A horizontal 52-N force is needed to slide a $50 - \mathrm { kg }$ 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?
A)0.11
B)0.09
C)0.10
D)0.13

Question

Quizplus · Accepted Answer

The coefficient of kinetic friction is given by:$$f_k = \frac{F}{N}$$where:* $f_k$ is the coefficient of kinetic friction* $F$ is the force of kinetic friction* $N$ is the normal forceIn this case, the force of kinetic friction is equal to the horizontal force needed to slide the box, which is 52 N. The normal force is equal to the weight of the box, which is $50 \times 9.8 = 490 \mathrm { N }$. Therefore, the coefficient of kinetic friction is:$$f_k = \frac{52 \mathrm { N }}{490 \mathrm { N }} = 0.106$$which is closest to choice A, 0.11.

A Horizontal 52-N Force Is Needed to Slide A 50−kg50 - \mathrm { kg }50−kg

A Horizontal 52-N Force Is Needed to Slide A $50 - \mathrm { kg }$