A spring with a spring constant of 200 N/m is compressed 0.1 m from its equilibrium position. A 0.5 kg block is placed against the spring and then released. What is the maximum speed of the block?

Core Concept

This is a conservation of energy problem. The elastic potential energy stored in the spring (½kx²) converts to kinetic energy of the block (½mv²). Setting them equal: ½kx² = ½mv². Solving for v: v = √(kx²/m) = √(200 × 0.1² / 0.5) = √(4) = 2.0 m/s. Option A is incorrect because it uses √(kx/m) instead of √(kx²/m). Option C is incorrect because it uses √(2kx²/m) instead of √(kx²/m). Option D is incorrect because it uses √(kx²) instead of √(kx²/m).