A $\mathbf{5}\mathbf{}\mathit{k}\mathit{g}$box with initial speed $\mathbf{4}\mathit{m}\mathbf{/}\mathit{s}$slides across the floor and comes to a stop after $\mathbf{0}\mathbf{.}\mathbf{7}\mathit{s}$(a) what is the coefficient of kinetic friction? (b) How far does the box move? (c) You put a $\mathbf{3}\mathit{k}\mathit{g}$block in the box, so the total mass is now $\mathbf{8}\mathit{k}\mathit{g}$, and you launch this heavier box with an initial speed of $\mathbf{4}\mathit{m}\mathbf{/}\mathit{s}$. How long does it take to stop?

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’s 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,

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

Where, F = Frictional Force

N = Normal Force

To find Frictional Force,

$F=ma$ …(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),

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

Substitute $“a”$value in Equation (2),

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

Frictional Force is$57.14N$

To find Normal Force $N$,

$N=mg$ …(4)

Where,

m= mass

$g=9.8m/s^2$

Substitute these values in Equation (4),

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

From Equation (1),

Coefficient of kinetic friction,

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

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

$\mathit{F}\mathbf{\times }\mathit{d}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ …(5)

Where, d =distance

From the Equation (5),

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

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$\mathbf{0}\mathbf{.}\mathbf{56}\mathit{s}$