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Simple Work problem with block and friction.

  1. Feb 20, 2009 #1
    Suppose that a 50kg block slides along a horizontal surface where the coefficient of kinetic friction between the block and the surface is 0.60. A force F = 400 N is now applied where the angle of the force above horizontal is 20°. (Like a string pulling the block from the right 20 degrees above the horizontal)

    What is the net work done on the block after the block is pulled 15 meters?

    A) 1.54 * 10^3 J
    B) 2.45 * 10^3 J
    C) 2.75 * 10^3 J
    D) 3.00 * 10^3 J
    E) 3.29 * 10^3 J

    ----

    The force applied across the horizontal is Fcos(20) = 400*cos(20) = 375.88N.
    The frictional force opposing this is (0.6)(50)(-9.8) = -294N.

    So the work done by the applied force is W = Fd = (375.88)(15) = 5638.2J.
    The work done by the frictional force is W = Fd = (-294)(15) = -4410J.

    The net force is then... W = (5638.2)+(-4410) = 1228.2J.

    I got 1.23 * 10^3J ... but that is not a choice. Where did I go wrong? Can anybody help me out here?
     
  2. jcsd
  3. Feb 21, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    You have to take the vertical component of the force, which reduces the weight of the block, and hence the frictional froce.
     
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