Question

A spring has a spring constant of 12.3 N/m. If it is stretched a distance of 1.08 m, how much elastic potential energy is stored in the spring?

Answer 1

By the definition of the elastic potential energy, we get:

$$PE=\dfrac{1}{2}kx^2=\dfrac{1}{2}\cdot 12.3\ \dfrac{N}{m}\cdot (1.08\ m)^2=7.17\ J.$$

Answer:

$PE=7.17\ J.$