Q 11-04-149JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
A system of two blocks of masses $m = 2$ kg and $M = 8$ kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is $0.5$. The maximum horizontal force $F$ that can be applied to the block of mass $M$ so that the blocks move together will be $(g = 9.8\ \text{m s}^{-2})$
Answer: (C) $49$ N
Only friction from $M$ accelerates $m$, so the largest common acceleration is
$$a_{\max} = \frac{\mu m g}{m} = \mu g = 0.5 \times 9.8 = 4.9\ \text{m s}^{-2}$$
For the whole system on the smooth table:
$$F_{\max} = (m + M)a_{\max} = 10 \times 4.9 = 49\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics