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Work and friction

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data

    This is a question about work:
    There is a dog pulling a sled across a patch of snow *with a force of 20 Newtons*, the dog's mouth is 15 degrees above horizontal and the sled weighs 5Kg. The snow has a coefficient of friction of 0.2, how much work has the dog done once it has covered 20 meters in distance (ignore the weight of the dog)?

    2. Relevant equations
    W = Fd cos(angle) d

    Where:
    W is work done
    Fd is force applied (20 N)
    angle is angle from horizontal (15 degrees)
    d is distance (20 meters)

    Ff = U * N

    Where:
    Ff is friction force
    U is coefficient of friction (0.2)
    N is normal force

    3. The attempt at a solution

    What I did is:
    - Calculated the Normal Force (N)
    - Using N, I calculated the force of friction
    - Then I calculated the work done by friction (assuming that the angle of the force is 180 degrees)
    - Then I calculated the work done by the dog pulling the sled (with 15 degree angle above horizontal)
    - Then I added the two (194 + 386)
    Result: 580 Joules

    Am I doing it right? I feel like I'm in the wrong direction.
    Thanks.
     
    Last edited: Sep 1, 2009
  2. jcsd
  3. Sep 1, 2009 #2
    Well, there seems to be something missing in your setup...just looking at the sliding motion over 20 meters, we'd have to know how much force the dog is applying, but we don't know this unless the sled is said to be moving at constant speed. Then the forces on the sled must balance in the horizontal direction. That means the frictional force [tex]F_{f}[/tex] is equal and opposite to the horizontal part of the force that the dog applies, which I'll call [tex]F_{x}[/tex]. That is, [tex]F_{f}+F_{x}=0[/tex]. But this horizontal part is the only part that matters in the work-by-dog equation: [tex]W=F_{x}d[/tex] since that's the direction of the sled's actual movement (can you see why this is same as using the cos(theta) equation? But in the way I wrote it, do we even need the 15 degree angle?). So now you should be able to piece all this together to get just the work done *by the dog*.
     
    Last edited: Sep 1, 2009
  4. Sep 1, 2009 #3
    Sorry, I forgot the force:
    It's 20 Newtons from the dog's mouth (15 degrees form horizontal).
     
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