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How is the formula Kinetic Energy=1/2mv^2 derived?

  1. Jun 27, 2006 #1
    I have returned to this forum after six months.
    How is the formula Kinetic Energy=1/2mv^2 derived?
  2. jcsd
  3. Jun 27, 2006 #2
    We have a particle at position and velocity [itex]\vec{x}_0, \vec{v}_0[/itex] at time [itex]t.[/itex] Under the application of some force, [itex]\vec{F},[/itex] a time [itex]dt[/itex] later, it has new position and velocity [itex]\vec{x}=\vec{x}_0+d\vec{s}, \vec{v}.[/itex] We can make the time [itex]dt[/itex] as small as we like so that the force [itex]F[/itex] is constant (to first order) over that time.

    Assuming the validity of

    [tex]\vec{v}^2 = \vec{v}_0^2 + 2\vec{a}\cdot\left(\vec{x}-\vec{x}_0\right),[/tex]

    under constant accelerations (which we argue that since [itex]\vec{F}[/itex] is constant, then so is [itex]m\vec{a}[/itex]) then we have

    [tex]\vec{v}^2 = \vec{v}_0^2 + 2\vec{a}\cdot\left(\vec{x}-\vec{x}_0\right)[/tex]
    [tex]\frac{1}{2}m\vec{v}^2 = \frac{1}{2}m\vec{v}_0^2 + m\vec{a}\cdot d\vec{s}[/tex]
    [tex]\frac{1}{2}m\vec{v}^2 - \frac{1}{2}m\vec{v}_0^2 = \vec{F}\cdot d\vec{s}[/tex]

    We now define [itex]\frac{1}{2}m\vec{v}^2[/itex] as kinetic energy, and [itex]\vec{F}\cdot d\vec{s}[/itex] as work. We can now say that for finite times, the change in kinetic energy is given by

    [tex]\int\vec{F}\cdot d\vec{s},[/tex]

    and in the absence of external forces, this quantity kinetic energy, remains constant.

    The above is the motivation for the classical form of kinetic energy. The expression for kinetic energy cannot be derived, since it is a definition - a function on co-ordinate space.
  4. Jun 27, 2006 #3
    My form of derivation for KE.

    Assume a mass at rest picks up speed after some time t. Let u and v be initial and final speed of the mass. Using kinematics,

    v^2 = u^2 + 2as

    u = 0 since mass is initially at rest. Substitute F=ma into equation to get

    v^2 = 0 + 2s (F/m)
    mv^2 = 2Fs
    Fs = 0.5 mv^2

    force (F) x distance (s) gives energy, so KE = 0.5 mv^2.

    The end.
  5. Jun 27, 2006 #4
    Even shorter:
    take x=.5at^2
    multiply both sides by a
  6. Jun 27, 2006 #5


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    A much better way of deriving it, is by using the concept of power (time-derivative of work W):
    We have, for a particle of constant mass m:
    Take the dot product of this equation with the particle velocity:
    This can then be rewritten as:
    [tex]\frac{dW}{dt}=\frac{d}{dt}(\frac{m}{2}\vec{v}^{2}), \frac{dW}{dt}\equiv\vec{F}\cdot\vec{v}[/tex]
    Or, even simpler:
  7. Jun 27, 2006 #6


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    Consider a particle moving from an initial point to a final point. Integrate [itex] \sum {\vec F } = m {\vec a } [/itex] over the trajectory. For the left hand side you get

    [tex] \int_{P_i}^{P_f} m {\vec a } \cdot {\vec ds} = \int_{P_i}^{P_f} m {{\vec dv } \over dt} \cdot {\vec ds}= \int_{P_i}^{P_f} m {\vec dv } \cdot {\vec v} [/tex]

    And this gives [itex] {1 \over 2} m v_f^2 - {1 \over 2} m v_i^2 [/itex].
    The integral on the right hand side (involving the sum of the forces) gives the sum of the work done by all the forces.

    This is the origin of the work-energy theorem. It's simply the integral along the trajectory of Newton's second law.

  8. Jul 10, 2006 #7
  9. Jul 10, 2006 #8


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    In my opinion, the approaches by arildno and nrqed are better because they are more general, making no assumption of constant acceleration.

    (Can the thread title be corrected?)
  10. Jul 10, 2006 #9
    Yes I would agree. (I tried to derive it from memory, and remembered the more elementary derivation I had come across earlier in my studies as opposed to the more general one that I had come across later).
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