# 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? [Friction Numerical]

Friction numerical

Question]

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?

**Solution:**

μ =0.45

Normal force on crate = F

_{N}= mg = (25 kg)(9.8 m/s^2) = 245 N

Friction = F_{F}= μF_{N}= (0.45)(245 N) = 110.25 NTo achieve constant velocity, the applied force is to be just equal (in magnitude) to the friction while moving.

Hence, the required force is 110.25 N