Q 11-13-089JEE MainJEE Main 2023 (1 Feb, Shift 2)Medium
A block is fastened to a horizontal spring. The block is pulled to a distance $x=10$ cm from its equilibrium position (at $x=0$) on a frictionless surface from rest. The energy of the block at $x=5$ cm is $0.25$ J. The spring constant of the spring is ______ $\text{N m}^{-1}$.
Numerical value type. Enter your answer.
Answer: 67
Taking the energy of the block as its kinetic energy at $x=5$ cm:
$$\tfrac12k(0.1^2-0.05^2)=0.25\ \Rightarrow\ k=\frac{0.5}{0.0075}\approx67\ \text{N m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics