A 80 N force is needed to slide a 50.0 kg box across a flat surface at a constant velocity.What is the coefficient of kinetic friction between the box and the floor? (Use g = 9.8 m/s². ) A)0.16 B)0.13 C)0.15 D)0.18

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Quizplus · Accepted Answer

The coefficient of kinetic friction is given by the equation:$$f_k = \mu_k * n$$where:* $f_k$ is the force of kinetic friction* $\mu_k$ is the coefficient of kinetic friction* $n$ is the normal forceIn this problem, the force of kinetic friction is 80 N, the normal force is equal to the weight of the box, which is 50.0 kg * 9.8 m/s^2 = 490 N, and the coefficient of kinetic friction is unknown.Substituting these values into the equation, we get:$$80 N = \mu_k * 490 N$$Solving for $\mu_k$, we get:$$\mu_k = 80 N / 490 N = 0.16$$Therefore, the coefficient of kinetic friction between the box and the floor is 0.16.

A 80 N Force Is Needed to Slide a 50