Q 11-05-121JEE MainJEE Main 2022 (25 Jun, Shift 1)Easy
A $0.5\ \text{kg}$ block moving at a speed of $12\ \text{m s}^{-1}$ compresses a spring through a distance $30\ \text{cm}$ when its speed is halved. The spring constant of the spring will be ______ $\text{N m}^{-1}$.
Numerical value type. Enter your answer.
Answer: 600
Loss of kinetic energy = spring energy stored:
$$\frac12(0.5)(12^2 - 6^2) = \frac12k(0.3)^2\ \Rightarrow\ 0.5\times108 = 0.09k\ \Rightarrow\ k = 600\ \text{N m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics