Q 11-04-037JEE MainMedium
A $5$ kg block rests on a horizontal floor ($\mu_s = 0.4$). A force $F$ is applied at $37°$ above the horizontal. The minimum value of $F$ that just starts the block moving is nearly (take $\sin 37° = 0.6$, $\cos 37° = 0.8$, $g = 10\ \text{m/s}^2$)
Answer: (B) $19.2$ N
Concept. The upward component reduces the normal force: $N = mg - F\sin37°$.
Step 1. Condition. $0.8F = 0.4(50 - 0.6F)$.
Step 2. Solve. $1.04F = 20 \Rightarrow F \approx 19.2$ N.
Answer. $19.2$ N (Option B).
Solution by Sreeraj P, M.Sc Physics