Question

A 2.1 kg mass is suspended at equilibrium from a spring with spring constant 17.0 N/m. Calculate the total elastic potential energy stored in the spring (in J), assuming the spring obey's Hooke's law.

Answer 1

Hooke's Law is \[F = -kx\], k is the spring constant, Suspended at equilibrium means that all forces acting on the mass cancel each other out. In this case there are only two forces, Gravitational force and the spring force. \[ m g = kx \\ \Rightarrow x = \frac{mg}{k}\]
The elastic potential energy stored in the spring is \[U = \frac{1}{2} k x^2 \]

Insert x into the last equation, enter all known values and you should get the answer.