Elastic PEMediumQuestion 6367
Question
For a spring (k=200 N/m) stretched by 0.1 m. Elastic PE:
- A1 JCorrect
- B20 J
- C0.1 J
- D2 J
Correct answer
1 J
Explanation
PE = ½kx² = ½×200×0.01 = 1 J.
Solve this quality-checked NEET multiple-choice question, then review the correct answer and explanation.
PE = ½kx² = ½×200×0.01 = 1 J.