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

Answer:

0.20

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:

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