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 force of kinetic friction is equal to the coefficient of kinetic friction times the normal force. The normal force is equal to the weight of the box, which is $50 \times 9.8 = 490 \mathrm { N }$. Therefore, the force of kinetic friction is $0.11 \times 490 = 53.9 \mathrm { N }$.

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 $50 - \mathrm { kg }$