1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    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

    User Avatar

    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..
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook