a spring with spring constant K when stretched 1 cm the potential energy is U if it stretched by 4 cm the potential energy will be
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Answer:
16U
Explanation:
Potential energy stored in spring of spring constant "k" is given as
- U = 0.5kx²
When the spring is stretched by 1 cm
- U = 0.5k(1)² = 0.5k
When the spring is stretched by 4 cm
- U' = 0.5k(4)² = 0.5k × 16 = 16U
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