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Coefficient friction

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Horizontal force of 210 N used to push 20 kg crate over 4 m with constant velocity. Looking for coefficient of kinetic friction.


    2. Relevant equations
    a=0
    Work done by friction force is:
    Net force=ma
    Fapp-Fk=ma
    210-Fk=0
    Friction force = 210

    3. The attempt at a solution

    friction force=(muk)mg
    210=(muk)(20)(9.8)
    muk=1.07

    Where did I go wrong?
    Scott
     
  2. jcsd
  3. Dec 7, 2008 #2

    PhanthomJay

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    Gold Member

    assuming a horizontal surface, looks OK. muk can be greater than 1 for certain rough or rubbery surfaces.
     
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