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Coefficient of kinetic friction between box and floor

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A 98 N box of oranges is being pushed across a horizontal clean steel surface until it stops. As it moves, it is slowing down at a constant rate of 1.1m/s each second. The push force has a horizontal component of 25N and a vertical component of 20N downward. Calculate the coefficient of kinetic friction between the box and the floor.


    2. Relevant equations



    3. The attempt at a solution
    summation of Fx = 25N - fk = ma
    = 25N - fk = (10kg)(1.1m/s^2)
    fk = 14N???
    summation of Fy= 0 = -Fw - Fdown +normal force
    normal force = 20N + 98N = 118N???

    fk = (uk)(Fn)
    (14N / 118N) = uk
    uk = 0.119
     
  2. jcsd
  3. Jul 26, 2007 #2

    PhanthomJay

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    You're on the right track, but not quite there yet. The box is slowing down. What's the direction of the acceleration and the net horizontal force acting on the box?
     
  4. Jul 26, 2007 #3
    thanks phantomjay
    summation of Fx = 25N - fk = ma
    = 25N - fk = (10kg)(-1.1m/s^2)
    fk = (1.1m/s^2)(10kg) +25N
    fk = 36N

    uk = (36N/118N)
    uk = 0.305

    is this right?
     
  5. Jul 26, 2007 #4

    PhanthomJay

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    Yep, looks real good now, although you should round off the uk value to 0.3. Nice work.
     
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