Energy Stored In A Compressed Spring

At maximum compression the block is momentarily at rest, so all of its initial kinetic energy has been transferred to elastic potential energy in the spring: $\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$. Solving for the compression gives $x=v\sqrt{m/k}$, which links the incoming speed to the spring stiffness. Substituting $v=4\text{ m/s}$, $m=0.5\text{ kg}$, and $k=800\text{ N/m}$ yields $x=4\sqrt{0.5/800}=4\sqrt{0.000625}=4\times 0.025=0.1\text{ m}$. The option 0.2 m doubles the result by dropping the square root. The option 0.05 m halves it through the same misstep in reverse. The option 0.4 m omits the stiffness scaling entirely. At maximum compression the block's velocity is instantaneously zero, which is exactly why every joule of kinetic energy has been momentarily stored in the spring; a moment later the spring pushes the block back and returns that energy. The compression depends jointly on how fast the block arrives and on how stiff the spring is, with a stiffer spring producing a smaller maximum deformation for the same impact. This applies the NCERT energy-conservation method for spring-block systems. As a plausibility check, recomputing the stored energy $\frac{1}{2}(800)(0.1{)}^2=4\text{ J}$ matches the initial kinetic energy $\frac{1}{2}(0.5)(4{)}^2=4\text{ J}$ exactly.