Question

A 325 N box moves with a constant velocity of 3.75 m/s when pushed with a force of 425 N exerted at a 35.2 angle below the horizontal. What is the coefficient of kinetic friction between the box and the floor?

Answer 1

Answer: 0.6

Explanation:

If we draw a free body diagram of the box we will have the following:

Net force in the x-axis:

$N-W-Fsin \theta=0$ (1)

Net force in the y-axis:

$-F_{r}+Fcos \theta=0$ (2)

Where:

$N$ is the normal force

$W=325 N$ is the weight of the box

$F=425 N$ is the force exerted on the box

$\theta=35.2\°$ is the angle below the horizontal

$F_{r}=\mu_{k}N$ is the friction force, being $\mu_{k}$ the coefficient of kinetic friction

Isolating $N$ from (1):

$N=W+Fsin \theta$ (3)

Substituting (3) in (2):

$-\mu_{k}(W+Fsin \theta)+Fcos \theta=0$ (4)

Finding $\mu_{k}$:

$\mu_{k}=\frac{Fcos \theta}{W+Fsin \theta}$ (5)

$\mu_{k}=\frac{425 N cos (35.2\°)}{325 N+425 N sin (35.2\°)}$ (6)

Finally:

$\mu_{k}=0.6$ (5)