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Man pushing piano

  1. Dec 5, 2005 #1
    Hi, this my first time asking a question, so here it goes:
    A 255 kg piano slides 4.6 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-35). The effective coefficient of kinetic friction is 0.40.

    [​IMG]

    (a) Calculate the force exerted by the man.
    So far, I have this answer, by subtracting the Fparallel(mg*sin(30)) by Ffriction(mu*mg*cos(30)). [383.8 N]

    (b) Calculate the work done by the man on the piano.
    Got this answer, too, by multiplying the answer above by 4.6m. [-1764.6 J]

    It's the next parts I got wrong.
    (c) Calculate the work done by the friction force.
    I figured Ffriction = mu * m * g * cos(30) = 865.7 * d = 3982.12 J, but that doesn't make sense. [?J]

    (d) What is the work done by the force of gravity?
    9.8 m/s^2 * 255g * 4.6m = 11495.4, it's incorrect [?J]

    (e) What is the net work done on the piano?
    This one I had no idea how to do altogether. J

    Please help me out, thank you in advance! :smile:
     
  2. jcsd
  3. Dec 5, 2005 #2

    Doc Al

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    Staff: Mentor

    This is OK except for the sign. Since the friction points up the ramp, while the piano slides down the ramp, the work done is negative.

    When calculating work, you must use the component of the force parallel to the displacement.

    What's the net force on the piano?
     
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