How this instrument works
Deform a spring and your work is not spent, it is banked. Atoms shift some thousandths of a nanometre off equilibrium spacing and that lattice holds them there, strained, until something lets go. William Rankine named that reservoir — potential energy — in an 1853 paper, and elastic storage stays its cleanest illustration, because the total depends only on where those coils finish, never on how fast or in what order you got them there. Squeeze slowly, hammer shut, stretch and let half return: same x, same joules.
Magnitudes fan out further than most people expect. The detent spring inside a ballpoint pen holds a few millijoules. Drawn to full length, a 60-pound recurve bow banks 50 to 70 joules, which is why arrows leave at roughly 60 m/s. One front suspension coil near 30 kN/m, squashed 80 mm crossing a speed bump, briefly sits on some 96 joules, while a fibreglass vaulting pole swallows an athlete's whole run-up — around 4 kilojoules — and pays most of it back as height. Laboratories seldom measure this quantity directly. They record force against displacement, then take an area under that trace, which is precisely what universal testing machines print.
Three assumptions ride along with ½kx². Stiffness must be linear, so k has to hold constant across your entire working range; beyond a proportional limit, or once coils bind solid, real curves stop being triangles and only an integral of measured data will serve. Second, material has to be genuinely elastic — willing to hand its store back. Rubber and bungee cord trace a hysteresis loop, returning noticeably less than they took, with the difference leaving as warmth. Third, and stranger: rubber's restoring force is entropic rather than bonded, which is why a pulled band heats and, famously, contracts when warmed. John Gough noticed both effects in 1805.
- Type stiffness into Spring constant (N/m). Datasheets often quote newtons per millimetre instead; that reading needs multiplying by 1000 before it belongs in this box.
- Put displacement from free, unloaded length into Extension or compression. Millimetres and inches sit on its unit menu; direction is irrelevant, since x gets squared.
- Read Stored energy in joules, or switch to kilojoules for heavy suspension work and calories when comparing against a thermal budget.
- For work done between two positions, run this sheet twice and subtract — energy at x₂ minus energy at x₁, never energy at their difference.
Worked example — a 200 N/m launcher and an 80 g ball
A bench launcher on the dynamics track: one coil rated at 200 N/m, held compressed 0.1 m against a stop, with an 80 gram steel ball resting on its plunger. Enter 200 into Spring constant (N/m), then 0.1 m into Extension or compression. E = ½ × 200 × 0.1² = ½ × 200 × 0.01 = 1 joule exactly, nothing rounded anywhere.
Release, and that joule turns into motion. Setting ½mv² equal to 1 with m = 0.080 kg gives v = 5 m/s, a figure any photogate will confirm. Now halve your pull: 50 mm stores ½ × 200 × 0.05² = 0.25 J and sends that ball off at 2.5 m/s. Speed tracks compression, energy tracks its square — so the second half of that 100 mm travel banked 0.75 J, three times what the first half managed. Springs always charge most for their last millimetre.
Questions
What units does elastic potential energy come out in?
Joules. Feed k in newtons per metre and x in metres, and N/m × m² leaves N·m, which is a joule — one kilogram metre squared per second squared. Trouble starts with mixed units: a catalogue figure of 2 N/mm entered as 2, alongside a 10 mm stretch entered as 10, returns 100 J where the truth is 0.1 J. Convert both first, or let the unit menu handle x while you fix k by hand.
Does compressing store the same energy as stretching?
For an ideal linear spring, yes — x appears squared, so its sign disappears and 20 mm either way stores identical joules. Real hardware is less even-handed. A helical coil eventually binds solid under compression, where stiffness climbs steeply and this formula stops applying, and a long slender one may buckle sideways instead. Extension springs meanwhile often carry initial tension from winding, so the first millimetre of pull costs more than the arithmetic here suggests.
How much work does stretching from 10 cm to 20 cm take?
Three times what the first ten centimetres cost, not the same again. Subtract the two stored figures rather than plugging in the gap: ΔE = ½k(x₂² − x₁²). With k = 200 N/m that reads ½ × 200 × (0.04 − 0.01) = 3 J, against 1 J for the opening stretch. Entering 0.1 m as though it were a fresh extension is the single most common error on this page, and it understates the answer threefold.
How is this different from Hooke's law?
Hooke's law gives you force at one position; this sheet gives accumulated work needed to reach it. On a force-against-displacement graph, F = kx is the line itself and E = ½kx² is the area beneath. That is why a spring pulled to 10 N of tension is not twice as dangerous as one at 5 N — it is holding four times the energy. Force is what a gauge reads; energy is what escapes when the hook slips.
Where does the energy go when a rubber band snaps back?
Some returns as motion, and a surprising fraction becomes heat. Elastomers run a hysteresis loop: their unloading curve sits below their loading curve, and that enclosed area is lost internally to friction between polymer chains. A rubber band can give back only 60 to 80 percent of what went in, which is why repeatedly stretching one warms it noticeably. Steel coils are far better behaved, typically returning above 95 percent, though even they shed a little to internal damping.
Can I use this on a progressive-rate or non-linear spring?
No. Progressive coils, conical washers, air springs and anything past its proportional limit all have a stiffness that changes with displacement, so ½kx² no longer matches the area under their curve. Measure force at a series of positions and integrate numerically instead — the trapezium rule over a load-test table is usually plenty. Where springs are only mildly non-linear over a narrow band, a local k fitted across that band gives decent estimates.