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Kinetic energy's dependence on velocity.

  1. Apr 5, 2009 #1
    This is not homework but rather a personal question.

    Why is kinetic energy dependent on the square of the velocity. Is there a physical answer or is it simply mathematically derived. If there are some real life examples that can explain this then it would be greatly appreciated.

    Similarly, why does it take more energy to accelerate an object at high velocities than low. For example, it takes more energy to raise an object's velocity from 100 - 200 m/s than it does to raise the velocity of the same object from 0 - 100 m/s.
     
  2. jcsd
  3. Apr 5, 2009 #2

    Astronuc

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    Staff: Mentor

    This is the way I remember explained - using the product of force over distance, which becomes
    [tex] W = m \int{v dv} [/tex], with an equivalence between work and energy.
    http://scienceworld.wolfram.com/physics/KineticEnergy.html

    So since kinetic energy is proportional to v2, and the change would be related to v dv, which changes linearly with v.

    Also, if one doubles v, e.g. vf = 2 vi, then Ef is proportional to vf2 = (2 vi)2 = 4 vi2 , or Ef = 4 Ei.
     
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