a
(a) Limiting friction between block and slab$ = {\mu _s}{m_A}g$ $ = 0.6 \times 10 \times 9.8 = 58.8N$
But applied force on block $A$ is $100\,N$. So the block will slip over a slab.
Now kinetic friction works between block and slab ${F_k} = {\mu _k}{m_A}g$ $ = 0.4 \times 10 \times 9.8 = 39.2\;N$
This kinetic friction helps to move the slab
$\therefore $ Acceleration of slab$ = \frac{{39.2}}{{{m_B}}} = \frac{{39.2}}{{40}} = 0.98\;m/{s^2}$