A block A of mass ${\mathrm{m}}_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass ${\mathrm{m}}_2$ is suspended. The coefficient of kinetic friction between the block and the table is ${\mathrm{\mu }}_{\mathrm{k}}$. When the block A is sliding on the table, the tension in the string is

Given situation is shown in the figure. 

$\begin{array}{l}\text{ Here, }\mathrm{N}={\mathrm{m}}_1\mathrm{g}\\ \mathrm{f}={\mathrm{\mu }}_{\mathrm{k}}\mathrm{N}={\mathrm{\mu }}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{g}\end{array}$                                   . . . . .(i)

Let a be the acceleration of blocks. 

Equation of motion for A and B

$$\begin{gathered} \mathrm{T}-\mathrm{f}={\mathrm{m}}_1\mathrm{a} . . . . .\left(\mathrm{i}\right) \\ {\mathrm{m}}_2\mathrm{g}-\mathrm{T}={\mathrm{m}}_2\mathrm{a} . . . . .\left(\mathrm{ii}\right) \end{gathered}$$

Adding equation
(ii) and (iii), we get.

$\begin{array}{l}{\mathrm{m}}_2\mathrm{g}-\mathrm{f}=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{a}\\ \mathrm{a}=\frac{{\mathrm{m}}_2\mathrm{g}-\mathrm{f}}{{\mathrm{m}}_1+{\mathrm{m}}_2}\end{array}$

Put this value of 'a' in equation (iii) 

$\mathrm{T}={\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2\frac{\left({\mathrm{m}}_2\mathrm{g}-\mathrm{f}\right)}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

$=\frac{{\mathrm{m}}_1{\mathrm{m}}_2\mathrm{g}+{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{\mu }}_{\mathrm{k}}\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2}=\frac{{\mathrm{m}}_1{\mathrm{m}}_2\left(1+{\mathrm{\mu }}_{\mathrm{k}}\right)\mathrm{g}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$                                   (using  . (i))