As shown in the figure, A is a man of mass 60 kg standing on a block of mass 40 kg kept on ground. The coefficient of friction between the feet of the man and the-block is
0.3 and that, between B and the ground is 0.2. If the person pulls the string with 125 N force, then

Maximum force of friction between A and B

${\left(f_1\right)}_{max}=0.3\times 60\times 10=180 N$

Since 125 N is less than 180 N, A will not slide on B . 

Maximum force of friction between B and ground

${\left(f_2\right)}_{max}=0.2\times \left(60+40\right)g=200 N$

Since the total horizontal external force acting on the system (125N+125N) is greater  than 200 N , Block B will not slide on the ground . 

So acceleration of the system  =$\frac{(125+125)-200}{(60+40)}m/s^2=0.5 m/s^2$