91Ó°ÊÓ

The mass of a box is 15kg.

The coefficient of static friction ${\mathrm{μ}}_s$is role="math" localid="1657711415813" $0.3$

The coefficient of kinetic friction ${\mathrm{μ}}_k$is$0.2.$

Acceleration is$2m/s/s$

The coefficient of static friction is known asthe limiting friction which has maximum value.

The coefficient of kinetic friction hasthe magnitude of the constant force which has less than the limiting friction.

The coefficient of static friction is equal or greater than the coefficient of kinetic friction.

The force required to start the box,

${\mathit{f}}_{\mathbf{f}\mathbf{r}\mathbf{i}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{=}{\mathit{μ}}_{\mathbf{s}}\mathbf{×}\mathit{N}$ …(1)

Where,

$$\begin{gathered} {\mathrm{μ}}_s=0.3 \\ N=mg \\ =15×9.8 \\ =147N \end{gathered}$$

Substitute these values in Equation (1),

$$\begin{gathered} f_{friction}={\mathrm{μ}}_s×N \\ =0.3×147 \\ =44.1N \end{gathered}$$

Hence, the force required to start the box is$\mathbf{44}\mathbf{.}\mathbf{1}\mathbf{}\mathit{N}$

${\mathit{f}}_{\mathbf{f}\mathbf{r}\mathbf{i}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{=}{\mathit{μ}}_{\mathbf{k}}\mathbf{×}\mathit{N}$ …(2)

Where,

$$\begin{gathered} {\mathrm{μ}}_k=0.2 \\ N=147N \end{gathered}$$

Substitute these values in Equation (2),

$$\begin{gathered} f_{friction}={\mathrm{μ}}_s×N \\ =0.2×147 \\ =29.4N \end{gathered}$$

Hence, the force required to start the box is $\mathbf{29}\mathbf{.}\mathbf{4}\mathbf{}\mathit{N}$

The force required to maintain an acceleration of $2m/s/s$,

$$\begin{gathered} \mathit{F}\mathbf{=}\mathit{m}\mathit{a} \\ =15×2 \\ =30N \end{gathered}$$

Hence, the force required to maintain an acceleration of $\mathbf{2}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}\mathbf{/}\mathit{s}$is $\mathbf{30}\mathbf{}\mathit{N}$.