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Calculating the Work done

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data

    A 260 kg piano slides 4.3 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). The effective coefficient of kinetic friction is 0.40.

    Figure 6-36

    (a) Calculate the force exerted by the man.
    391.3469084 N
    (b) Calculate the work done by the man on the piano.
    -1682.791706 J
    (c) Calculate the work done by the friction force.
    -3795.408294 J
    (d) What is the work done by the force of gravity?
    ? J
    (e) What is the net work done on the piano?
    ? J



    2. Relevant equations
    Work=Force*Distance
    gravity=9.8
    Fnet=Sum of all forces

    3. The attempt at a solution
    Letters a through c are correct, but I have not been able to solve for d and e. I have tried doing W=FX for letter d, but with no luck. I cannot solve for e without solving for d.

    Any help is appreciated. Thank you!
     
  2. jcsd
  3. Oct 2, 2009 #2
    Kinetic friction cannot do any work.
     
  4. Oct 2, 2009 #3
    Think of the formula

    Work = force * distance

    Now it's kept from accelerating, what does that mean about your net force?

    Once you have that you can assume that Fnet= Fpiano + FFriction + Fperson.

    So all you need to find to solve this problem is the horizontal force applied by the piano. After that you just plug stuff in. So think about what is making the piano move down, then what component of that the person would be resisting.
     
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