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A spring with spring constant k is compressed by x. The elastic potential energy stored is:
Ak x
B½ k x
Ck x²
D½ k x²
Answer & Solution
Correct answer: D. ½ k x²
1. Spring force F = k x (Hooke's law).
2. Work done to compress from 0 to x is ∫₀ˣ k x' dx' = ½ k x².
3. This is stored as elastic PE.
_Source: NCERT Class 11 Physics Ch 5, §5.8 PE of a Spring, p. 83_
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