Audited ·Last updated 27 Jul 2026·6 citations·Tier 1·0 uses

Spring Force Calculator (Hooke's Law)

Free Hooke's law calculator. Solve F = -kx for spring force, displacement, or spring constant and get elastic potential energy U = 1/2 k x^2 instantly.

Spring Force Calculator

Solve For
Spring force (F)
-10
Restoring force F = -kx. Negative value means the force points opposite to the displacement, pulling the mass back toward equilibrium.
Spring constant (k)
100 N/m
Displacement (x)
0.1 m
Elastic potential energy (U)
0.5 J

Background.

This hooke's law calculator solves the spring-force equation F = -kx for any one unknown, given the other two, and reports the elastic potential energy U = (1/2) k x^2 stored in the spring. It is the workhorse expression of linear elasticity in physics and mechanical engineering, named after Robert Hooke's 1678 essay De Potentia Restitutiva ("On the Power of Springs"), where he summarised his finding as the anagram ceiiinosssttuv — "ut tensio, sic vis", as the extension, so the force.

The relationship is staggeringly useful because it is linear, the restoring force is simply proportional to how far the spring has been pulled or pushed from its natural length, with a constant of proportionality k called the spring constant or stiffness. That linearity is what makes the whole machinery of simple harmonic motion, vibrational normal modes, acoustic waves, and even the small-oscillation limit of every smooth potential well in classical mechanics fall out so cleanly. Drop a mass on the end of a Hookean spring, displace it slightly, let go, and it oscillates sinusoidally forever (in the idealised, frictionless case) with period T = 2*pi*sqrt(m/k). Take any well-behaved potential energy curve, expand it in a Taylor series about a stable minimum, and the leading non-trivial term is automatically (1/2) k x^2 — every smooth oscillator looks Hookean if you zoom in close enough to its equilibrium point.

This calculator handles three solve modes. Solve for force when you know how stiff the spring is (k) and how far it has moved (x). Solve for displacement when you know the spring constant and the force currently acting on (or being delivered by) the spring. Solve for spring constant when you have measured a force at a known displacement, which is exactly the procedure used in undergraduate physics labs to characterise a real coil spring before pushing it into a dynamics experiment. The output also includes elastic potential energy U = (1/2) k x^2, which is the area under the F-vs-x line from 0 to x — energy that is recoverable as kinetic energy when the spring is released.

Real springs are only Hookean over a finite range. Below a material's proportional limit the F-x curve is a straight line and Hooke's law is exact; between the proportional and elastic limits the spring still returns to its original shape when unloaded but the curve bends; beyond the elastic limit the spring is permanently deformed (plastic deformation) and Hooke's law no longer applies. This calculator silently assumes you are inside the elastic, proportional regime — if your displacement is so large that the spring has been visibly stretched out of shape, the numbers it returns are mathematical, not physical.

The applications are everywhere. Car suspensions use coil springs sized so that the static deflection under vehicle weight gives a comfortable ride frequency of about 1 Hz, which fixes k once you know the mass. Mattress springs, watch hairsprings, archery bow limbs, pole-vault poles, and trampoline springs all live or die by their spring constant. In precision instrumentation an atomic-force microscope uses a tiny cantilever whose Hookean k is calibrated to the nano-newton-per-nanometre range so that deflections of the cantilever tip, read out by a laser, translate directly into the inter-atomic force between tip and sample. Vibration analysts model entire buildings, bridges, and engine mounts as networks of lumped masses and Hookean springs to find natural frequencies and avoid resonant catastrophe.

Even chemistry leans on it: the harmonic-oscillator approximation models molecular bond stretches as little springs whose k controls the infrared vibrational frequencies that you see in IR spectra. Get the spring constant right and a vast amount of physics falls into your lap.

What is spring force calculator?

Hooke's law states that the restoring force F exerted by an ideal linear spring is proportional to its displacement x from the spring's natural (unstretched) length, and points opposite to that displacement. In the standard sign convention used in Halliday, Resnick & Walker and most introductory physics texts, F = -k*x. The constant of proportionality k is the spring constant, measured in newtons per metre (N/m), and is always positive — it encodes how stiff the spring is. A small k means a soft, easily stretched spring; a large k means a stiff one. The negative sign is critical: it says the force is a restoring force, always pulling the spring back toward equilibrium. Stretch the spring (x > 0) and the force points back toward x = 0 (F < 0). Compress it (x < 0) and the force pushes outward (F > 0). The energy stored in the deformed spring is U = (1/2) k x^2, a non-negative quantity equal to the work you had to do against the spring to displace it. Hooke's law is the defining equation of a linear elastic system and is the small-displacement limit of essentially every stable mechanical system in nature.

How to use this calculator.

  1. Decide which quantity you want to compute: the spring force F, the displacement x, or the spring constant k. Select it from the Solve For dropdown.
  2. Enter the spring constant k in N/m. This is always a positive number — a steel coil spring used in a physics lab is typically 10 to a few thousand N/m; a car suspension spring is roughly 20,000 to 40,000 N/m per corner.
  3. Enter the displacement x in metres, with a sign. Positive x means the spring has been stretched past its natural length; negative x means it has been compressed. A 5 cm stretch is x = 0.05 m.
  4. If you are solving for displacement or spring constant, enter the force F in newtons instead, also signed. A force you apply to stretch the spring is positive; the spring's restoring force pulling back on your hand is negative.
  5. Read the calculated value alongside the elastic potential energy U = (1/2) k x^2. Confirm the sign of F is opposite to the sign of x — if it isn't, you have either swapped a sign or pushed the spring past its elastic limit and Hooke's law no longer applies.
  6. Sanity-check the magnitude. For a typical pen-sized lab spring (k around 50 N/m) stretched 10 cm, you should expect a few newtons of force and tens of millijoules of stored energy. If your result is wildly larger or smaller, recheck the units of k (N/m, not N/cm or N/mm).

The formula.

F = −k × x

Hooke's law in its standard form is F = -k*x, where F is the restoring force the spring exerts (in newtons), k is the spring constant (in N/m, always positive), and x is the signed displacement from the spring's natural length (in metres). Rearranged: x = -F/k for the displacement form, and k = |F/x| for the spring-constant form, with the absolute value reflecting that k itself is always positive. The negative sign in F = -k*x is not optional cosmetic notation, it carries the physics — it tells you the force always opposes the displacement, which is what makes a Hookean spring a restoring system rather than a runaway one. The elastic potential energy stored in the spring is U = (1/2) * k * x^2, which is the integral of the magnitude of the restoring force from 0 to x: U = integral from 0 to x of k*x' dx' = (1/2) k x^2. Because x is squared, U is always non-negative; a stretched and a compressed spring at equal |x| store identical energy. From Hooke's law plus Newton's second law, m * (d^2 x / dt^2) = -k*x, you get the differential equation of simple harmonic motion (SHM). Its solution is x(t) = A * cos(omega*t + phi) with angular frequency omega = sqrt(k/m), so the period is T = 2*pi*sqrt(m/k) and the frequency is f = 1/(2*pi) * sqrt(k/m). Notice that the period depends only on the mass and the spring constant, not on the amplitude — that amplitude-independence is the hallmark of a linear (Hookean) oscillator and is why pendulum clocks and watch balance wheels can keep time. For springs combined in series, the effective spring constant is 1/k_eff = 1/k_1 + 1/k_2 + ..., which is always less than the smallest k_i (series springs are softer than the softest individual spring). For springs in parallel — both attached between the same two endpoints, sharing the load — the effective constant adds: k_eff = k_1 + k_2 + ..., so parallel springs are stiffer than any one of them. These rules let you decompose complex elastic systems into networks of equivalent Hookean springs.

A worked example.

Example

Suppose you have a coil spring with stiffness k = 100 N/m clamped vertically to a bench, and you stretch the free end downward by 10 cm so that x = 0.1 m. Plug into Hooke's law: F = -k*x = -(100 N/m) * (0.1 m) = -10 N. The minus sign says the spring pulls up on whatever is holding it at x = 0.1 m with a 10 N force, opposing the stretch. Equivalently, if a 1.02 kg mass is hanging in equilibrium on this spring (since 1.02 kg * 9.81 m/s^2 ~ 10 N), the spring's restoring pull of 10 N exactly balances gravity. The elastic potential energy stored in the stretched spring is U = (1/2) * k * x^2 = 0.5 * 100 * (0.1)^2 = 0.5 J. That half-joule is the work you did against the spring while stretching it, and it is exactly what you would recover as kinetic energy if you let the mass slingshot back through x = 0 unimpeded. If you doubled the displacement to x = 0.2 m, the force would double to -20 N (linear in x) but the stored energy would quadruple to 2.0 J (quadratic in x) — a tidy reminder that energy grows much faster than force as you stretch a spring further.

spring Constant100
displacement0.1
solve Forforce

Frequently asked questions.

Why is there a negative sign in F = -kx?
The minus sign encodes the fact that a Hookean spring always exerts a restoring force — its force vector points opposite to the displacement vector. If you pull the spring to the right (positive x), it pulls back to the left (negative F); if you compress it to the left (negative x), it pushes back to the right (positive F). Without the minus sign you'd describe a runaway anti-spring that accelerates itself away from equilibrium, which is unphysical for an ordinary spring. Some textbooks write F = kx where F is interpreted as the applied force (the force you apply to the spring) rather than the restoring force the spring exerts on you. Both forms are correct so long as you are clear about which force you mean — this calculator uses the restoring-force convention F = -kx from Halliday, Resnick & Walker.
What exactly is the spring constant k, and what are typical values?
The spring constant k is the stiffness of the spring, measured in newtons per metre (N/m). It tells you how many newtons of restoring force the spring exerts for every metre of displacement. A k of 100 N/m means a 1 N force stretches the spring by 1 cm. Typical values: a soft lab demonstration spring is 5–50 N/m; a stiff coil spring in a ballpoint pen is 100–500 N/m; the suspension spring at each corner of a car is around 20,000–40,000 N/m; an atomic-force microscope cantilever is around 0.01–40 N/m; a steel guitar string under tension behaves as an effective spring with k in the thousands of N/m. The spring constant depends on the material's Young's modulus, the wire diameter, the coil diameter, and the number of active coils — for a helical spring, k = G*d^4 / (8*D^3*n), where G is the shear modulus, d is the wire diameter, D is the coil diameter, and n is the number of active coils.
How do I find the equivalent spring constant of springs in series and in parallel?
Springs in series share the same force but their displacements add. The reciprocal of the equivalent spring constant equals the sum of the reciprocals: 1/k_eq = 1/k_1 + 1/k_2 + ... For two identical springs of stiffness k in series, k_eq = k/2 — the combination is half as stiff as either spring alone. Springs in parallel share the same displacement but their forces add. The equivalent spring constant is simply the sum: k_eq = k_1 + k_2 + ... For two identical k springs in parallel, k_eq = 2k — the combination is twice as stiff. A useful mnemonic: series springs behave like capacitors (reciprocals add), parallel springs behave like resistors (values add). Note that 'series' and 'parallel' here are mechanical, not electrical: two springs are in series if you can trace a single force path through both end-to-end, in parallel if both anchor to the same two endpoints and share the load.
What happens beyond the proportional or elastic limit?
Hooke's law applies only inside the proportional limit of the material — the region where the force-vs-displacement graph is a straight line. Push the spring further and you enter the elastic regime where the spring still returns to its original shape on unloading but the F-x curve bends (typically softening). Push further still and you cross the elastic limit (also called the yield point), beyond which the spring undergoes plastic deformation — it acquires a permanent set and never returns to its original length. Beyond that lies the ultimate strength and eventual fracture. ASTM standards like ASTM E8 (tension testing of metallic materials) prescribe how to measure these limits in the lab. For practical spring design, engineers stay well inside the proportional limit, typically using only 30–50% of the yield strength to allow for fatigue and dynamic loads. If you find that doubling the displacement of a real spring does not double the force, you have likely exited the Hookean regime and need a non-linear constitutive model (or a new spring).
Why is the elastic potential energy always positive even when the spring is compressed?
Elastic potential energy is U = (1/2) k x^2, and because x is squared, U is always non-negative regardless of whether the spring is stretched (positive x) or compressed (negative x). Physically this makes sense: in both cases you have done positive work against the spring's restoring force to deform it from equilibrium, and that work is stored as energy you can later recover. A spring compressed by 5 cm stores exactly the same energy as the same spring stretched by 5 cm. The minimum of U(x) is at x = 0, the spring's natural length, which is the stable equilibrium point — any displacement in either direction increases the stored energy, which is precisely what makes that point stable in the sense of Lagrangian mechanics.
What is the period of a mass oscillating on a Hookean spring?
Combining Hooke's law with Newton's second law gives m * d^2x/dt^2 = -k*x, a differential equation whose solution is sinusoidal motion: x(t) = A * cos(omega*t + phi) with angular frequency omega = sqrt(k/m). The period — the time for one full oscillation — is T = 2*pi*sqrt(m/k), and the frequency is f = 1/T = (1/(2*pi)) * sqrt(k/m). Crucially, the period depends only on the mass and the spring constant, not on the amplitude of the motion. That amplitude-independence (called isochronism) is what makes Hookean springs the basis of mechanical clocks: a watch balance wheel oscillating on a hairspring keeps the same period regardless of how vigorously it is wound. A 1 kg mass on a 100 N/m spring has period T = 2*pi*sqrt(1/100) ~ 0.628 s, or about 1.59 Hz.
Does Hooke's law apply to rubber bands, bungee cords, or stretchy fabrics?
Only loosely, and only over a small range. Rubber and similar elastomers are markedly non-linear: their F-x curves typically start out steep (initial stiffness), flatten out in a plateau region, then steepen again as the polymer chains fully extend, an S-shaped response sometimes modelled with the Mooney–Rivlin or Ogden hyperelastic equations rather than Hooke's law. They also exhibit hysteresis (the unloading curve differs from the loading curve, dissipating energy as heat) and Mullins-effect softening on repeated cycling. For a quick engineering estimate over a narrow displacement range you can fit a local k to a rubber band and treat it as Hookean, but for accurate work — bungee jumping safety calculations, tyre modelling, soft-tissue biomechanics — you need a non-linear constitutive law. Metal coil springs, by contrast, are very nearly perfectly Hookean well below their yield point and are the canonical example for which F = -kx really does hold.
How is an atomic-force microscope cantilever related to Hooke's law?
An atomic-force microscope (AFM) uses a tiny silicon or silicon-nitride cantilever, typically 100 to a few hundred micrometres long, with a sharp tip at the free end. The cantilever behaves as a Hookean spring with k typically in the range 0.01 to 40 N/m. When the tip is brought near a sample, inter-atomic forces (van der Waals, electrostatic, chemical) deflect the cantilever; a laser bounced off the cantilever's back onto a position-sensitive photodiode measures the deflection with sub-angstrom resolution. Multiplying the measured deflection by the calibrated spring constant gives the force directly via F = -k*x — that is how AFM measures forces in the piconewton to nanonewton range. Spring-constant calibration (using thermal noise, added-mass, or reference-cantilever methods) is therefore one of the most carefully controlled steps in modern AFM metrology, because every reported force depends linearly on it.
Why does every smooth potential well look Hookean for small displacements?
Take any well-behaved potential energy function U(x) that has a stable minimum at x = 0. Taylor-expand it: U(x) = U(0) + U'(0)*x + (1/2)*U''(0)*x^2 + higher-order terms. Because x = 0 is a minimum, U'(0) = 0 (no first-order term) and U''(0) > 0. Dropping the constant U(0) (which doesn't affect forces) and ignoring the cubic-and-higher terms for small x, you get U(x) ~ (1/2)*U''(0)*x^2, which is exactly the Hookean form (1/2)*k*x^2 with k = U''(0). The corresponding force F = -dU/dx = -U''(0)*x = -k*x is Hooke's law. This is why the simple harmonic oscillator is the universal small-amplitude model in physics — from molecular bond vibrations and crystal lattice phonons to gravitational-wave detector test masses, every stable oscillator behaves Hookeanly close to its equilibrium. The Feynman Lectures, Volume I, Chapter 21, develops this argument in detail and shows how it underpins almost all of vibrational physics.

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