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Tension of Rope, Friction, And Work Done

  1. Nov 17, 2009 #1
    A man drags a 25.0kg crate, by a rope, across the floor whose coefficient of kinetic friction is 0.250. The tension in the rope is 200.0 N at an angle of 30.0° to the vertical. What is the net work done on the box as it travels 10.0 m?


    well I broke the tension force into the x and y components with, these equations.
    cos60x200 = 100N
    sin60x200=173.205N
    also the Frictional force in the opposite direction of +x movement is fk=ukFN
    fk= 0.25 x 25kg x 9.8m/s^2 = 61.25N

    This is where I'm stuck, I thought I could subtract the x components and create a new triangle, find the new angle and determine the work done with W= Force x Distance cos theta. The answer I'm getting seems to be wrong.

    Please answer this as soon as possible, I have a midterm tommorow and I'm struggling to teach myself. Thank you.
     
  2. jcsd
  3. Nov 17, 2009 #2
    oh I finally got it, I was forgetting that the y comp of tension was making the object lighter, so the answer is 820 Joules.
     
  4. Nov 17, 2009 #3

    PhanthomJay

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    OK,(100 N is the x component, and 173 N up is the y component)
    The Normal force and the weight are not the same. Sum forces in y direction to find it.
    rather, you should find the net force in the x direction
    try using W_net = F_x(net) times distance. Is there any F_net in the y direction?
     
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