Question

Roy pushes a box of mass 16.6 kilograms across a floor by exerting a horizontal force of 99.1 newtons. The coefficient of kinetic friction between the box and the floor is 0.324, and Roy pushes the box 11.4 meters. What work does friction do? Include units in your answer.

Answer 1

The normal force acting on the mass can be given as,

$N=mg$

The frictional force acting on the box is,

$f=\mu_kN$

Substitute the known expression,

$f=\mu_kmg$

The work done by frictional force can be given as,

$W=fd$

Substitute the known expression,

$W=\mu_kmgd$

Substitute the known values in the equation,

$\begin{gathered} W=(0.324)(16.6kg)(9.8m/s^2)(11.4\text{ m)(}\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ \approx601\text{ J} \end{gathered}$

Thus, the work done by the frictional force is 601 J.