A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?0.11 a) 0.10 b) 0.13 c) 0.09 d) 0.11

Question

Reginald Hill

Physics

5 August, 06:30

A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?0.11 a) 0.10 b) 0.13 c) 0.09 d) 0.11

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Answers (2)

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Nancy Simon · Answer 1

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down, that means that the horizontal forces on it add up to zero.

Louie · Answer 2

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,

that means that the horizontal forces on it add up to zero.

Since you're pushing on it with 53N in that direction, friction must be pulling

on it with 53N in the other direction.

The 53N of friction is (the weight) x (the coefficient of kinetic friction).

53N = (490N) x (coefficient).

Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082.

Rounded to the nearest hundredth, that's 0.11. (choice 'd')