A crate with a mass of 100 kg sits on a flat horizontal surface. If a person slides the crate with a constant velocity across the floor with a force of 325 N, what is the coefficient of kinetic friction between the crate and the floor? Please show work :)

The create is moving at a constant velocity, so there is no acceleration and the crate is in equilibrium. This means the net force acting on the crate is 0. By Newton's second law, we have

n - w = 0

s - f = 0

where n denotes the magnitude of the normal force acting on the crate, w denotes its weight, s is the magnitude of the force applied by the person (s for sliding), and f is the magnitude of the kinetic friction force.

The friction force is proportional ot the normal force by a factor of µ, the coefficient of kinetic friction. So we have

n = w = (100 kg) (9.80 m/s²) = 980 N

s = f = µ n  →  325 N = µ (980 N)   →   µ = (325 N) / (980 N)

So the coeficient of kinetic friction is approximately µ ≈ 0.332.

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

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