JEE Main Physics question
When a spring is stretched by 2 cm, it stores 100 J of energy. If it is stretched further by 2 cm, the stored energy will be increased by
Answer: (C)
Solution
½k(2)² = 100 (given). Stretching from 2 cm to 4 cm: W = ½k[(4)² − (2)²] = 3 × ½k(2)² = 3 × 100 = 300 J.