Question 14
ElasticityThe energy contained in a wire when it is extended by 0.02 m by a force of 500 N is
- A.5 J
- B.10 J
- C.103 J
- D.104 J
Show the answer and explanation
Answer: A (5 J)
Explanation
Energy stored in a stretched wire = ½ × force × extension = ½ × 500 × 0.02 = 5 J.