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3 years ago

4

A horizontal 52 Newton force is needed to slide a 50-kg box across a flat surface of a constant velocity of 3.5 m s what is the

coefficient of kinetic friction between the box and the floor

1 answer:

andreyandreev [35.5K]3 years ago

4 0

Answer:

μk = 0.106

Explanation:

When the box is moving across the flat surface with a constant velocity, the force applied on it must be equal to the kinetic frictional force applied on the object:

F = μk*W = μk*mg

Where,

F = Force applied on box = 52 N

m = mass of box = 50 kg

g = 9.8 m/s²

μk = coefficient of kinetic friction between box and floor = ?

Therefore,

52 N = μk(50 kg)(9.8 m/s²)

μk = (52 N)/(490 N)

<u>μk = 0.106</u>

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