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

Step-by-step 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)=(Fcos \theta)/(W+Fsin \theta)` (5)

`\mu_(k)=(425 N cos (35.2\°))/(325 N+425 N sin (35.2\°))` (6)

Finally:

`\mu_(k)=0.6` (5)