JEE Main Physics question
When a spring is extended by 2 cm, energy stored is 100 J. When extended by a further 2 cm, the energy increases by
Answer: (B)
Solution
U ∝ x², so U(4 cm) = (4/2)² × 100 = 400 J. Additional energy = 400 − 100 = 300 J.
When a spring is extended by 2 cm, energy stored is 100 J. When extended by a further 2 cm, the energy increases by
U ∝ x², so U(4 cm) = (4/2)² × 100 = 400 J. Additional energy = 400 − 100 = 300 J.