Hooke's Law Explained: F = kx, Real Spring Constants, and Where It Stops Holding

F = kx: which force, which sign, which units

In F = kx, x is not the length of the spring. It is the displacement from its natural, unloaded length, and measuring from the bench or from the top of the coil instead is the most common set-up error in the whole topic. The constant k is measured in newtons per metre: a spring with k = 200 N/m needs 200 N to be stretched a full metre, or 2.00 N to be stretched 1.00 cm. Hang a 1.00 kg mass on it and the applied force is that mass's weight, mg = 1.00 × 9.80665 = 9.80665 N with the NIST standard acceleration of gravity, so the spring settles at x = 9.80665 / 200 = 0.0490 m, just under 5 cm.

You will see the law written both as F = kx and as F = −kx, and both are right because they name different forces. F = kx is the force you apply to hold the spring at extension x; F = −kx is the restoring force the spring exerts back, and the minus sign says only that it points toward the relaxed position. The stored energy follows from adding up that force over the distance travelled: U = ½kx². The factor of one half matters, and so does the square. On the same 200 N/m spring, x = 0.05 m stores 0.25 J, x = 0.10 m stores 1.00 J, x = 0.20 m stores 4.00 J and x = 0.30 m stores 9.00 J. Force grows in proportion to the stretch; energy grows with its square, which is why drawing a bow twice as far roughly quadruples the energy it can deliver.

What a spring constant is made of, and how springs combine

k is not a property of a material; it is a property of one particular object. For a straight bar pulled along its length, k = EA/L. A 1.00 m steel rod of 1 mm² cross-section, with a Young's modulus near 200 GPa, has k = 200 × 10⁹ × 1 × 10⁻⁶ / 1.00 = 2.00 × 10⁵ N/m, so a 1.00 kg mass stretches it by 0.049 mm. Cut it to 0.50 m and k doubles to 4.00 × 10⁵ N/m: same steel, twice the stiffness. A helical spring follows k = Gd⁴/(8D³n), with d the wire diameter, D the mean coil diameter and n the number of active coils. Steel wire of 1.0 mm wound into 10 mm coils, 20 turns, with a shear modulus of 79.3 GPa, gives k = 496 N/m. Double the wire to 2.0 mm and k becomes 7 930 N/m — exactly sixteen times stiffer, because d enters to the fourth power.

Two springs joined end to end, in series, are softer than either of them alone, because each contributes its own stretch: 1/k = 1/k₁ + 1/k₂. Join 100 N/m to 300 N/m and the pair behaves as 75 N/m. Check it with a 1.00 kg load: the same 9.80665 N passes through both, so the first stretches 9.80665/100 = 9.81 cm and the second 9.80665/300 = 3.27 cm, a total of 13.08 cm — which is exactly 9.80665/75. Side by side, in parallel, they share the load and their constants add: k = k₁ + k₂ = 400 N/m, the pair stretches 9.80665/400 = 2.45 cm, and the two springs carry 2.452 N and 7.355 N, summing back to 9.80665 N. Two identical 100 N/m springs give 50 N/m in series and 200 N/m in parallel, which is the quickest sanity check on which rule you just used.

Where Hooke's law stops being true

Hooke's law is a first-order approximation to a curve, not a law of nature. Plot force against extension for a real spring and the trace is straight only over a limited range. The point where it begins to bend is the proportional limit; a little beyond lies the elastic limit, the largest extension from which the spring still returns to its exact original length. Past that the material yields and takes a permanent set — release the load and the spring is longer than it started. Nothing in F = kx warns you about this, which is why manufacturers publish a maximum working deflection alongside a value for k, and why a laboratory experiment that keeps adding masses eventually produces points that curve away from the fitted line.

Compression springs have a second, blunter limit at the other end: once the coils touch, the spring is at solid height and its stiffness becomes that of a short steel cylinder, which is enormous and has nothing to do with k. The workable habit is to treat k as valid inside a stated range of x, and to say what that range is. The same caution applies when Hooke's law stands in for material stiffness in general. The linear stress–strain relation behind k = EA/L is a good description of metals and ceramics at small strains, but rubber, polymers and biological tissue bend away from a straight line almost immediately, so fitting a single k to them is a modelling choice you should declare rather than a measurement you can quote.

Frequently asked questions

Is Hooke's law F = kx or F = −kx?

Both, for different forces. F = kx gives the magnitude of the force you apply to hold the spring at displacement x, which is what a calculator or a lab write-up normally wants. F = −kx gives the restoring force the spring exerts on whatever is attached to it, and the minus sign records that this force points back toward the relaxed position — it is what you substitute into Newton's second law to derive simple harmonic motion. If a question asks how hard you must pull, drop the sign; if it asks for the force on the mass, keep it.

What are the units of the spring constant?

Newtons per metre, N/m, which in SI base units is kg/s². Since F = kx, k = F/x, so a force in newtons divided by a length in metres. Two conversions trip people up: 1 N/mm = 1 000 N/m, the form car and machine catalogues use, and 1 kgf/cm ≈ 980.665 N/m, still seen on older data sheets. If you feed a spring constant expressed in N/mm into a formula expecting N/m, every force comes out a thousand times too small, so check the unit on the data sheet before the arithmetic rather than after.

Does Hooke's law apply to rubber bands?

Only over a very short initial stretch, and even there it is a rough fit. A rubber band's force-extension curve is S-shaped, so a single k cannot describe it: the band is compliant at first, then stiffens sharply as the polymer chains straighten. It also shows hysteresis — the curve on the way back is lower than on the way out, so some of the energy you put in comes back as heat rather than motion. That is why a rubber band is a poor choice for a Hooke's law experiment and why energy calculations of the form ½kx² do not transfer to it. Use a steel coil spring if the point of the experiment is the linear law.