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Why is kinetic energy 1/2mv^2 instead of mv^2?

  1. Mar 26, 2009 #1
    The definition of joules, th unit of energy is the 1 J = 1 kg * m2/v2.

    And that 1 joules is the amount of energy needed for the work done by one newton traveling one meter.

    From all this, I got the impression that to get energy, you would have to multiply force times distance. Is work, then just a measurement of energy.

    Assuming my first assumption that energy = F x D, lets sat a 5kg object is accelerating towards earth from 20 feet above at 9.8 m/s2 . To get the force, you would have to do 5 kg x 9m/s2 Then to get energy, that force times distance. So if it traveled 5 meters. The kinetic energy should be F(5kg x 9.8 m/s2) x D(5 meters.)

    Which would end up being 5kg x 49m2/s2.

    Sooo, can someone explain why ke is 1/2 mv2 instead of just mv2.
     
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  3. Mar 26, 2009 #2

    robphy

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    Work Energy Theorem
     
  4. Mar 26, 2009 #3

    Pengwuino

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    I hope you'll notice that nowhere did you determine its actual speed after falling the 5 meters. If you use your kinematic equations and solve for the time it took to reach the 5 meters and subsequently found the velocity at that 5meter point, you'd see that the energy is infact, mv^2/2
     
  5. Mar 26, 2009 #4
    Assuming my first assumption that energy = F x D

    Right, work is the energy added to or taken away from a system when it encounters a force over some distance. The definition of work is actually the integral,
    W=[tex]\int[/tex]F[tex]\bullet[/tex]dr
    Where dr is the change in position, the D in your equation.
    The equation for the kinetic energy of an object, K.E.=1/2mv^2 is derived by making a series of substitutions in the above equation and integrating.
     
    Last edited: Mar 26, 2009
  6. Mar 26, 2009 #5

    Fredrik

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    [tex]W=\int_{x_1}^{x_2} F dx=\int_{t_1}^{t_2} m\ddot x \dot x dt=\int_{t_1}^{t_2}\frac{d}{dt}(\frac 1 2 m\dot x^2)dt=\frac 1 2 mv_2^2-\frac 1 2 mv_1^2[/tex]
     
  7. Mar 26, 2009 #6

    arildno

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    Well, there would be no conceptual problems whatsoever to define a quantity mv^2, and use this instead of "kinetic energy".

    This has already been done; mv^2 is called "vis viva".

    But, the work-"vis viva" theorem is less aesthetic than the equivalent work-energy theorem, due to the explicit inclusion of the factor of 1/2 in the formula.
     
  8. Mar 26, 2009 #7
    Thx for the replies, everyone. l'll check them all out.
     
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