91Ó°ÊÓ

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,

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

Where,

F = frictional Force

N = Normal Force

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

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

$$\begin{gathered} \mathrm{μ}=\frac{F}{N} \\ F=\mathrm{μ}N \\ =0.6×78.48 \\ =47.08N \end{gathered}$$

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

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,

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

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

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,

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

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