NEET (UG) 2024 — Question 17

Treat the two blocks as a single system of mass $M = m_A + m_B = 2 + 3 = 5\,\text{kg}$. The net horizontal force on the system is $F = 10\,\text{N}$. Using Newton's second law: $F = M a$, so $a = \frac{F}{M} = \frac{10}{5} = 2\,\text{m/s}^2$. Now consider block $B$ alone. The only horizontal force on $B$ is the contact force $N$ exerted by block $A$. Applying Newton's second law to block $B$: $N = m_B a = 3 \times 2 = 6\,\text{N}$. Thus the force exerted by block $A$ on block $B$ is $6\,\text{N}$.