91影视

It is given that the mass of block as m, initial linear speed as $v_1$ , initial radius as$r_1$ and final radius as$r_2$ .

The tension force provide a centripetal force then $T=\frac{m{v}^2}{r}$and the angular momentum of the force is $L=m{v}_1{r}_1$.

Find speed from the angular momentum as follows:

$$\begin{gathered} mvr=m{v}_1{r}_1 \\ 鈬�v=\frac{v_1{r}_1}{r} \end{gathered}$$

Substitute the value of v in T and simplify.

$\begin{array}{r}T=\frac{m{\left(\frac{v_1{r}_1}{r}\right)}^2}{r}\\ =\frac{m{v}_1^2{r}_1^2}{r^3}\end{array}$

Therefore, the required tension force is $T=\frac{m{v}_1^2{r}_1^2}{r^3}$.

It is given that the work done by the tension force as $W={鈭�}_{r_1}^{r_2}鈥�\stackrel{鈫�}{T}\left(r\right)鈰�d\stackrel{鈫�}{r}$.

Since, $\stackrel{鈫�}{T}\left(r\right)$ and $d\stackrel{鈫�}{r}$are always opposite in direction implies $\stackrel{鈫�}{T}\left(r\right).d\stackrel{鈫�}{r}=-T\left(r\right)dr$. Thus the work done can be written as follows:

$\begin{array}{r}W=鈭�{鈭�}_{r_1}^{r_2}鈥�T\left(r\right)鈰�dr\\ ={鈭�}_{r_1}^{r_2}鈥�\frac{m{v}_1^2{r}_1^2}{r^3}dr\\ =m{v}_1^2{r}_1^2{鈭�}_{r_1}^{r_2}鈥�\frac{1}{r^3}dr\\ =m{v}_1^2{r}_1^2{\left(\frac{鈭�2}{r^2}\right)}_{r_1}^{r_2}\end{array}$

Further, simplify as follows:

$\begin{array}{r}W=鈭�m{v}_1^2{r}_1^2{\left(\frac{鈭�1}{2r^2}\right)}_{r_1}^{r_2}\\ =m{v}_1^2{r}_1^2{\left(\frac{1}{2r^2}\right)}_{r_1}^{r_2}\\ =m{v}_1^2{r}_1^2\left(\frac{1}{2r_2^2}鈭�\frac{1}{2r_1^2}\right)\\ =\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}鈭�\frac{1}{r_1^2}\right)\end{array}$

Therefore, the work done by the tension force is $W=\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}鈭�\frac{1}{r_1^2}\right)$.

The change in kinetic energy is given by $鈭�K=K_2-K_1$. Here,$K_1=\frac{m{v}_1^2}{2}$and$K_2=\frac{m{v}_2^2}{2}$then,$鈭�K=\frac{m{v}_2^2}{2}-\frac{m{v}_1^2}{2}$

Find speed from the angular momentum as follows:

$\begin{array}{r}m{v}_1{r}_1=m{v}_2{r}_2\\ 鈬�v_2=\frac{v_1{r}_1}{r_2}\end{array}$.

Substitute $v_2=\frac{v_1{r}_1}{r_2}$in $鈭�K$and simplify.

$\begin{array}{r}\mathrm{螖}K=\frac{m{\left(\frac{v_1{r}_1}{r_2}\right)}^2}{2}鈭�\frac{m{v}_1^2}{2}\\ =\frac{m}{2}\left(\frac{v_1^2{r}_1^2}{r_2^2}鈭�v_1^2\right)\\ =\frac{m{v}_1^2}{2}\left(\frac{r_1^2}{r_2^2}鈭�1\right)\\ =\frac{m{v}_1^2{r}_1^2}{2}\left(\frac{1}{r_2^2}鈭�\frac{1}{r_1^2}\right)\end{array}$

Therefore, the change in kinetic energy is equal to work done that is $鈭�K=W$.