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Beginnger Coefficient of Kinetic Friction Problem

  1. Jan 14, 2009 #1
    1. The problem statement, all variables and given/known data

    If you exert 28N of horizontal force while pushing a 10.2 kg box across the surface of a floor at constant velocity, then what is the kinetic friction coefficient between floor and the box?

    2. Relevant equations

    EF = ma
    Ffk = uK * Fn


    3. The attempt at a solution

    Fg = 10.2 * -9.8 = -100 = 100 = FN
    Ef = ma
    28N + Ffk = 10.2 * 0 m/s^2 (constant velocity)
    Ffk = -28N

    uk = -28/100 = -.28 coefficient is this correct?
     
  2. jcsd
  3. Jan 14, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Perfectly correct, except for the minus sign. μN just gives the magnitude of the friction force; μ is always positive.
     
  4. Jan 14, 2009 #3
    thanks for the help
     
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