Q 11-04-127JEE MainJEE Main 2023 (1 Feb, Shift 2)Medium
As shown in the figure a block of mass $10$ kg lying on a horizontal surface is pulled by a force $F$ acting at an angle $30^\circ$ with horizontal. For $\mu_s=0.25$, the block will just start to move for the value of $F$: [Given $g=10\ \text{m s}^{-2}$]
Answer: (B) $25.2\ \text{N}$
$N=mg-F\sin30^\circ$. Just moving: $F\cos30^\circ=\mu_s(mg-F\sin30^\circ)$
$$0.866F=0.25(100-0.5F)\ \Rightarrow\ 0.991F=25\ \Rightarrow\ F\approx25.2\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics