Hooke's Law Calculator

spring force = spring constant × displacement

k

The spring constant, or stiffness, in N/m. It is the force needed per metre of stretch — a retractable pen is around 10 N/m, a spring scale 100 N/m, a car suspension spring 30 N/mm, an industrial die spring 100 kN/m. The field also takes kN/m, N/mm and lbf/in, and the table below writes one stiffness in three of those at once. It must be positive: a stiffness of zero is not a spring

x

The displacement, in centimetres — how far the spring has been moved from its resting length. The field also takes metres, millimetres and inches. It is a size rather than a signed quantity, so compression and extension are not distinguished: a spring pulled 10 cm and one pushed 10 cm pull back with the same 10 N, in opposite directions. Zero is a legal value and means the spring is at rest

F

The spring force, in newtons, kilonewtons, pounds-force and kilogram-force. It is the force the spring exerts, which is also the force you have to supply to hold it where it is. Leave this field empty and the page solves the other two for it; fill it in together with one of the others and it solves the third

Use this page when you want the reading rather than the energy: how hard a spring scale pulls at a given extension, whether a given spring is stiff enough to hold a door open, what force a bowstring is putting on its limbs at full draw, how much a suspension spring compresses under a known corner weight, or what stiffness a spring needs to be if it has to produce a stated force at a stated travel. The last of those is the direction this page is built for and most calculators are not: two of the three fields can be left empty, and which two you fill in decides what gets solved. The two habits worth forming are about units and about range. On units, the field defaults to centimetres and the formula needs metres, so the page converts internally — but the stiffness may arrive as N/mm or lbf/in from a supplier's datasheet, and those differ from N/m by factors of a thousand and 175 respectively, which is exactly what the table below is for. On range, Hooke's law is only true up to the spring's elastic limit, and every real spring leaves the straight line somewhere; a spring pulled past that point gives a force lower than this page predicts and does not come back to its original length. The energy stored in the same spring as it is stretched is worked out on the elastic potential energy page, which is the same two inputs and the same spring.

1. The default spring, pulled out 10 cm

   1. Stiffness: 100 N/m. Displacement: 10 cm, which is 0.1 m — the formula needs metres
   2. Spring force: 100 × 0.1 = 10 N
   3. The other two rows simply echo the inputs, because nothing had to be solved for
   4. Compare with a 1 kg mass hanging on the same spring: it would pull with 9.81 N and stretch it 9.81 cm

   Ten newtons is the weight of a litre of water, and that comparison is the fastest way to a feel for what 100 N/m means: hang a litre bottle off this spring and it stretches about a centimetre per 100 grams. The other thing to take from this row is the unit trap it walks through — 10 cm had to become 0.1 m before multiplying, and doing it the other way round gives 1000 N, a hundred times too much. That is the single most common arithmetic error on this page, and the reason the field is labelled in centimetres while the formula is not.

2. A force is known, the stiffness is not: 25 N at 10 cm

   1. Leave the stiffness field empty and fill in the other two: displacement 10 cm, force 25 N
   2. Rearrange F = kx for k: k = F ÷ x
   3. Convert first: 10 cm = 0.1 m
   4. Stiffness: 25 ÷ 0.1 = 250 N/m

   This is the direction the page exists for. A force and a distance are both easy to measure with a kitchen scale and a ruler, and stiffness is the number you cannot measure directly — so pulling a spring and reading the force is how the constant is actually obtained in practice. The same arithmetic in imperial units is the source of the lbf/in spring rates in every automotive catalogue: 25 lbf at 10 inches would be 2.5 lbf/in. Note that the two fields left filled are displacement and force, and the page has still solved for all three quantities, because F = kx has three variables and any two of them fix the third.

3. A known spring, a known force: how far does it move?

   1. Stiffness 1000 N/m — a car suspension spring is around 30 times this, an industrial die spring about 100 times
   2. Rearrange F = kx for x: x = F ÷ k
   3. Displacement: 30 ÷ 1000 = 0.03 m
   4. Converted back for display: 3 cm

   The answer comes out in metres and the page shows it in centimetres, which is the conversion the first example did in the other direction. It is worth noticing how small the travel is: 30 N — the weight of three litres of water — moves this spring only three centimetres, whereas the same 30 N would move the 100 N/m spring of the first two examples thirty centimetres. Stiffness is exactly that ratio, and it is why a stiff spring feels unyielding under a small load and a soft one bottoms out under a big one. Neither is better; the choice is set by how much travel the mechanism has.

Hooke's law is a linear approximation and every real spring stops obeying it. Beyond the elastic limit the force rises more slowly than kx and the spring does not return to its original length, so an answer computed from a datasheet stiffness is only valid inside the rated travel — which for a die spring may be a few millimetres and for a bowstring is a large fraction of its length. Coil springs are also not perfectly linear within their range: the ends of the coil close up as it compresses, so the effective stiffness rises towards full compression, and a spring extended far enough straightens out entirely. The page treats the spring as massless and frictionless, which is wrong for a spring that is itself heavy or that has to slide over a guide rod. It distinguishes neither compression from extension nor one direction from the reverse, because the model genuinely does not: if the application depends on that difference, this equation is not where it lives. There is a practical trap in how the fields work, too. Because the first two fields start filled in, anyone who wants to solve for the stiffness has to clear one of them first — if they type a force on top of the existing defaults, the page will report that the numbers disagree rather than silently use the new one, which is deliberate but does surprise people. And the answer has no energy in it: the work done in stretching the spring to this point is ½kx², which is larger than F × x would suggest because the force was not at its final value the whole way.

What is the Hooke's law formula?

F = kx: the spring force equals the spring constant times the displacement. A 100 N/m spring pulled 10 cm (0.1 m) pulls back with 100 × 0.1 = 10 N. The relation is linear, so twice the displacement is twice the force, and it holds in both directions — pulling and pushing — which is why a spring can be used as a scale. The famous square in spring calculations belongs to the energy stored, ½kx², not to the force.

How do I find a spring constant if I cannot measure it directly?

Hang a known weight on the spring, measure how far it stretches, and divide. A 2 kg mass gives 19.62 N of force; if the spring stretches 4 cm, then k = 19.62 ÷ 0.04 = 490.5 N/m. Two cautions. Measure between the resting length and the loaded length, not from the top of the spring, and take the reading after the spring has settled rather than while it is still swinging. If the spring is too stiff to stretch measurably with a weight you have, add more weight or use a lever — the arithmetic is identical, and the uncertainty in a 1 mm measurement over a 4 cm stretch is what limits the answer.

Which two fields should I fill in?

Any two of the three, and the page solves the third. Stiffness and displacement give you the force, which is the ordinary direction. Displacement and force give you the stiffness, which is how a spring rate is measured. Stiffness and force give you the travel, which is what sizing a mechanism usually needs. Fill in all three only if you want them checked against each other: if they disagree the page says so instead of picking one, because showing a force in the result panel that contradicts a force in an input box on the same screen is worse than showing an error.

What is the difference between N/m, N/mm and lbf/in?

They are three ways of writing the same stiffness, and the factors between them are large. 1 N/mm equals 1000 N/m, because both the force and the length scale by a thousand. 1 lbf/in equals about 175.13 N/m, and it is not a round number because neither the pound-force nor the inch is defined by the metric system. The practical consequence is that a spring quoted as 30 N/mm is 30000 N/m — thirty times stiffer than the 1000 N/m spring in the third example, not thirty times softer — and a rate copied from an imperial catalogue without converting is wrong by a factor of 175.

Does it matter whether the spring is compressed or stretched?

Not to this equation. F = kx uses the size of the displacement, so a spring pushed in 10 cm and one pulled out 10 cm both pull back with the same magnitude — the force just points the other way. Real springs are not perfectly symmetric: a coil spring can be wound so that it behaves slightly differently in compression, and a spring compressed far enough will bind with its coils touching, at which point it becomes effectively rigid and this equation stops applying. Within the rated travel, though, treating the two directions as identical is what the model does and is accurate enough for most work.

Why is the energy ½kx² and not kx²?

Because the force is not at its final value while the spring is being stretched — it starts at zero and rises evenly, so the average force over the whole travel is half the final force, and work is average force times distance. Multiply that out and you get ½kx². It is the same reason a triangle's area is half its base times its height. This page deliberately does not show that number: the energy stored in a spring is its own calculation with its own worked examples, and putting a squared term next to a linear one on the same screen invites the reader to think the two are two results from one formula.

• Hooke's Law: Stress and Strain Revisited (College Physics 2e, §16.1) — the linear region, the spring constant, and where real springs leave the straight line — OpenStax
• Hooke's law — the force-displacement relation, the elastic limit, and why the constant is a property of the spring rather than of the material — Wikipedia
• NIST Guide to the SI — the newton, the pound-force and kilogram-force factors, and the unit prefixes used by the stiffness field and the reference table — National Institute of Standards and Technology