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Simple problem, but having a bit of trouble

  1. Oct 30, 2004 #1
    A 14.0 kg block is dragged over a rough, horizontal surface by a 87.0 N force acting at 20.0° above the horizontal. The block is displaced 6.50 m, and the coefficient of kinetic friction is 0.300.

    (b) Find the work done on the block by the normal force.
    J
    (c) Find the work done on the block by the gravitational force.
    J

    I thought normal force was mg which turned out wrong.

    For gravitational force, I tried force times delta r times cos theta, but that also turned out wrong as well.
     
  2. jcsd
  3. Oct 30, 2004 #2
    [tex]W = Fdcos(\theta)[/tex]
    What is the displacement in the direction of both of those forces? Think about it another way. The work is equal to the change in potential energy.
     
    Last edited by a moderator: Oct 30, 2004
  4. Oct 31, 2004 #3
    hmm...

    not sure of what u mean lol
     
  5. Oct 31, 2004 #4
    The magnitude of the work done by gravity is equal to the magnitude of the change in gravitational potential energy. I can't say anymore without flat out telling you :)
     
  6. Oct 31, 2004 #5

    cepheid

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    vsage, his problem says that the block is dragged over a rough, horizontal surface. Why should the change in gravitational potential energy enter into things at all? Is it not clearly zero, as is the answer to part c? Or is that what you were trying to get him to realise? It just seemed a roundabout way to attack the problem, that's all.

    ramin, my two cents:

    It's very important to realise that the normal force in any given problem is not always equal to the weight of the object. Granted, by Newton's third law, the ground should exert an upward force balancing the object's weight. However, what about the applied force? It is at an angle..., that means that some component of it is downward as well. How does this affect the normal force? Finally, what is the key point to remember in parts b and c? I'd say its that a force only does work if it causes some displacement in the same direction as it.
     
  7. Oct 31, 2004 #6

    Doc Al

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    Because the question asked for the work done by gravity! vsage's suggestion is correct.
    The point is to provide help and understanding, not just the answer! :smile:

    Right.
    You mean Newton's 2nd law.
    In this example, the applied force has an upward component.
    Right! (As vsage tried to point out in his first post.)
     
  8. Oct 31, 2004 #7
    lol...I had a feeling it was zero, but for some reason I thought it'd be simple then.

    I'm having a problem with the rest of the question thought:

    (d) What is increase in internal energy of the block-surface system due to friction?
    J
    (e) Find the total change in the block's energy.

    Not sure what to do here, I'm guessing you have to add up all the energies.
     
  9. Oct 31, 2004 #8

    Doc Al

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    Find the work done against the frictional force.
    Start by finding the work done by the applied force. Some of that work goes into the energy you found in part (d). What happens to the rest of it?
     
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