Physicshard · Past Paper
A spring is stretched by 2 cm. If the spring constant is 100 N/m, the energy stored is:
A0.02 J
B0.2 J
C2 J
D200 J
✓ Correct Answer: A — 0.02 J
Elastic PE = 1/2 kx^2. Convert x to meters: 2 cm = 0.02 m. PE = 1/2 * 100 * (0.02)^2 = 50 * 0.0004 = 0.02 J.
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