A $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{\text{kg}}$rock is sliding on a rough, horizontal surface at $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is $\mathbf{0}\mathbf{.}\mathbf{200}$. What average power is produced by friction as the rock stops?

Average power is defined as the ratio of the total work done by the body to the total time taken by the body.

Consider the given data as below.

Mass of the rock, $m=20.0\text{kg}$

Initial velocity, role="math" localid="1660060507788" $u=8.00\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.$

Final velocity, $v=0$

Acceleration due to gravity, role="math" localid="1660060467758" $g=9.8\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$

Coefficient of kinetic friction,${\mu }_{\text{k}}=0.20$

Force is define by using the following equation.

$f={\mu }_{\text{k}}mg$

Now from newton’s second law:

$$\begin{gathered} \sum f=ma \\ {\mu }_{\text{k}}mg=ma \\ a={\mu }_{\text{k}}g \end{gathered}$$

Substitute known values in the above equation.

$\begin{array}{rcl}a& =& 0.20\times 9.8\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & =& 1.96\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

From the first kinetic equation of motion.

$v=u+at$

Here, $t$is the time.

Substitute known numerical values in the above equation.

$$\begin{gathered} 0=8.0\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.-1.96\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\times \left(t\right) \\ 1.96\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\times \left(t\right)=8.0\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right. \end{gathered}$$

$\begin{array}{rcl}t& =& \frac{8.0{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.}}{1.96{\displaystyle \raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\\ & =& 4.08\text{s}\end{array}$

Calculate the average power as below.

$P=\frac{KE}{t}$

Here, $KE$is the kinetic energy.

$\begin{array}{rcl}P& =& \frac{\frac{1}{2}m{u}^2}{t}\\ & =& \frac{m{u}^2}{2t}\end{array}$

Putting known values in the above equation.

$\begin{array}{rcl}P& =& \frac{{\displaystyle \left(\text{20.0 kg}\right)}\times {\left(8.00{\displaystyle \raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{s}}$}\right.}\right)}^2}{{\displaystyle 2\times \left(\text{4.08 s}\right)}}\\ & =& 157\text{W}\end{array}$

Hence, the average power is produced by friction as the rock stops is $157\text{W}$.