Question

A 4 kg block is attached to a vertical spring with a spring constant 800 N/m. The spring stretches 5 cm down. How much elastic potential energy is stored in the spring?

Answer 1

The elastic potential energy is

$$E_e=\frac{1}{2}kx^2=\frac{1}{2}800\cdot0.05^2=1\text{ J}.$$