91Ó°ÊÓ

Mass of the block is $m=\text{2.5}×{\text{10}}^{\text{-2}}\text{kg}$,

Initial radius is $R=\text{0.300}\text{m}$,

Initial angular speed ${\mathrm{Ó\neg }}_0=\text{2.85}\text{rad/s}$

Final radius is $r=\text{0.150}\text{m}$.

The law of conservation of angular momentum states that the product of moment of inertia and angular speed of the body, remains constant.

${\mathbf{\text{I}}}_{\mathbf{i}}{\mathbf{Ó\neg }}_{\mathbf{i}}\mathbf{=}{\mathbf{I}}_{\mathbf{f}}{\mathbf{Ó\neg }}_{\mathbf{f}}\mathbf{=}\mathbf{constant}$

Yes, the law of conservation of momentum states that the angular momentum of a systems remains conserved.

Solve as follows:

$$\begin{gathered} m{r}^2\mathrm{Ó\neg }=m{R}^2{\mathrm{Ó\neg }}_0 \\ {\left(\text{0.15}\text{m}\right)}^{\text{2}}\mathrm{Ó\neg }={\left(\text{0.3}\text{m}\right)}^{\text{2}}×\text{2.85}\text{rad/s} \\ \mathrm{Ó\neg }=\frac{{\left(\text{0.3}\text{m}\right)}^{\text{2}}×\text{2.85} \text{rad/s}}{{\left(\text{0.15}\text{m}\right)}^{\text{2}}} \\ \mathrm{Ó\neg }=\text{11.40}\text{rad/s} \end{gathered}$$

Thus, the angular velocity is$\text{11.40}\text{rad/s}$.

Change in kinetic energy of the block as follows:

$$\begin{gathered} \mathrm{Δ}KE=\frac{1}{2}{m}_1{v}_1^2-\frac{1}{2}{m}_2{v}_2^2 \\ =\frac{1}{2}m\left(v_1^2-v_2^2\right) \\ =\frac{1}{2}m\left({\left(R{\mathrm{Ó\neg }}_0\right)}^2-{\left(r\mathrm{Ó\neg }\right)}^2\right) \end{gathered}$$

For the given values

$$\begin{gathered} \mathrm{Δ}KE=\frac{1}{2}×\left(2.5×{10}^{-2}\text{kg}\right)\left({\left(0.855\text{m}/\text{s}\right)}^2-{\left(1.77\text{m}/\text{s}\right)}^2\right) \\ =\frac{1}{2}×\left(2.5×{10}^{-2}\text{kg}\right)\left(0.7310{\text{m}}^2/{\text{s}}^2-3.1329{\text{m}}^2/{\text{s}}^2\right) \\ =\left(1.25×{10}^{-2}\text{kg}\right)\left(-2.40{\text{m}}^2/{\text{s}}^2\right) \\ =-3.00×{10}^{-2}\text{J} \end{gathered}$$

Negative sign is showing that the value is decreasing.

Hence, the kinetic energy changes by $\text{3.00}×{\text{10}}^{\text{-2}}\text{J}$.

Work done in pulling the cord is equal to change in kinetic energy.

$$\begin{gathered} Workdone=\mathrm{Δ}KE \\ =\text{0.03}\text{J} \end{gathered}$$

Thus, the work done in pulling the cord is $\text{0.03}\text{J}$.