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 = (F)/(mg)= (99 N)/((50 kg)(9.81 m/s^2)) =0.20`