JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Laws of Motion

The upper block is carried along only by the friction the lower block exerts on it, so the greatest acceleration the pair can share before that block slips is set by its limiting friction, ${\mu }_{s}mg=m{a}_{\max }$, which gives $a_{\max }={\mu }_{s}g=0.5\times 9.8=4.9\mathrm{m}{\mathrm{s}}^{-2}$ independent of the upper mass. As long as the whole system accelerates at this value or less the blocks stay together, and since the table is smooth the applied force simply accelerates the combined mass, $F_{\max }=(m+M)a_{\max }=(2+8)\times 4.9=49\mathrm{N}$.