Static friction force between \(10 \mathrm{~kg}\) and \(40 \mathrm{~kg}\) block,


\(\begin{aligned}
F_s & =\mu_s R=0.6 \times m g \\
& =0.6 \times 10 \times 9.8=58.8 \mathrm{~N}
\end{aligned}\)
Here, we see that the applied force \((F=100 \mathrm{~N})\) is greater than friction force, hence \(10 \mathrm{~kg}\) block will start motion due to application of \(100 \mathrm{~N}\) force. Due to motion, kinetic friction force
\(\begin{aligned}
f_k & =\mu_k R=0.4 \mathrm{mg} \\
& =0.4 \times 10 \times 9.8=39.2 \mathrm{~N}
\end{aligned}\)
\(40 \mathrm{~kg}\) body experiences a force of \(f_k=39.2 \mathrm{~N}\)
\(\therefore\) Acceleration of \(40 \mathrm{~kg}\) slab
\(a=\frac{f_k}{40}=\frac{39.2}{40}=0.98 \mathrm{~ms}^{-2}\)