Skip to main content
Calculators/Hooke's Law
Classical Mechanics

Hooke's Law Calculator

Calculate spring force, spring constant or extension using F = kx. Also computes spring potential energy PE = ½kx². Live spring animation updates with your result.

Solve for

F = k × x
PE = ½kx²
Spring examples
Spring visualisation
Enter positive values to calculate.

Hooke's Law equations

Hooke's Law
F = kx
Spring constant
k = F/x
Extension
x = F/k
Spring PE
PE = ½kx²
PE from force
PE = F²/(2k)
Force from PE
F = √(2kPE)

Hooke's Law

Hooke's Law states that the force exerted by a spring is proportional to its extension from its natural length: F = kx. The spring constant k (also called the stiffness) has units of N/m — a stiffer spring has a higher k. The negative sign in the formal expression (F = −kx) indicates the restoring force acts opposite to the extension direction.

The elastic potential energy stored in a spring is PE = ½kx². This energy is released when the spring returns to its natural length. Hooke's Law holds until the elastic limit — beyond that, the spring deforms permanently.

Hooke's Law — The Complete Physics Guide

Hooke's Law states that the extension of an elastic material is proportional to the applied force, provided the elastic limit is not exceeded: F = kx. Published by Robert Hooke in 1676, it is the foundation of all elasticity theory and applies far beyond springs — to rubber bands, steel beams, biological tissues, and even atoms vibrating in a crystal lattice. Any system that returns to equilibrium after small displacements follows the same mathematics.

Hooke encoded his discovery as a Latin anagram "ceiiinosssttuv" before publishing the solution "ut tensio, sic vis" — as the extension, so the force. This characteristically secretive approach was common in the 17th century, when natural philosophers wished to establish priority without revealing results prematurely. The anagram predates the 1678 publication by six years.

Understanding F = kx

The spring constant k (measured in newtons per metre, N/m) quantifies stiffness — how much force is required per unit of extension. A higher k means a stiffer spring. Typical values: a very soft spring (child's toy) ~1 N/m; soft mattress spring ~100 N/m; car suspension spring ~15,000–25,000 N/m; industrial press spring ~500,000 N/m; automotive valve spring ~50,000 N/m.

The formal expression includes a minus sign: F = −kx, where F is the restoring force (the force the spring exerts on whatever stretched it) and x is the displacement from equilibrium. The negative sign indicates the force is always directed opposite to the displacement — a restoring force, always pulling the spring back toward its natural length. When we calculate F = kx in problems, we typically work with magnitudes and apply the direction physically.

On a force-extension graph, Hooke's Law predicts a straight line through the origin with gradient equal to k. The graph is only linear up to the elastic limit — beyond this, the graph curves (the Hookean region ends) and eventually the material fractures. The area under the F-x graph up to any extension x equals the elastic potential energy stored: EPE = ½kx² — the area of a right triangle with base x and height kx.

Elastic Potential Energy

The work done in stretching a spring from 0 to extension x is: W = ∫₀ˣ F dx' = ∫₀ˣ kx' dx' = ½kx². This work is stored as elastic potential energy (EPE), ready to be released as kinetic energy when the spring is allowed to contract. The formula EPE = ½kx² is completely analogous to KE = ½mv² — both are quadratic in the "displacement" quantity (x or v) and both involve a characteristic constant (k or m).

When a compressed spring launches a mass m from extension x₀ on a frictionless surface, conservation of energy gives: ½kx₀² = ½mv² → v = x₀√(k/m). This is the operating principle of spring launchers, pneumatic nail guns, trebuchets, and many mechanical timing devices.

EPE can also be expressed in terms of force: since F = kx, we have x = F/k, and EPE = ½kx² = ½k(F/k)² = F²/(2k). This form is useful when the force is known but the extension is not — for example, calculating the energy stored in a car suspension spring when you know the weight it supports.

Worked Example 1 — Spring Extension and Energy

Problem: A spring with k = 350 N/m is extended by 12 cm. Find (a) the restoring force, (b) the elastic potential energy stored, and (c) the speed a 200 g mass would reach if launched from this extension on a frictionless surface.

(a) Force: F = kx = 350 × 0.12 = 42 N

(b) EPE: ½kx² = ½ × 350 × 0.12² = 175 × 0.0144 = 2.52 J

(c) Speed: ½mv² = EPE → v = √(2 × 2.52/0.2) = √25.2 = 5.02 m/s

Worked Example 2 — Finding k Experimentally

Problem: A 250 g mass is attached to a spring and produces an extension of 4.0 cm. A 500 g mass produces an extension of 8.1 cm. Calculate k from each measurement and comment on the results.

F₁ = 0.25 × 9.81 = 2.453 N → k₁ = 2.453/0.040 = 61.3 N/m

F₂ = 0.5 × 9.81 = 4.905 N → k₂ = 4.905/0.081 = 60.6 N/m

Both measurements give k ≈ 61 N/m — consistent with Hooke's Law (proportional extension). The small difference (1.1%) reflects experimental uncertainty in measuring extension. The best estimate would use the gradient of an F vs x graph through several data points.

Springs in Series and Parallel

Springs in series (end to end): The same force F acts on each spring; total extension is the sum. 1/k_series = 1/k₁ + 1/k₂ + ... The combined stiffness is less than either spring alone. This is the spring analogue of resistors in series — more compliance (less stiffness) in series.

Springs in parallel (side by side): Each spring supports part of the load; same extension for each. k_parallel = k₁ + k₂ + ... The combined stiffness is greater than either spring alone. This is the spring analogue of resistors in parallel — more stiffness in parallel.

Real engineering applications: a vehicle suspension uses springs in a combination of series and parallel arrangements to tune ride quality and stiffness. Anti-roll bars act as a parallel spring addition. Rubber bump-stops add series stiffness at extreme compression. Understanding these combinations is fundamental to mechanical engineering design.

Hooke's Law and Simple Harmonic Motion

Hooke's Law is the defining condition for simple harmonic motion (SHM). Any system where the restoring force is proportional to displacement — F = −kx — will oscillate with period T = 2π√(m/k). The spring constant k plays the role of "restoring stiffness"; the mass m plays the role of "inertia."

This SHM connection extends far beyond mechanical springs. A pendulum (for small angles) obeys F = −(mg/L)x, giving effective k = mg/L and period T = 2π√(L/g). An LC electrical circuit obeys the same mathematics with inductance L playing the role of mass and 1/C (reciprocal capacitance) playing the role of spring constant. Molecular bonds vibrate as quantum harmonic oscillators with the same F = −kx model.

The universality of Hooke's Law — the fact that essentially all small oscillations around a stable equilibrium obey it — makes it one of the most widely applicable results in all of physics. Taylor expanding any potential energy well around its minimum: U(x) ≈ U(0) + ½U''(0)x². The second derivative U''(0) plays the role of spring constant. Every stable equilibrium in nature is a local minimum, and every local minimum behaves like a Hookean spring for small displacements.

Young's Modulus — Hooke's Law for Continuous Materials

For a bar of material with cross-sectional area A, length L, and Young's modulus E, the spring constant is k = EA/L. Young's modulus E (Pa = N/m²) is a material property, independent of shape: steel E ≈ 200 GPa, aluminium E ≈ 70 GPa, bone E ≈ 20 GPa, rubber E ≈ 0.01–0.1 GPa.

The same Hooke's Law physics governs the deformation of every structural element — beams, columns, bolts, cables. Structural engineering calculations always begin with the elastic regime where Hooke's Law applies. Safety factors ensure that materials never approach the elastic limit under design loads, keeping structures in the linear Hookean regime throughout their service life.

Frequently Asked Questions

What is Hooke's Law?
Hooke's Law states that the force exerted by a spring is proportional to its extension from natural length: F = kx. The spring constant k (N/m) measures stiffness. The law holds only within the elastic limit — beyond this, permanent deformation occurs. It applies to any elastic material, not just springs.
What is the elastic limit?
The elastic limit is the maximum extension at which the material fully recovers when the force is removed. Below it: elastic (reversible) deformation, Hooke's Law holds, F-x graph is linear. Above it: plastic (permanent) deformation, the material is stretched beyond recovery. Well-designed structures always operate well below the elastic limit.
How is the spring constant measured?
Apply known forces and measure extensions. Plot F versus x — the gradient of the straight-line region is k. Alternatively, measure the oscillation period T of a known mass m on the spring: k = 4π²m/T². This is a more precise method because period can be timed accurately.
What is elastic potential energy?
EPE = ½kx² is the energy stored in a spring at extension x. It equals the triangular area under the F-x graph. When released, EPE converts to kinetic energy. It can also be written as F²/(2k) when force is known, or ½Fx when both F and x are known.
Why does everything follow Hooke's Law for small displacements?
Any stable equilibrium is a local minimum of potential energy. Expanding the potential energy in a Taylor series around the minimum: U(x) ≈ ½kx² where k = d²U/dx². This quadratic potential always gives a linear restoring force F = −dU/dx = −kx — Hooke's Law — for small enough displacements. This is why SHM is ubiquitous in nature.

Physics Fundamentals

Channel · Updates only

👋 Get concise physics updates — new articles, calculators, and tools. Your number stays private. No spam. No group chats. Just worthy content.

Your number is never shared or visible to others

Join the Channel