Question

When a 20 g mass hangs attached to one end of
light spring of length 10cm, the spring
stretches by 2 cm. The mass is pulled down
until the total length of the spring is 14 cm. The
elastic energy, in Joule stored in the spring is-

Answer 1

Answer:

Force due to the ball= force exerted by the spring

Force due to ball= m×g = 20×10

Force exerted by the spring= k×x

k×2= 200

k=100

Now the spring is extended to 14 cm, elongation = 14-10=4cm

Energy stored=k×x²÷2 =50×x² 50×16

=800J