A light helical spring extends by 4.0 cm when loaded with a mass of 0.8 kg. Assuming Hooke's law is obeyed, calculate the elastic potential energy stored in the stretched spring. (Take g = 10 m/s²)
Answer and explanation
Correct answer: B. 0.16 J
- Step 1: Convert extension to meters: e = 0.04 m. Step 2. Calculate stretching force: F = mg = 0.8 × 10 = 8.0 N. Step 3. Compute energy stored: E = 0.5 F e = 0.5(8.0)(0.04) = 0.16 J. Answer: 0.16 J. Why not the others: 0.32 J forgets the factor of 0.5; 0.08 J halves the correct answer; 1.60 J fails to convert centimeters to meters.
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