Question

A spring is hung from the ceiling. When a 0.450-kg block is attached to the free end of the spring and released from rest, the block drops 5.0 cm before momentarily coming to rest, after which it moves back upward. What is the spring constant of the spring?

Answer 1

Law of conservation of energy says

$$mgh=\frac{kx^2}{2},\quad h=x$$

Thus, the spring constant of the spring

$$k=\frac{2mg}{x}=\frac{2*0.50*9.8}{0.05}=196\:\rm N/m$$