with time then the time derivative of d(mT)/dt=F dot product m and i'm sure this can be correct shown in my attempt at a solution below.

    2. Relevant equations


    3. The attempt at a solution

    If we are to assume that both kinetic energy and m and no constant with time then d(mT)/dt can be given using the product rule as = m dot T + m T dot.

    m T dot will be as T dot is F dot product v = m F.v = F.p

    Which is supposed to be the answer, but the first term will be non zero, given m dot is non zero, the first term goes to:

    m dot 0.5 m r dot dot product r dot. This is only zero is the m dot is zero or mass or r dot is zero, but this is not a fair assumption. Hence how can his answer be correct. It seems as if he's just multipled dT/dt=F.v by m on both sides and then said well m is independent of time and hence it can go in the derivative. But his question is with m changing. Am i wrong?

    Thanks

    Evan
     
  2. jcsd
  3. Mar 16, 2015 #2

    Svein

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    [itex]\frac{d}{dt}(\frac{1}{2}mv^{2})=\frac{1}{2}\frac{dm}{dt}\cdot v^{2}+\frac{1}{2}m\frac{dv^{2}}{dv}\frac{dv}{dt} [/itex]
     
  4. Mar 16, 2015 #3

    vela

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    The forum software provides you with typesetting tools. You should use them as your post as written is not easy to follow. Proofreading your post and fixing typos wouldn't hurt either.

    How can you dot F and m? The mass m is a scalar.
    You derived the expression for ##\dot{T}## under the assumption the mass was constant, so you can't use it here.

    Another approach you could take is to start by expressing the kinetic energy in terms of momentum instead of velocity.
     
  5. Mar 16, 2015 #4
    True let me have another go :)
     
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