When a 1.0kg 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, if g = 10m/s2
Force constant of a spring
k=Fx = mgx = 1×102×10−2⇒k = 500 N/m
Increment in the length = 60 - 50 = 10 cm
U = 12Kx2 = 12500(10×10−2)2 = 2.5 J