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Conservation of Energy

  1. Nov 7, 2007 #1
    1. A 70kg pole vaulter runs with a top speed of 9.5 m/s. During the jump the pole is highly curved then straigtens out. The vaulter is able to clear a bar height of 5.4m
    a) Using conservation of energy, determine the maximum height the vaulter should be able to clear. The vaulters center of gravity point is 90cm above his feet.
    b) Discuss the energy conversions that take place.

    2. K=.5mv^2, Ug=mgh

    3. (a) .5mv^2=mgh. I got h is equal to 4.9m but I added the center of gravity point to make it 5.8 m. Is this right? why do we add the COG point?
    (b) Kinetic energy transforms to energy into the bending the pole, the pole then converts the energy into gravitational potential energy by raising you a height H.

  2. jcsd
  3. Nov 7, 2007 #2

    Because the energy equations allow you to predict how far the running vaulter, given a perfect vaulting technique, has the energy to lift their COG.

    Taking the vaulter as a point (in which case how can their COG be above their feet?!), adding the height of the COG gives you an indication of the height the vaulter should be able to clear.

    Taking the vaulter as having size and an adjustable shape, the vaulter can adjust their shape and thus put their COG below the part of their body that is passing over the bar. This allows the vaulter to clear higher than predicted by converting KE to GPE and adding the height of COG above ground when running. The question does not give any data allowing you to compute this third term of the height the vaulter should be able to clear.
    Last edited: Nov 7, 2007
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