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Minimum work and work-energy theorem

  1. Nov 26, 2013 #1
    Hi guys,

    I've got a doubt concerning to the minimum mechanical work and the work-energy theorem. Consider the following Tippens' problem (8.4):

    A 5-kg hammer is lifted to a height of 3 m. What is the minimum required work?

    The answer looks very simple and inocent, W = weight times distance. However, deeping inside we can calculate the negative work done by the gravity, it would be: W = -weight times distance. So, the total work is equal to zero and if we assume that the hammer starts from rest then its final velocity have to be equal to zero too (work-energy theorem).

    It's very confusing for me how this could be possible, how a displacement could be obtained if at any time the velocity is equal to zero. Also is confusing how being the total force equal to zero (weight - weight = 0) the motion is possible.

    Thanks in advance for your help.
    Last edited: Nov 26, 2013
  2. jcsd
  3. Nov 26, 2013 #2


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    Gold Member

    I believe the question is supposed to be:
    What is the minimum work you would need to lift a hammer 3m?
    ...and not the net value of the work due to gravity plus your own lifting.

    The minimum amount of work you need to lift the hammer is equal to the minimum amount of force you need to lift the hammer times the change of height.

    If the force you apply is larger than the weight of the hammer, the hammer will accelerate upwards faster and faster, so that when it reaches 3m, it is still moving, and has extra kinetic energy on top of the needed gravitational potential energy.

    The minimum amount of work will be when the amount of force approaches the weight of the hammer, so that when the hammer reaches a height of 3 meters, there will be no extra kinetic energy.
  4. Nov 27, 2013 #3
    Thanks for answering jfizzix. That's what I wanted to discuss.

    Best Regards!
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