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Energy change and work involved in lifting a ball

  1. Dec 18, 2011 #1
    I was thinking about the work-energy theorem today and how it states that:

    Wnet = ΔEkinetic

    If this is true, then when a ball is moved upward a distance of d, the net work done is equal to zero because there is no change in kinetic energy.

    Because: Work = Force x Displacement

    for every infinitely small distance, dr, that the ball moves in the upward direction, the work done by the upward force is equal to F * dr and the work done by gravitational force is equal to -mg * dr.

    However, according to the previous statement, net work done when the ball moves a distance of dr is zero, which means that F*dr = mg * dr and F = mg

    Because Fnet = F - mg, there is no net force. If there is no net force, why does the ball move up?

    I feel like I am missing a very crucial part of logic, but I can't seem to figure it out. Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 18, 2011 #2

    rcgldr

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    For total energy, you need to include a potential term.

    W = ΔEkinetic - ΔEpotential
     
  4. Dec 18, 2011 #3
    This is wrong. The net work will always be the change in kinetic energy. No potential energy
    terms are involved.
     
  5. Dec 18, 2011 #4

    rcgldr

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    Note I dropped "net" from the work term in the equation I posted. I was relating total work done to total energy. If a 1kg ball is raised 1 meter, than 9.8 newton meters of total work is done. If I then release the ball, it's kinetic energy at the moment it falls back to it's original position will be 9.8 newton meters.
     
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