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Homework Help: Dynamics Question

  1. Jun 11, 2010 #1
    A 50.0 kg crate is dragged at a constant velocity of 1.5 m/s for a distance of 8.0 m. If the force of friction on the crate is 225 N, how much work was done by the person to drag the crate over the 8.0 m distance if the rope used to drag the crate was directed at 20degrees angle above the horizontal?

    my work:
    mass 50kg
    distance 8m
    angle 20 degrees
    Force of Friction 225N
    since we have a constant velocity Fnet = 0
    Fnet = T - Ff

    Solve for T
    0 = T - 225N
    T = 225N

    therefore T x distance moved by the rope x cosx = work done
    so I got 225N x 8m x cos20 = 1690 J

    However the answer sheet says it's 1800 J............
     
  2. jcsd
  3. Jun 11, 2010 #2

    berkeman

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    Staff: Mentor

    The tension T is greater than 225N, since the horizontal force (the cos component of T) is 225N to oppose the friction force.
     
  4. Jun 11, 2010 #3
    That's only if the object is "not moving at constant velocity". In this question it is moving at constant velocity so Fnet must equal 0.
     
  5. Jun 11, 2010 #4

    berkeman

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    Staff: Mentor

    Correct. The Fnet in the horizontal direction has to equal zero (225 -225 = 0).
     
  6. Jun 11, 2010 #5
    I still get 1690J for work done. Driving me nuts...
     
  7. Jun 11, 2010 #6
    You may confuse last part.
    net F=0
    Tcos20 = f =225 N
    W=(Tcos20)*8=1800 J
     
  8. Jun 11, 2010 #7
    umm this might seem stupid but I think the question is asking what is the work done by the "person" not the "rope" so then it's T 225 N x 8m = 1800J ?
     
  9. Jun 11, 2010 #8

    berkeman

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    Staff: Mentor

    The person is doing the work, but you've got the correct answer. The horizontal component of T is 225N, to oppose the frictional force. The actual tension T is greater than 225N, since there is also a vertical component to the Tension.
     
  10. Jun 11, 2010 #9
    thanks I can finally rest now..
     
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