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Determine the work done by the applied force

  1. Oct 14, 2006 #1
    Okay I am confuse on how i came to this conclusion and how i came to that answer.

    A block of mass 2.50 kg is pushed 2.80 m along a frictionless horizontal table by a constant 12.0 N force directed 25.0° below the horizontal.

    (a) Determine the work done by the applied force.
    (b) Determine the work done by the normal force exerted by the table.
    0 J
    (c) Determine the work done by the force of gravity.
    (d) Determine the work done by the net force on the block.

    I am confuse on b, c, d. I know that to determine the applied force i would use W = (Fcostheta)(deltaX).

    For b i would use W = (mgcos90)deltaX is that correct? because since the block is on a table which is flat and normal force is pointing up so from table to normal force it makes a 90 degree angle.

    For c I did a guess since I was not sure but i thought that it would be zero since there is no incline and gravity is not acting on this block because its not falling.

    D i came up with Wnet = W_F + W_f + W_N + W_mg
    W_F = 30.45
    W_f = 0 (no friction so it is zero)
    W_N = 0 (not sure why)
    W_mg = 0 (not sure why)

    sorry its a bit long but can anyone explain this to me.
  2. jcsd
  3. Oct 14, 2006 #2
    can u clarify the direction of the force on the object? what u mean by a constant force of 12 N directed below the horizontal? So the force is acting 25 degrees upwards on the block or downwards?
  4. Oct 15, 2006 #3
    that is all the info was given and thats the exact question given.

    I think the force is acting 25 degrees downwards from the block.
  5. Oct 15, 2006 #4
    For a, b and c, the answers u provided are right.. if the force on the block is really acting downwards from the block. for a) All u have to do is to take the horizontal component of the force on the block and multiply it by the distance travelled which i think u got it. For both b and c, the WD equals zero because there is no distance travelled in the direction of the force. for d, all u got to do is to sum up the WD on the block since they are scalars which will actually give u the wd done by moving the block, 30.45.
  6. Oct 15, 2006 #5
    Thanks, that make much more sense now.
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