Energy conservation in vertical spring-mass system

Steps:

• At maximum extension δ, velocity is zero; gravitational potential energy loss equals spring potential energy gain.
• Set initial position (unstretched) as gravitational PE zero: mgδ = (1/2)kδ².
• Solve for δ: δ = 2mg/k.
• Substitute m=0.5 kg, g=10 m/s², k=30 N/m: δ = 2(0.5)(10)/30 = 10/30 = 1/3 m ≈ 0.3 m.

Why D is correct:

• D matches δ = 2mg/k from equating energies at release and maximum stretch (Hooke's law and conservation of energy).

Why the others are wrong:

• A is 1/300 of actual δ, possibly confusing k with 3000 N/m.
• B approximates mg/k (equilibrium extension) but ignores full oscillation amplitude.
• C is about 1.5 times mg/k, underestimating the doubling from energy balance.

Final answer: D