PhysicsElasticityJAMB
A spring of force constant 200 N/m is stretched by 0.1 m. Find the energy stored in it.
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Answer: D. 1 J
E = ½kx² = ½ × 200 × 0.01 = 1 J.
An AdmitMe practice question, written for the JAMB syllabus. Not an official past question.