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Work and work-energy theorum

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data
    In trying to lift a sandbag B attached to the rope as in the attached figure, a worker at the top of the scaffolding dumps his tools and lunch kit into the box A so that it is now 80kg. if the sandbag is 65kg, find its speed as it hits the scaffolding. Assume that the sandbag starts from rest.

    2. Relevant equations

    Win = change in kinetic + change in potential + Wout

    3. The attempt at a solution
    I started by drawing 2 free body diagrams.

    Can someone guide me? Not too sure where to start.

    Attached Files:

  2. jcsd
  3. Jul 12, 2010 #2
    This is easily done using energy conservation.

    Write down the totl energy (kinetic & potential energy) of the system (both the blocks) at the initial position & the final position.

    Then equate the two to find v. (the two blocks & the rope move have the same acceleration & so will have the same speed at any instant)
  4. Jul 12, 2010 #3
    SO do I use KE + PE = 0?

    Do I add the total energies of both objects?? or keep them separate?
  5. Jul 12, 2010 #4
    For this problem you want to use the conservation equation as follows:

    mghi + [tex]\frac{1}{2}[/tex]mv2i = mghf + [tex]\frac{1}{2}[/tex]mv2f

    So the equation is more so [tex]\Delta[/tex]Ug + [tex]\Delta[/tex]KE=0

    As graphene mention being that the objects are connected via the same rope they will have the same velocities. Also if you write down what all your initial and final P.E.'s and K.E.'s it will be easier for you to see how they apply to the above equation. Hope that helps.

  6. Jul 12, 2010 #5
    is the inital height 0 and the final height is 4.2?
  7. Jul 12, 2010 #6
    Ok, i got my answer. I hope I did this right. If I made a mistake, please let me know where I did wrong. Thanks.

    I found the total energy for the sandbag and the box.

    1/2mv2^2 - 1/2mv1^2 + mg(h2 - h1) = 0
    1/2(65)v2^2 - 0 + (65)(9.8)(4.2 - 0) = 0
    32.5 v2^2 + 2675.4 = 0 (equation 1)

    1/2mv2^2 - 1/2mv1^2 + mg(h2 - h1) = 0
    1/2(80)v2^2 - 0 + (80)(9.8)(0 - 4.2) = 0
    40v2^2 - 3292.8 = 0 (equation 2)

    I added both equations and came up with this:
    72.5v2^2 - 617.4 = 0

    solved for v2 --> 2.9m/s

    Is that correct?
  8. Jul 13, 2010 #7
    can someone look over my work and tell me if i did it right? Thanks in advance :smile:
  9. Jul 14, 2010 #8

    Doc Al

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    Staff: Mentor

    You got the right answer, but only because you added the equations. The separate equations for sandbag and box are not correct. It's only the total mechanical energy of both that is conserved, not the energy of each separately.

    Think of it this way: ΔPEsandbag + ΔPEbox + ΔKEsandbag + ΔKEbox = 0

    But ΔPEsandbag + ΔKEsandbag = 0 is not true! Same for the box.
  10. Jul 14, 2010 #9
    Thanks. Is that true for pulleys, the total mechanical energy of both is conserved??
  11. Jul 14, 2010 #10

    Doc Al

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    Staff: Mentor

    If you mean like an Atwood's Machine (masses hanging over a pulley), then yes--as long as there's no friction.
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