91Ó°ÊÓ

The given data is listed below as,

• The total mass of the box and its contents is, $M$
• The force required to pull the string is,$F$
• The initial speed of the box is,$V_i$
• Initially, the distance moved by the box is, $W$
• The distance moved by the hand is,$d$
• The total distance moved by the hand and the box is$w+d$
• The mass of the three masses is,$m$
• The distance at which the masses are attached is,$r$
• The angular speed relative to the axle is,${\mathrm{Ó\neg }}_i$

According to the work-energy theorem, the work done on an object is the same as that of the change in kinetic energy of the object.

The law of angular motion states that a body will continue to rotate with a constant angular speed unless an external torque acts on it.

The work-energy theorem gives the final speed and the law of angular motion gives the final angular speed of the box.

The equation of work done in the box is expressed as,

$$\begin{gathered} \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}^{\mathbf{2}}\mathbf{}{\mathbf{}}_{\mathbf{f}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}_{\mathbf{i}}^{\mathbf{2}} \\ \mathbf{F}\mathbf{.}\mathbf{d}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{M}\left(v^2{}_f-v_i^2\right) \end{gathered}$$

Here,$M$ is the total mass of the box,$v_f$ is the final velocity of the box, $v_I$and is the initial velocity of the box.

Substitute w for $d$in the above expression.

$$\begin{gathered} F.w=\frac{1}{2}M\left(v_f^2-v_i^2\right) \\ F.w+\frac{1}{2}M{v}_i^2=\frac{1}{2}M{v}_i^2 \\ {v}_f^2=\frac{F.w+\frac{1}{2}M{v}_i^2}{\frac{1}{2}M} \\ {v}_f=\sqrt{\frac{F.w+\frac{1}{2}M{v}_i^2}{\frac{1}{2}M}} \\ =\sqrt{\frac{2Fw}{M}}+v_i^2 \end{gathered}$$

Thus, the speed of the box is $\sqrt{\frac{2Fw}{M}}+v_i^2$.

The expression for the final angular speed of the box can be expressed as,

$\mathrm{Ó\neg }=\frac{v}{r}$

Here,$v$ is the final velocity and $r$is the distance of the masses from the axle.

Substitute all the values in the above expression

$\mathrm{Ó\neg }=\frac{\sqrt{{\displaystyle \frac{2Fw}{M}+v_i^2}}}{r}$

Substitute for and for in the above expression.

$$\begin{gathered} \mathrm{Ó\neg }=\frac{\sqrt{{\displaystyle \frac{2Fw}{M}+v_i^2}}}{r} \\ =\sqrt{\frac{2Fd}{3m{r}^2}+\frac{v_i^2}{r_i^2}} \\ =\sqrt{\frac{2Fd}{3m{r}^2}+}{\mathrm{Ó\neg }}_i^2 \end{gathered}$$

Thus, the final angular speed relative to the axis is$\sqrt{\frac{2Fd}{3m{r}^2}+}{\mathrm{Ó\neg }}_i^2$.