Q 11-04-188JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
Two blocks $A$ and $B$ of masses $m_A = 1\ \text{kg}$ and $m_B = 3\ \text{kg}$ are kept on a table, with $A$ resting on top of $B$. The coefficient of friction between $A$ and $B$ is $0.2$ and between $B$ and the surface of the table is also $0.2$. The maximum force $F$ that can be applied on $B$ horizontally, so that the block $A$ does not slide over the block $B$ is: [Take $g = 10\ \text{m/s}^2$]
Answer: (A) $16\ \text{N}$
Friction from $B$ is the only horizontal force on $A$, so its maximum acceleration is $\mu g = 2\ \text{m/s}^2$.
For both blocks moving together, the table's friction on $B$ is $0.2\times(1+3)\times10 = 8\ \text{N}$:
$$F - 8 = (1+3)\times2 \;\Rightarrow\; F = 16\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics