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Weird mechanical advantage/energy transfer problem

  1. Jul 19, 2007 #1

    cxz

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    1. The problem statement, all variables and given/known data
    There is an engineer with a mass of 100kg standing on a platform raised 10m above ground. On the ground is a platform that holds a box with mass m. Connecting the two platforms is a machine of unknown design (possibly pulleys) that generates a mechanical advantage of 2. When the engineer steps on platform1, platform 2 is raised accordingly. The mechanical advantage of the machine cannot be cannoted.

    Problem: Assume mass(box) 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 5m (maximum height) will be equal to:
    (note: the answer must be given in terms of the original potential energy and final kinetic energy of the engineer)

    2. Relevant equations

    Conservation of mechanical energy of the entire system.

    ΔKEsystem + ΔPEsystem = 0

    Mechnical work done on the box by the engineer:
    Work = Fd = ΔKEbox + ΔPEbox


    3. The attempt at a solution

    "The net force on both masses is zero so velocity from the push remains constant. The work done on the mass is

    m2*g*h2 + 1/2*m2*v2 ^2

    2*m1 = m2, 2*v2 = v1, and 2*h2 = h1 (v=h/t where t is equal for both platforms).

    Substituting those variables in, the equation for the work done on the box becomes m1*g*h1 + 1/2 * (1/2*m1*v1^2), or the original potential energy of the engineer + 1/2 of the final kinetic energy of the engineer."

    That's the solution from the book, and it makes perfect sense to me. What I'm confused about is this:

    The initial energy of the engineer is his potential energy at a height of 10m + the kinetic energy from the push. This expression is equal to the total energy present in the system initially.

    When the engineer reaches the bottom, his potential energy is 0 and his kinetic energy is 1/2*m1*v1^2, where v1 is the final velocity of the engineer.

    Now let's call the engineer's original velocity v0. The total energy lost by the engineer is: m1*g*h1 + (1/2*m1*v1^2 - 1/2*m1*v0^2).

    Equating that expression to the expression for the work done on the mass (the solution), we have (1/2*m1*v1^2 - 1/2*m1*v0^2) = 1/2*(1/2*m1*v1^2), which can be interpreted to mean that half of the kinetic energy from the original "push" was lost as the engineer descended down the height of the machine.

    This is where my confusion lies. This concept is not intuitive to me at all. I can see how the math works out from the conservation angle, but I can't picture how the kinetic energy is transfered, or why exactly 1/2 of it is transfered. Maybe there's a flaw in my logic. Can someone nudge me in the right direction? Thank you very much.
     
  2. jcsd
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