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Work Done by Force

  1. Jun 23, 2006 #1
    I have spent the last hour on this problem and I just can't seem to come up with the right answer:A 2.8 kg is pushed 1.52 m up a vertical wall with constant force of magnitude F applied at an angle of 63.7 degrees with the horizontal. The acceleration of gravity is 9.8 m/s (seconds squared) If the coefficient of kinetic friction between the block and the was is .586 find the:
    a) work done by F
    b) Find the work done by the force of gravity
    c) Find the work done by the normal force between the block and the wall
    d) by how much does the gravitational potential energy increase during this motion?

    I assumed that I was on the right track when I was trying to find the force by using : F=mu*mg/ cos theta + mu sin theta
     
  2. jcsd
  3. Jun 23, 2006 #2

    FredGarvin

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    I think you need to revisit your free body diagram...

    The force driving the object will break down into two component forces, one in the direction of the motion (the one that does the work) and one perpendicular to the wall (the normal force for the calculation of the frictional force).

    By the looks of things, I think you just made an algebra error
     
  4. Jun 23, 2006 #3

    Andrew Mason

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    But you are not asked to find F. You are given F. You are asked to find the work done by F.

    As Fred points out, your analysis of the forces is incorrect. What are the horizontal forces? What is the net horizontal force? What is the downward force? What is the upward force? What is the acceleration?

    AM
     
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