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\( = \frac{{39.2}}{{{m_B}}} = \frac{{39.2}}{{40}} = 0.98\;m/{s^2}\)