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Homework Help: Problem related to work Confused between two answers.

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    How much work must you perform to lift a 5 kg bag from the floor to the countertop, 1 meter above the floor?

    2. The attempt at a solution

    A. You could say that any force greater than the weight of the bag would lift it up. Thus, the weight is 50 N and the force should at least 50 N and above.

    since W= F * d * cos(theta)
    W = 50 * 1 * cos0 = 50 Joules

    B. My classmates insist that this is the correct way. The net work is equal to the change in potential energy, and the total work includes: Work done by weight, and work done by the force...

    Work of weight = 50 * 1 * cos180 = -50 Joules
    deltaPotential energy = PE2 -PE1 = 50 - 0 = 50 Joules

    W(total) = W(weight) + W(force) = deltaPE
    -50 + W(force) = 50 joules ==> W(force) = 100 Joules

    Can anyone tell me which answer is the correct one? It seems to me that the second answer would be correct if you pulled the bag up suddenly and let it go and it would then stop mid-air at 1 meter. Can anyone verify this? Thanks in advance :)
  2. jcsd
  3. Apr 25, 2012 #2


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    Gold Member

    Your answer is correct. If you are looking for total or net work, that's the change in KE (which is zero). But if you are looking for work done by you, which the question asks, that's the change in PE plus the change in KE which is 50 +0 = +50 J. If you are looking for the work done by the weight, that is the negative of the potential energy change, or -50 J.
  4. Apr 25, 2012 #3
    Thanks for the reply :D

    However I am confused as to why total work equals to the change in KE...

    I would really appreciate it if you could explain it.
  5. Apr 25, 2012 #4
    This involves a lot of formula manipulation...
    We know that W = Fd and F = ma.
    ∴ W = mad

    Solving for a from v22 = v12 + 2ad gives us
    a = (v22 - v12)/2d

    plugging that back into W = mad gives us
    W = m[(v22 - v12)/2d]d
    = [itex]\frac{1}{2}[/itex]m(v22 - v12)
    = [itex]\frac{1}{2}[/itex]mv22 - [itex]\frac{1}{2}[/itex]mv12

    Notice anything familiar? [itex]\frac{1}{2}[/itex]mv2 is our generic equation for KE.
    [itex]\frac{1}{2}[/itex]mv22 - [itex]\frac{1}{2}[/itex]mv12 gives us ΔKE.

    ∴ W = ΔKE
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