Question

When 2 kg mass hangs 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, if g = 10 m/s² 

(A) 10 J 

(B) 2 J 

(C) 2.5 J 

(D) 5 J

Answer 1

Correct option: (D) 5 J

Explanation:

m = 2 kg

x = 2 cm = extension in wire

F = mg also F = Kx

i.e mg = Kx

K = (mg / x) = {(2 × 10) / (2 × 10^–2)}

Energy stored in spring = (1/2) Kx² 

= (1/2) × 10³ × [(60 – 50) × 10^–2]² 

= (1/2) × 10³ × 10^–2 = 5 J