Ahorizontal 52-n force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 m/s. what is the coefficient of kinetic friction between the box and the floor?

0.61

Explanation:

This is calculation based on friction.

Since the box rests on a flat surface, the force that exists between them is known as frictional force.

Since the friction is dynamic (velocity is not zero)

The frictional force = kinetic energy gained by the body.

Ff = 1/2mv^2

coefficient of kinetic friction × normal reaction = 1/2mv^2

Since normal reaction is equal to the weight(force acting along the vertical component)

Normal reaction= mg = 50 × 10 = 500N. Therefore,

coefficient of kinetic friction × 500 = 1/2×50×3.5^2

coefficient of kinetic friction = 50×3.5^2/1000

coefficient of kinetic friction= 0.61

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