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Finding work done by friction on a box

  1. Nov 25, 2015 #1
    1.
    A box of mass
    m = 59.0 kg
    (initially at rest) is pushed a distance
    d = 83.0 m
    across a rough warehouse floor by an applied force of
    FA = 204 N
    directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the work done by the force of friction
    7-p-001.gif

    2. Relevant equations
    W = fdcosΘ
    Fƒ=μN
    3. The attempt at a solution
    I used the two above equations and I found that the applied force of friction would be 57.82 because that is .1*59*9.8 and then if you put that into the first equation, the work would be 57.82*83*cos30 which equals 4156.108 and then since it is opposite the applied force, it would be -4156.108. At least I thought. That's wrong and I don't know why. Please help!
     
  2. jcsd
  3. Nov 25, 2015 #2
    But in this problem, the normal force is larger than the weight.
     
  4. Nov 25, 2015 #3
    The vertical component of F add to weight.
    Remember when resolving vector, here force, noted all the components. Here the X and Y components.
     
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