Question

A box of mass 50 kg is pushed hard enough to set it in motion across a flat surface. Then a 99-N pushing force is needed to keep the box moving at a constant velocity. What is the coefficient of kinetic friction between the box and the floor?

0.23

0.20

0.18

0.17

Answer 1

0.20

Step-by-step explanation:

The box is moving at constant velocity, which means that its acceleration is zero; so, the net force acting on the box is zero as well.

There are two forces acting in the horizontal direction:

- The pushing force: F = 99 N, forward

- The frictional force:
`F_f=\mu mg`, backward, with

`\mu` coefficient of kinetic friction

m = 50 kg mass of the box

g = 9.8 m/s^2 gravitational acceleration

The net force must be zero, so we have

`F-F_f = 0`

which we can solve to find the coefficient of kinetic friction:

`F-\mu mg=0\\\mu = (F)/(mg)=(99 N)/((50 kg)(9.8 m/s^2))=0.20`