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 0.18 0.20

Answer 1

The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force $F_f$ which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
$F-F_f=0$
so
$F=F_f$

The frictional force can be rewritten as
$F_f = \mu m g$
where $m=50 kg$, $g=9.81 m/s^2$. Re-arranging, we can solve this equation to find $\mu$, the coefficient of dynamic friction:
$\mu = \frac{F}{mg}= \frac{99 N}{(50 kg)(9.81 m/s^2)} =0.20$