It is sometimes claimed that friction forces always slow an object down, but this is not true. If you place a box of mass $\mathbf{8}\mathbf{}\mathit{k}\mathit{g}$on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the belt, which is $\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. The coefficient of kinetic friction between box and belt is $\mathbf{0}\mathbf{.}\mathbf{6}$. (a) How much time does it take for the box to reach this final speed? (b) What is the distance (relative to the floor) that the box moves before reaching the final speed of$\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$?

The mass of a box is$8kg$

The speed of the belt is$5m/s$

The coefficient of kinetic friction between box and belt is$0.6$

Coefficient of kinetic friction,

$\mathit{\mu }\mathbf{=}\frac{\mathbf{F}}{\mathbf{N}}$ …(1)

Where,

F = frictional Force

N = Normal Force

role="math" localid="1657729671494" $$\begin{gathered} N=mg \\ =8\times 9.81 \\ =78.48N \end{gathered}$$

Substitute N value in Equation (1) to find F value,

$$\begin{gathered} \mu =\frac{F}{N} \\ F=\mu N \\ =0.6\times 78.48 \\ =47.08N \end{gathered}$$

To find the time takes for the box to reach this final speed, we need to find acceleration first.

$$\begin{gathered} a=\frac{F}{m} \\ =\frac{47.08}{8} \\ =5.89m/s^2 \end{gathered}$$

To find Time,

$$\begin{gathered} v^2=u^2+2at \\ {v}^2-u^2=2at \\ t=\frac{v^2-u^2}{2a} \end{gathered}$$

Where,

$v=5m/s$

$u=0$

$a=5.89m/s^2$

Substitute these values in above Equation to find $t$,

role="math" localid="1657730489038" $$\begin{gathered} \mathit{t}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}}{\mathbf{2}\mathbf{a}} \\ =\frac{5^2-0}{2\times 5.89} \\ =2.12s \end{gathered}$$

Hence, the time takes for the box to reach this final speed is $\mathbf{2}\mathbf{.}\mathbf{12}\mathbf{}\mathit{s}$.

Work done by friction,

$\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{=}{\mathit{f}}_{\mathbf{f}\mathbf{r}\mathbf{i}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{\times }\mathit{d}$

$\mathrm{d}\left(\mathrm{di}s\tan ce\right)=\frac{{\mathrm{mv}}^2}{2\times {\mathrm{f}}_{friction}}$

Where,

$$\begin{gathered} m=8kg \\ V=5m/s \end{gathered}$$

$$\begin{gathered} m=8kg \\ V=5m/s \end{gathered}$$

$f_{friction}=47.08N$

Substitute these values in the above Equation to find Distance,

$$\begin{gathered} \mathit{d}\mathbf{\left(}\mathit{d}\mathit{i}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{c}\mathit{e}\mathbf{\right)}\mathbf{}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}\mathbf{\times }{\mathbf{f}}_{\mathbf{f}\mathbf{r}\mathbf{i}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{8\times 5^2}{2\times 47.08} \\ =2.12m \end{gathered}$$

Hence, the distance that the box moves before reaching the final speed of $\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$is $\mathbf{2}\mathbf{.}\mathbf{12}\mathbf{}\mathit{m}$