Static friction force between $10\, kg$ and $40\, kg$ block,
$F_{s}=\mu_{s} R=0.6 \times m g$
$=0.6 \times 10 \times 9.8=58.8\, N$
Here, we see that the applied force $(F=100\, N)$ is greater than friction force, hence $10\, kg$ block will start motion due to application of $100\, N$ force. Due to motion, kinetic friction force
$f_{k}=\mu_{k} R=0.4\, mg$
$=0.4 \times 10 \times 9.8=39.2\, N$
$40\, kg$ body experiences a force of $f_{k}=39.2\, N$
$\therefore $ Acceleration of $40\, kg$ slab
$a=\frac{f_{k}}{40}=\frac{39.2}{40}=0.98\, ms ^{-2}$
