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Derivation of Kinetic energy formula and energy principle

  1. Jun 9, 2014 #1
    Hi all,

    Here is the derivation of kinetic energy from Work:

    W = ∫Fds
    From the second law of motion F = dp/dt, which is equal to mdv/dt, so:

    W = m∫dvdx/dt which = m∫dv x v because dx/dt = v
    Therefore W = 1/2mv2, when integrated.
    However from simple algebra derivation, W = Δ1/2mv2.

    Did I skip something in the derivation? I know that the actual integration would be 1/2mv2
    + C, where C is a constant. However I omitted the C. Is it because of that that I did not end up with the proper equation? If so, why would C = -1/2vi2. (vi because Δv = v - vi and the rest comes from kinetic energy equation.) Or is it because I wrongly integrated?

    Any answers would be appreciated. Thanks
     
  2. jcsd
  3. Jun 9, 2014 #2

    jtbell

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    Staff: Mentor

    Do it as a definite integral between vi and vf, which corresponds to doing the original integral of Fds as a definite integral between an initial position si and final position sf.
     
  4. Jun 9, 2014 #3

    stevendaryl

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    Staff Emeritus
    Science Advisor

    The formula for work is not an indefinite integral, it's a definite integral. That means that there is no arbitrary constant [itex]C[/itex]. The definite integral gives:

    [itex]W = 1/2 m (v^2 - v_i^2)[/itex]

    where [itex]v_i[/itex] is the initial velocity. The work is only equal to the kinetic energy in the case of a force accelerating a particle from rest ([itex]v_i = 0)[/itex].
     
  5. Jun 9, 2014 #4
    You integrated correctly and C = -1/2vi2. To see that just keep in mind that the final time tf can be any time at all and in particular we can chose tf = ti in which case no time has elapsed no displacement has occurred and no work was done, so you have

    0 = W = 1/2 m vf2 + C = 1/2 m vi2 + C,

    hence C = -1/2 m vi2
     
  6. Jun 9, 2014 #5
    Thanks for the answers. However, there is something else I wish to clear up now. Why would work be a definite integral?
    Thanks
     
  7. Jun 9, 2014 #6

    UltrafastPED

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    Science Advisor
    Gold Member

    Because it is the work done in moving from point A to point B.
     
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