A constant force F = m2 g / 2 is applied on the block of mass m, as shown in fig. The string and the pulleyare light and the surface of the table is smooth The acceleration of m1 is

Let a be the acceleration of mass m2 in the downward direction. Then T -m2(g/2) = m1a   …(1)and m2 g -T = m2 a   …(2)Adding eqs. (1) and (2), we getm1+m2a=m2g−m2(g/2)=m2g/2∴ a=m2g2m1+m2.