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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 = FN = mg = (25 kg)(9.8 m/s^2) = 245 N

Friction = FF= μFN = (0.45)(245 N) = 110.25 N

To 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

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