Physics

A 1.8 kg block slides on a horizontal surface with a speed of v=0.80 m/s and an acceleration of magnitude a= 2.8m/s 2.
a) What is the coefficient of kinetic friction between the block and the surface?
b) When the speed of the block slows to 0.40m/s, is the magnitude of the acceleration greater then, less than or equal to 2.8m/s2? Explain.

asked by Anonymous
  1. I guess there is no pull on this block so it is coasting to a stop.
    The Force F is all friction
    F = m a = -(1.8)2.8 = - 5.04 Newtons of friction force
    (by the way for part B this is constant until we stop)
    but Friction F = - mu m g
    so
    mu m g = m a
    a = mu g
    2.8 = mu *9.81
    mu = 2.8/9.81

    posted by Damon

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