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Mechanical advantage ideal systems

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    An engineer with mass 100 kg has designed an ideal mechanical advantage machine shown below. Platforms 1 and 2 are attached to the machine. When he steps on platform 1, platform 2 rises straight up. The maximum weight that he can lift using his machine in this manner is twice his own. The mechanical advantage of the machine cannot be adjusted. Platform 1 can be lowered a maximum of 10 m.

    Question:

    Assume mass m is twice the mass of the engineer and the engineer gives himself a push downwards to get moving. Including the push, the work done on the mass as platform 2 rises to its top height of 5 m will be equal to:
    (A) the original potential energy of the engineer
    (B) the final kinetic energy of the engineer
    (C) the original potential energy of the engineer plus half of the final kinetic energy of the engineer
    (D) the original potential energy of the engineer minus the final kinetic energy of the engineer.

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    I have no idea how to approach this problem but it appears that there is a tradeoff between PE and KE.
     

    Attached Files:

  2. jcsd
  3. Apr 7, 2015 #2

    SammyS

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    We ask that you not put your work into a proprietary file format. For one thing, some people don't have Word and can't see what is in that file. It would be easy enough to upload that image some way, as I did below.
    Capture_4.PNG
    It does involve conservation of energy, also the work-energy theorem.

    Show us an attempt. You have over 100 posts, so you should know how things work at PF.
     
  4. Apr 8, 2015 #3
    I apologize about the Word format. I posted before and it wasn't a problem. I will copy and paste from now on.

    So, back to the question above, I separated the problem in to 2 components. Work done by the man and work done by the mass. Acceleration 10 m/s^2 is used for this problem.

    Man: W = PE + KE = Fd + 1/2 mv^2 = 10000 J + 50 v^2 } v = 14 m/s
    Mass: W = PE + KE = Fd + 1/2 mv^2 = 10000 J + 100 v^2 } v = 10 m/s

    Interpretation: The PE's are the same in each problem, 10000 J, but the velocity resulting from the kinetic energy is almost half that of the mass (of the box). So, A & B are out because the work done on the mass (box) is not equal to the PE and KE of the engineer respectively. C and D are candidates but I chose C (which is the right answer) because original PE is equal for both and the velocity of the engineer is approx. half that of the engineer.

    I need a second pair of eyes to check and see if I am working this problem out the right way. Thanks!
     
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