Elastic Potential Energy

Compressing or stretching a spring puts energy into it, and $U = \tfrac{1}{2}kx^2$ says how much. The half is not a fudge factor. By Hooke's law the force you must apply grows steadily from zero at the start of the stretch to $kx$ at the end, so the average force over the whole displacement is $\tfrac{1}{2}kx$, and work is average force times distance. Draw the Hooke's law line on a force-versus-displacement graph and the stored energy is the area beneath it — a triangle, and the area of a triangle carries a half.

A 400 N/m spring compressed 50 mm stores $U = 0.5 \times 400 \times 0.05^2 = 0.5$ J. Compress the same spring 100 mm and it stores 2 J, not 1 — the square again. That is the arithmetic behind a mousetrap, a valve spring, and the reason a bow drawn to full draw stores so much more than one drawn halfway.

The shape of this expression is worth recognising because it recurs everywhere. Kinetic energy is $\tfrac{1}{2}mv^2$; rotational kinetic energy is $\tfrac{1}{2}I\omega^2$; the energy in a capacitor is $\tfrac{1}{2}CV^2$ and in an inductor $\tfrac{1}{2}LI^2$. Every one of them is the integral of a quantity that grows linearly, and every one of them therefore comes out as a half times a coefficient times a square. Spot the pattern once and four formulas stop needing to be memorised separately.

The mistake that matters in real machinery is measuring $x$ from the wrong place. The $x$ in this formula is displacement from the spring's free length — its length when nothing is touching it — not from its installed length and not its total length. A spring installed with 20 mm of preload and then compressed a further 10 mm has not stored $\tfrac{1}{2}k(0.010)^2$. It has gone from 20 mm to 30 mm of deflection, so the energy added is $\tfrac{1}{2}k(0.030^2 - 0.020^2)$ — five times as much. Preloaded springs are everywhere in mechanisms, and this catches people every time. Two lesser traps: spring rates are quoted in N/m and in N/mm, and mixing them is a factor of a thousand in $k$; and the formula holds only inside the elastic limit, so a spring stretched until it takes a permanent set has absorbed energy this equation cannot account for, because some of it went into deforming the metal rather than into recoverable storage.