Answer:
70.15 Joule
Step-by-step explanation:
mass of man, m = 70 kg
intial length, l = 11 m
extension, Δl = 1.5 m
Let K is the spring constant.
In the equilibrium position
mg = K l
70 x 9.8 = K x 11
K = 62.36 N/m
Potential energy stored, U = 0.5 x K x Δl²
U = 0.5 x 62.36 x 1.5 x 1.5
U = 70.15 Joule