3. A 100 kg crate is sliding at a rate of 30.0 m/s, when a force of 250 N is applied opposite the direction of motion, in addition to the friction, until the crate stops 5.0 s later. What is the
coefficient of kinetic friction between the crate and the floor?

Take the direction of motion to be the positive direction. The crate slows to a rest from 30.0 m/s in a matter of 5.0 s, so it has acceleration a such that

0 = 30.0 m/s + a (5.0 s)   →   a = -6.0 m/s²

At the moment its speed is 0, the crate has a net force of s + f acting in negative direction, where s and f denote the magnitudes of the stopping force (s = 250 N) and the friction force, respectively. By Newton's second law, we have

(-s) + (-f) = (100 kg) (-6.0 m/s²)

250 N + f = 600 N

f = 350 N

The friction force is proportional to the normal force by a factor of µ, the coefficient of kinetic friction. There is no movement in the up- and downward directions, so Newton's second law says

(-w) + n = 0

where w is the weight of the crate and n is the magnitude of the normal force. So

n = w = (100 kg) (9.80 m/s²) = 980 N

Then

f = µ n

350 N = µ (980 N)

µ = (350 N) / (980 N) ≈ 0.357

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