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More Work and Energy Problems

  1. Oct 25, 2003 #1
    I'm having trouble solving a problem related to work done on an object with friction present. Any help would be appreciated. Here's the question:

    A 15.0 kg block is dragged over a rough, horizontal surface by a 82.0 N force acting at 20.0° above the horizontal. The block is displaced 5.50 m, and the coefficient of kinetic friction is 0.300.

    a) Find the work done on the block by the 82 N force.

    This was easy enough. Since the force is constant, I used the equation [sum]W = F*d*cos([the]). I got the right answer, which was
    423.8014 J.

    b) Find the work done on the block by the normal force


    c) Find the work done on the block by the gravitational force.

    Since the displacement in the vertical direction was 0, no work was done by these forces.

    Here's the part that I'm having trouble with:

    d) What is the increase in internal energy of the block-surface system due to friction?

    I figured that since friction is not a conservative force, I could say that the increase in internal energy would be equal to the work done by the frictional force. I was wrong, and I'm not sure what concept I'm missing here.
    Last edited: Oct 25, 2003
  2. jcsd
  3. Oct 25, 2003 #2
    Now I'm really baffled. I've just read several sources which tell me the energy loss due to friction should equal the work done by friction, which would be, if I'm not mistaken, μk*n*d.

    μk - coefficient of kinetic friction

    n - the normal force, which I believe would be m*g - F*sin(θ)

    d - distance over which the force has acted

    Can someone shed light on this for me?

    edit - I'm really loving these math symbols, I don't see these on many forums.

    Lord I'm such a dork.
    Last edited: Oct 25, 2003
  4. Oct 25, 2003 #3
    "the increase in internal energy would be equal to the work done by the frictional force"

    I'm no physics prof. (as you know) but this makes perfect sense to me. :)
  5. Oct 26, 2003 #4


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    I'll second that.

    Are you turning it into a computer? If so, go through the calculations once more and check your sig figs. Computers are snippy with those.
  6. Oct 26, 2003 #5
    I found the problem. I was using an incorrect theta. I plug all this stuff into matlab so I had to change the degrees into radians... I made a slight calculation error.

    Thanks for the help!
  7. Oct 29, 2003 #6
    Wait why is m*g - F*sin(θ) = n ???

    I thought it was m*g*cos(θ)
    Last edited: Oct 29, 2003
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