A spring stores 18 J energy when stretched 3 cm. Force constant k is: MCQ with Answer and Explanation

A spring stores 18 J energy when stretched 3 cm. Force constant k is:
A. 2 N/m
B. 6 N/m
C. 40000 N/m
D. 200 N/m
Answer: Option C
Solution (By JKSSB Mock Tests)
Elastic PE: U = ½kx² ⇒ k = 2U/x². x = 3 cm = 0.03 m. k = 2×18/(0.03)² = 36/0.0009 = 40000 N/m. Unit conversion critical: cm to m. Memory aid: 'k = 2U/x²; always use SI units'. Tests spring energy formula application with unit conversion, frequently appearing in competitive exams.

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