Squeeze a spring and it answers in proportion: $F = kx$ — F equals k x. $F$ is the spring force in newtons, $x$ is the displacement from the spring's rest length in metres (stretch or squeeze, the law does not care), and $k$ is the spring constant in newtons per metre — the spring's price list, how many newtons each metre of travel will cost you.

But a spring keeps a second ledger, and this chapter cares about that one: $U = \tfrac{1}{2} k x^{2}$, the elastic potential energy in joules, from the same $k$ and the same $x$. Where does the ½ come from? The force is not constant — it climbs from zero at rest to $kx$ at full squeeze, so the AVERAGE force through the travel is $\tfrac{1}{2}kx$, and average force times distance gives $\tfrac{1}{2}kx^{2}$. Check the units and it survives: $\mathrm{N/m} \times \mathrm{m^2} = \mathrm{N \cdot m} = \mathrm{J}$.

Two traps live here, both computed into the wrong answers below: forgot the half (an answer exactly double) and forgot the square (an answer that is not even an energy). And a third, quieter one: springs are small, so problems hand you centimetres while $k$ thinks in metres. Convert BEFORE the socket — and note that in $U$ the error SQUARES, so a stray centimetre lands you ten thousand times off, not a hundred.