JEE Main Physics question
When a 1.0 kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition? [MP PET 2001]
Answer: (C)
Solution
k = mg/x = (1×10)/(0.02) = 500 N/m. Total extension now = 60 − 50 = 10 cm = 0.1 m. U = ½kx² = ½(500)(0.1)² = 2.5 J.