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Derivation of Kinetic Energy

  1. Jun 23, 2009 #1
    Hi all,
    I was doing an A-level Mechanics paper the other day and one of the quesitons was to show that, starting with Hookes law T=(lambda)(x)/(l), show that the energy stored in an elastic rope is (lambda)(e2)/(2l).
    This was ok, I just said that energy stored would be equal to the sum of the work done stretching the rope a small distance δx and as δ tended to zero it would be dx. Which could be re-written as integral with limits e and 0 dx.
    Which leads to the equation for elastic potential energy.
    After doing this i realised that kinetic energy is in a similar form, i.e power of 2 and has a multiplying factor of 1/2 which leads me to my question, is 1/2mv2 the result of an integral? Has it also got something to do with work done? but with respect to v?
    Any help would be greatly appreciated!
     
  2. jcsd
  3. Jun 23, 2009 #2

    rl.bhat

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    dW = f*dx = m*a*dx = m*dv/dt*dx = m*v*dv ( since dx/dt = v)
    Take the integration between 0 to v.
     
  4. Jun 23, 2009 #3
    ok brilliant thanks! what is dW? respect to work done?
     
  5. Jun 23, 2009 #4

    rl.bhat

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