91影视

The mass of a box is5 kg.

The initial speed is4m/s.

Time taken to stop the box is 0.7s.

Frictional force is defined asthe force required resisting the motion of a box when one object鈥檚 surface contacts with other ones object.

The coefficient of kinetic friction is defined as the ratio of kinetic friction force to normal force.

Coefficient of kinetic friction,

$\mathit{渭}\mathbf{=}\frac{\mathbf{F}}{\mathbf{N}}$ ..(1)

Where, F = Frictional Force

N = Normal Force

To find Frictional Force,

$F=ma$ 鈥�(2)

Where, m = mass of a box

a = acceleration

To find acceleration a,

$v^2=u^2+2as$ ..(3)

Where,$v=0$

$u=4m/s$

$s=0.7s$

Substitute these values in Equation (3),

$$\begin{gathered} 0={\left(4\right)}^2+\left(2脳a脳0.7\right) \\ 0=16+1.4a \\ 1.4a=-16 \\ a=11.43m/s^2 \end{gathered}$$

Substitute $鈥�a鈥�$value in Equation (2),

$$\begin{gathered} F=ma \\ F=5脳11.43 \\ F=57.14N \end{gathered}$$

Frictional Force is$57.14N$

To find Normal Force $N$,

$N=mg$ 鈥�(4)

Where,

m= mass

$g=9.8m/s^2$

Substitute these values in Equation (4),

$$\begin{gathered} N=mg \\ N=5脳9.8 \\ N=49N \end{gathered}$$

From Equation (1),

Coefficient of kinetic friction,

$$\begin{gathered} 渭=\frac{F}{N} \\ =\frac{57.14}{49} \\ =1.17 \end{gathered}$$

Hence, the Coefficient of kinetic friction isrole="math" $\mathbf{1}\mathbf{.}\mathbf{17}$

$\mathit{F}\mathbf{脳}\mathit{d}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ 鈥�(5)

Where, d =distance

From the Equation (5),

$$\begin{gathered} F脳d=\frac{1}{2}m{v}^2 \\ d=\frac{1}{2}\frac{m{v}^2}{F} \\ =\frac{1}{2}脳\frac{\left(5脳4^2\right)}{57.14} \\ =0.7m \end{gathered}$$

Hence, the box moves $\mathbf{0}\mathbf{.}\mathbf{7}\mathit{m}$.

Time taken to stop the box,

$$\begin{gathered} \mathit{F}\mathbf{=}\frac{\mathbf{m}\mathbf{(}\mathbf{v}\mathbf{-}\mathbf{u}\mathbf{)}}{\mathbf{t}} \\ \mathit{t}\mathbf{=}\frac{\mathbf{m}\mathbf{(}\mathbf{v}\mathbf{-}\mathbf{u}\mathbf{)}}{\mathbf{F}} \\ \mathbf{}\mathbf{=}\frac{\mathbf{8}\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{4}\mathbf{)}}{\mathbf{57}\mathbf{.}\mathbf{14}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{56}\mathit{s} \end{gathered}$$

Hence, the Time taken to stop the box is$\mathbf{0}\mathbf{.}\mathbf{56}\mathit{s}$