Q 11-04-099JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
A block of mass $5\ \text{kg}$ is placed on a rough plane inclined at $30^\circ$ to the horizontal, with coefficient of friction $\mu = 0.1$. If $\vec F_1$ is the force (parallel to the incline) required to just move the block up the inclined plane and $\vec F_2$ is the force required to just prevent the block from sliding down, then the value of $|\vec F_1| - |\vec F_2|$ is: [Use $g = 10\ \text{m s}^{-2}$]
Answer: (B) $5\sqrt3\ \text{N}$
$F_1 = mg\sin30^\circ + \mu mg\cos30^\circ$ and $F_2 = mg\sin30^\circ - \mu mg\cos30^\circ$.
$$F_1 - F_2 = 2\mu mg\cos30^\circ = 2\times0.1\times50\times\frac{\sqrt3}{2} = 5\sqrt3\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics