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Derivation of Common Formula

  1. Nov 11, 2003 #1
    Can someone please explain the derivation of the common formula for average kinetic energy: K = 0.5mv^2?
     
  2. jcsd
  3. Nov 11, 2003 #2
    Derivation of KE?

    a = v^2 - v0^2 / 2d

    ΣWork = ΣF * d = mad = 0.5m(v^2 - v0^2) = ke
     
  4. Nov 11, 2003 #3

    HallsofIvy

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    First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!
     
  5. Nov 16, 2003 #4
    First, K = 0.5mv^2 is not "average" kinetic energy- it is kinetic energy- in fact it is the definition of kinetic energy- there is no "derivation". At some time, back at the beginning of "physics" someone notice that, in many dynamics problems that quantity was conserved- so they gave it a name!

    Quantities, mv^2, 35mv^2, are also conserved!
     
  6. Nov 17, 2003 #5
    So ummmm ... This is relative kinetic energy not average kinetic energy?

    That is, relative to the kinetic energy of the object when v0 = 0

    Interesting derivation. Thanks.
     
  7. Nov 17, 2003 #6
    energy is the ability to do work.

    work is force times distance.

    kinetic energy is that energy a body has due to its motion.

    so lets see:

    [tex]
    W=\int_A^B \mathbf{F}\cdot d\mathbf{s}
    [/tex]

    but by Newton
    [tex]
    \mathbf{F}=m\mathbf{a}
    [/tex]
    so
    [tex]
    W=\int_A^Bm\mathbf{a}\cdot d\mathbf{s} = m\int_{t_{A}}^{t_{B}}\mathbf{a}\cdot\mathbf{v}dt\\
    = \frac{m}{2}\int_{t_{A}}^{t_{B}}\frac{d(v^2)}{dt}\! dt = \frac{1}{2}mv_B^2 - \frac{1}{2}mv_A^2
    [/tex]

    so you see that the work done is equal to the change in kinetic energy between the two points, and this is a derivation of the formula for that kinetic energy.
     
    Last edited: Nov 17, 2003
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