Q54E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: Average power:

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

Step 2: A given data:

Consider the given data as below.

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

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

Final velocity, $v=0$

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

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

Step 3: Define acceleration and time:

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}}$}\right.}}\\ & =& 4.08\text{ s}\end{array}$

Step 4: Average power produced by friction as the rock stops:

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 $157\text{ W}$.