Energy Stored In A Compressed Spring

Question

A 0.5 kg block slides along a frictionless surface at 4 m/s and runs into a horizontal spring of force constant 800 N/m fixed to a wall. What is the maximum compression of the spring during the interaction?

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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}mv^2 = 	frac{1}{2}kx^2$. Solving for the compression gives $x = v\sqrt{m/k}$, which links the incoming speed to the spring stiffness. Substituting $v = 4	ext{ m/s}$, $m = 0.5	ext{ kg}$, and $k = 800	ext{ N/m}$ yields $x = 4\sqrt{0.5/800} = 4\sqrt{0.000625} = 4 	imes 0.025 = 0.1	ext{ m}$. 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	ext{ J}$ matches the initial kinetic energy $	frac{1}{2}(0.5)(4)^2 = 4	ext{ J}$ exactly.

Energy Stored In A Compressed Spring

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