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A sled is pulled with a horizontal force of 17 N along a level trail, and the acceleration is found to be 0.40 m/s2. An extra mass m = 4.4 kg is placed on the sled. If the same force is just barely able to keep the sled moving, what is the coefficient of kinetic friction between the sled and the trail?

  • Physics -

    Let the friction force be f without the added mass. It increases to
    f (M+4.4)/M with the added mass.
    You have two unknowns (f and M) and need two equations.
    17 - f = M*a = M*0.40
    17 - f*(M+4.4)/M = (M+4.4)*0 = 0 (zero acceleration with added mass)
    f*(M+4.4)/M -Ff = 0.4 M
    f*4.4/M = 0.4M
    f = M^2/11
    17-f = 17 - M^2/11 = 0.4M
    Solve for M and use that to solve for f. Take the postitive root when solving for M.

    The coefficient of kinetic friction is
    muk = f/(M*g)

  • Physics -

    That makes a lot more sense now. Thanks!

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