Question

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.

Answer 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