Physics – 1.7.2 Work | e-Consult
1.7.2 Work (1 questions)
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Given:
- Extension (x) = 0.1 m
- Spring constant (k) = 500 N/m
Equation: Elastic Potential Energy (ΔE) = ½ k x²
Calculation:
ΔE = ½ × 500 N/m × (0.1 m)² = ½ × 500 × 0.01 J = 2.5 J
Answer: The amount of elastic potential energy stored in the spring is 2.5 J.