A block $A$ of mass $2$ kg rests on a block $B$ of mass $4$ kg, which lies on a smooth horizontal floor. The coefficient of friction between $A$ and $B$ is $0.4$ (take the static and kinetic values equal). A horizontal force of $18$ N is applied to $A$. The accelerations of $A$ and $B$ are, respectively, ($g = 10\ \text{m/s}^2$)
Answer: (A) $5\ \text{m/s}^2$ and $2\ \text{m/s}^2$
Concept. First assume the blocks move together and check whether the friction needed is within the limit. If not, they slip and kinetic friction $\mu m_Ag$ acts.
Step 1. Assume no slipping. $a = \dfrac{18}{6} = 3\ \text{m/s}^2$. Friction needed to drive $B$: $4 \times 3 = 12$ N.
Step 2. Compare with the limit. $f_{\max} = 0.4 \times 2 \times 10 = 8$ N $< 12$ N, so $A$ slides on $B$.
Step 3. Accelerations.
$$a_A = \frac{18 - 8}{2} = 5\ \text{m/s}^2,\qquad a_B = \frac{8}{4} = 2\ \text{m/s}^2$$
Answer. $5\ \text{m/s}^2$ and $2\ \text{m/s}^2$ (Option A). Option B is the trap of assuming common motion.
Solution by Sreeraj P, M.Sc Physics