Q 11-05-081JEE MainJEE Main 2024 (8 Apr, Shift 2)Medium
A block is simply released from the top of an inclined plane as shown in the figure. The maximum compression in the spring when the block hits the spring is:
Answer: (D) $2\ \text{m}$
Height of the incline: $h = 10\sin30^\circ = 5\ \text{m}$, so the block starts with $mgh = 5\times10\times5 = 250\ \text{J}$ (the incline is smooth).
On the rough $2\ \text{m}$ stretch, friction removes
$$\mu mg\,d = 0.5\times50\times2 = 50\ \text{J}$$
The remaining $200\ \text{J}$ is stored in the spring at maximum compression:
$$\frac12(100)x^2 = 200 \;\Rightarrow\; x = 2\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics