If the coefficient of sliding friction between a 25 kg crate and the floor is 0.45, how much force is required to move the crate at a constant velocity across the floor?

At constant velocity, the crate would be in equilibrium, so that Newton's second law tells us

p + (-f ) = 0

where p and f denote the magnitudes of the added pushing force and friction force, respectively.

The friction force is proportional to the normal force by a factor of µ, the coefficient of friction. There's no vertical movement going on, so Newton's second law says

n + (-w) = 0

where n and w are the magnitude of the normal force and the weight of the crate, respectively.

Compute n :

n = w = (25 kg) (9.80 m/s²) = 245 N

Use this and µ = 0.45 to compute f :

f = µ n = 0.45 (245 N) = 110.25 N

Solve for p :

p + (-110.25 N) = 0

p = 110.25 N ≈ 110 N

the answer is a)

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