Work done in elastic extension under Hooke's law Steps:

• Convert units: force F = 7.00 MN = 7.00 × 10⁶ N; extension ΔL = 5.00 mm = 0.005 m.
• For Hooke's law, force varies linearly with extension, so work W = (1/2) F ΔL.
• Calculate: (1/2) × 7.00 × 10⁶ × 0.005 = (1/2) × 35,000 = 17,500 J = 17.5 kJ.
• Result matches option C. Why C is correct:
• Work equals area under linear force-extension curve: (1/2) F ΔL = 17.5 kJ, per Hooke's law. Why the others are wrong:
• A: 17.5 J uses correct formula but wrong units (J instead of kJ).
• B: 35.0 J calculates full F ΔL (ignores linear variation) in wrong units.
• D: 35.0 kJ calculates full F ΔL (ignores average force of 1/2 F).

Final answer: C