What Hooke's law actually claims
Hooke's law says that the force a spring exerts is proportional to how far you have moved it from its free length. Double the stretch and you double the force. Robert Hooke published it in 1678 as the anagram ut tensio, sic vis — as the extension, so the force — and it remains the first equation you reach for whenever something elastic resists being deformed.
The constant of proportionality k is the spring rate. It carries units of force per unit length: newtons per metre in SI work, newtons per millimetre on most European spring drawings, pounds-force per inch in North American catalogues, and kilograms-force per millimetre in older Japanese suspension data. A rate of 100 N/m means every extra metre of stretch adds 100 N of pull; a car coil spring at 30 N/mm means every extra millimetre of compression adds 30 N.
The negative sign in the textbook form F = −kx is a statement about direction, not magnitude. It says the spring force always points back towards the free length: pull the spring and it pulls back, compress it and it pushes back. That restoring behaviour is what makes a spring the engine of every oscillator, which is why the same k reappears in the period of a mass on a spring and in the resonant frequency of a suspension.
What Hooke's law does not claim is that every spring behaves this way everywhere. It is the first term of a series, an excellent approximation over the working range of a properly designed spring and a poor one outside it.
Reading each term, and where the energy goes
Rearranging gives you the three jobs this calculator does. F = kx predicts the load a known spring will produce at a known deflection — the sizing problem. k = F/x reduces a load test to a rate, which is how you find the stiffness of an unmarked spring: hang a known mass, measure the stretch, divide. x = F/k predicts travel under a known load, which is the packaging problem — will the spring fit in the space you have left?
The energy stored follows from the area under the force–deflection line. Because the line is straight and starts at the origin, the area up to deflection x is a triangle of base x and height kx, so U = ½kx². The factor of one half catches people out: a spring at twice the deflection holds four times the energy, not twice. That quadratic is why a fully compressed spring released without control is genuinely dangerous, and why the stored energy in a valve spring dominates the dynamics of a high-revving engine.
Combining springs follows from asking what is shared. Springs in series — one hanging from the next — each feel the full load, and the deflections add, so the compliances add: 1/k_eq = 1/k₁ + 1/k₂ + …. For n identical springs that is k_eq = k/n. Springs in parallel — side by side between the same two plates — all share one deflection, and the forces add, so the rates add: k_eq = nk. The pattern is the reverse of resistors, and mixing the two up is the single most common error in a spring calculation.
Worked example: finding the rate of an unmarked extension spring
You have an extension spring with no part number. You hang a 2.00 kg mass from it and measure 78.5 mm of stretch. What is its rate, and how much energy is stored?
- Turn the mass into a force. F = mg = 2.00 kg × 9.80665 m/s² = 19.613 N.
- Put the deflection in metres. 78.5 mm = 0.0785 m.
- Divide. k = F / x = 19.613 / 0.0785 = 249.85 N/m, which is 0.24985 N/mm, or about 1.427 lbf/in.
- Energy at that deflection. U = ½ × 249.85 × 0.0785² = ½ × 249.85 × 0.0061623 = 0.770 J.
- Check it against the load. The average force through the stretch is half the final force, 9.81 N, and 9.81 N × 0.0785 m = 0.770 J. The two routes agree, which confirms the factor of one half.
Now suppose you need 40 N at that same 78.5 mm. A single 250 N/m spring gives only 19.6 N, so you fit two of them in parallel: keq = 2 × 249.85 = 499.7 N/m, and F = 499.7 × 0.0785 = 39.2 N. Each spring still stretches 78.5 mm and each still carries 19.6 N, so neither is working harder than it was in the test — you have doubled the load capacity without changing the stress in any one spring.
How to judge whether the answer is trustworthy
Start with the linearity assumption. As a design rule of thumb, a helical spring is treated as linear over roughly the middle 80% of its travel, and the two ends are where that assumption fails. Near zero deflection an extension spring with initial tension does not follow the line at all — it needs a threshold force before it opens at all, and the true relation is F = F₀ + kx. Near full travel a compression spring approaches solid height, coils begin to touch, and the effective rate climbs steeply. If your deflection sits in either region, the number this calculator gives is optimistic in one direction and pessimistic in the other.
Then check the stress, which Hooke's law never mentions. A spring can be perfectly linear and still be seconds from failure. Manufacturers publish a maximum working load or maximum deflection for exactly this reason, and it is the figure that governs fatigue life. Hooke's law tells you what force you will get; the catalogue tells you whether you are allowed to ask for it.
Finally, sanity-check the magnitude against something familiar. A ballpoint pen spring is a few hundred N/m. A garage door extension spring is in the low thousands. A passenger-car coil spring is 20–40 N/mm, meaning 20,000–40,000 N/m. If your computed rate is three orders of magnitude away from the family your spring belongs to, you have almost certainly mixed millimetres and metres — the most common arithmetic slip in this calculation, and one that produces an answer wrong by exactly 1,000.
Spring rate in every unit you are likely to meet
| N/m | N/mm | kgf/mm | lbf/in | Typical example |
|---|---|---|---|---|
| 100 | 0.100 | 0.0102 | 0.571 | Light laboratory demonstration spring |
| 500 | 0.500 | 0.0510 | 2.855 | Ballpoint pen / small return spring |
| 1,000 | 1.000 | 0.1020 | 5.710 | Screen-door closer spring |
| 5,000 | 5.000 | 0.5099 | 28.55 | Trampoline or garage-door spring |
| 17,513 | 17.513 | 1.7859 | 100.0 | Round-number 100 lbf/in reference |
| 30,000 | 30.00 | 3.0592 | 171.3 | Passenger-car front coil spring |
| 100,000 | 100.0 | 10.197 | 571.0 | Heavy machine die spring |
Conversions use 1 lbf/in = 175.126835 N/m and 1 kgf = 9.80665 N. The examples are order-of-magnitude guides for sanity-checking a result, not catalogue values.
Mistakes that produce a wrong spring number
- Mixing millimetres with metres. Dividing newtons by millimetres gives N/mm, not N/m. The answer is off by a factor of exactly 1,000, and it looks plausible because spring rates span that whole range.
- Swapping the series and parallel rules. Springs add like capacitors, not like resistors: parallel rates add, series compliances add. Stacking two springs end to end makes the assembly softer, which surprises people every time.
- Ignoring initial tension in an extension spring. Close-wound extension springs are made with the coils pressed together, so the first few newtons produce no movement. Fitting a straight line through data that includes that region gives a rate that is too low.
- Using weight in kilograms as a force. A 2 kg mass exerts 19.6 N, not 2 N. Multiply by g before dividing by deflection, or work consistently in kgf.
- Applying the law past the proportional limit. Once the wire yields, the spring takes a permanent set, the free length changes, and every subsequent prediction from the old rate is wrong.
- Forgetting the factor of one half in the energy. Using
U = kx²doubles the stored energy and doubles the predicted launch speed of anything the spring throws.
Where the linear model stops and what to use instead
Hooke's law is the small-displacement limit of elasticity, and the same idea generalises well beyond coil springs. For a bar in tension the equivalent statement is σ = Eε, where Young's modulus E plays the role of k for a unit cube of material; the axial stiffness of a rod is then k = EA/L. Beams, torsion bars and even the bonds in a crystal lattice all show the same linear-then-not behaviour, because any smooth energy well looks like a parabola near its minimum.
Where linearity fails, you need a different model. Rubber and elastomer springs are markedly non-linear and depend on temperature and rate. Air springs follow a gas law rather than Hooke's law, so their rate rises with compression. Progressive-rate coil springs are wound with variable pitch so that coils go solid one at a time, producing a deliberately rising rate — no single k describes them.
Once you have the rate, the natural next question is dynamic. A mass m on a spring of rate k oscillates with angular frequency ω = √(k/m), giving a period T = 2π√(m/k) — structurally identical to the simple pendulum period, where gravity supplies the restoring force instead of a spring. The elastic energy from this page converts entirely into kinetic energy at the equilibrium point, which is the standard route into wave problems, since a chain of masses and springs is exactly what carries a mechanical wave. If your spring is part of a fluid or pressure system, the load side of the problem often comes from a hydrostatic pressure acting on a piston area, and thermal effects on the wire modulus connect to the heat energy question of how much the spring warms in service.
Compressed springs store real energy
The U = ½kx² figure is not academic. A 30 N/mm car spring compressed 100 mm holds ½ × 30,000 × 0.1² = 150 J, comparable to a 1 kg mass dropped from 15 m. Always use a proper compressor and never release a captive spring by cutting it.