A horizontal 52-N force must be exerted for a $50 - \mathrm { kg }$ box to slide 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 ($\mu_k$) is calculated using the formula $\mu_k = \frac{F}{N}$, where $F$ is the force of friction (52 N in this case) and $N$ is the normal force. For an object on a flat surface, the normal force equals the weight of the object, which is $mg = 50 \times 9.8 = 490 \, \text{N}$. Thus, $\mu_k = \frac{52}{490} \approx 0.106$, which rounds to 0.11.

A Horizontal 52-N Force Must Be Exerted for A 50−kg50 - \mathrm { kg }50−kg

A Horizontal 52-N Force Must Be Exerted for A $50 - \mathrm { kg }$