A block A of mass m1 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 m2 is suspended. The coefficient of kinetic friction between the block and the table is μk. When the block A is sliding on the table, the tension in the string is

Given situation is shown in the figure.  Here, N=m1g   f=μkN=μkm1g                                   . . . . .(i)Let α be the acceleration of blocks. Equation of motion for A and BT-f=m1a                   . . . . .(i) m2g-T=m2a              . . . . .(ii)Adding equation(ii) and (iii), we get.m2g-f=m1+m2aa=m2g-fm1+m2Put this value of α in equation (iii) T=m2g-m2m2g-fm1+m2=m1m2g+m1m2μkgm1+m2=m1m21+μkgm1+m2                                   (using  . (i))