A spring with a spring constant k of 44 N/m is stretched a distance of 30 cm (0.3 m) from its original unstretched position. What is the increase in potential energy of the spring?

Answer 1

P.E. = (1/2) k X^2

Use X = 0.3 meters

The original value of X (deviation from the equilibrium position) was zero.
The answer will be in Joules

Answer 2

1.98