Hooke's Law states that the restoring force of a spring is proportional to its displacement: F = k × x. F is force in Newtons, k is the spring constant in N/m, and x is displacement in meters. Robert Hooke published this in 1678. It applies within the elastic limit — once you exceed that, the spring permanently deforms and the linear relationship breaks down.

What is the Hooke's Law formula?

F = k × x. Rearranged to find spring constant: k = F ÷ x. Rearranged to find displacement: x = F ÷ k. Elastic potential energy stored: E = 0.5 × k × x² in joules. Spring-mass oscillation period: T = 2π × √(m/k) seconds.

How do you calculate the spring constant k?

k = F ÷ x. Apply a known force in Newtons and measure the displacement in meters. Example: hanging a 5 kg mass (F = 49.05 N) causes 5 cm (0.05 m) extension. k = 49.05 ÷ 0.05 = 981 N/m. Repeat at several loads and average the results for accuracy.

What is elastic potential energy in a spring?

E = 0.5 × k × x² in joules. For k = 1000 N/m and x = 0.05 m: E = 0.5 × 1000 × 0.0025 = 1.25 J. Doubling displacement quadruples stored energy because x is squared. When released, this energy converts entirely to kinetic energy (in the absence of friction).

What is the elastic limit in Hooke's Law?

The elastic limit is the maximum displacement at which a spring returns to its natural length after the force is removed. Beyond this point, permanent plastic deformation occurs and F = kx no longer holds. Engineering springs are typically designed to operate at 60–70% of yield stress to ensure reliable performance over millions of cycles.

How do springs in series and parallel combine?

Springs in series (end to end): 1/k_total = 1/k1 + 1/k2. Result is softer — extends more for the same force. Springs in parallel (side by side): k_total = k1 + k2. Result is stiffer. Two identical springs in series give k/2; in parallel give 2k.

What is Hooke's Law used for in engineering?

Vehicle suspension springs, shock absorbers, engine valve springs, clock mechanisms, weighing scales, force gauges, seismic isolation bearings, atomic force microscope tips, and any elastic element that stores and releases energy predictably. Structural deflection calculations also follow Hooke's Law at the material level through Young's modulus.

What is the relationship between Hooke's Law and Young's modulus?

Young's modulus E is the material-level spring constant: stress = E × strain (σ = E × ε). For a rod of length L, cross-section A, and modulus E, the spring constant is k = E × A ÷ L. Hooke's Law for springs (F = kx) and for elastic materials (σ = Eε) describe the same linear elastic behaviour at different scales.

What is the period of oscillation for a spring-mass system?

T = 2π × √(m/k). Frequency f = (1/2π) × √(k/m). For k = 1000 N/m and m = 1 kg: T = 2π × √(0.001) = 0.199 s, f = 5.03 Hz. Stiffer springs oscillate faster; heavier masses oscillate slower.

What is the difference between compression and tension springs?

Compression springs have open coils and resist being compressed — they push back when shortened. Tension springs have closed coils with hooks and resist being stretched — they pull back when extended. Both follow F = kx within their elastic range. Torsion springs resist rotation: torque = k_torsion × angle in radians.

How do I convert spring constant units?

1 N/mm = 1000 N/m. 1 lbf/in = 175.127 N/m. 1 kgf/mm = 9806.65 N/m. Engineering springs are commonly rated in N/mm for automotive and industrial use. Physics problems typically use N/m. This calculator converts all inputs to SI base units internally.

What are typical spring constant values in real applications?

Watch spring: 10–50 N/m. Pen click: 100–500 N/m. Kitchen scale: 1,000–5,000 N/m. Trampoline spring: 5,000–10,000 N/m. Car suspension: 15,000–30,000 N/m. Engine valve spring: 20,000–50,000 N/m. Seismic isolator: 1,000,000+ N/m. The range spans six orders of magnitude from watchmaking to structural engineering.

Hooke's Law Calculator — Spring Force F=kx

Spring Constant (k)

N/m

Enter a valid spring constant greater than zero.

k Unit

Displacement (x)

Enter a valid displacement greater than zero.

x Unit

Common Examples

💡 F = k × x
F in Newtons, k in N/m, x in meters.
F is the restoring force opposing displacement.

Applied Force (F)

Enter a valid force greater than zero.

Force Unit

Displacement (x)

Enter a valid displacement greater than zero.

x Unit

💡 k = F / x
Measure with a known weight (F = mass × 9.81) and a ruler. Average multiple measurements for accuracy.

Applied Force (F)

Enter a valid force greater than zero.

Force Unit

Spring Constant (k)

N/m

Enter a valid spring constant greater than zero.

k Unit

💡 x = F / k
Displacement is in the direction of the applied force. Both extension and compression use the same formula.

Spring Constant (k)

N/m

Enter a valid spring constant greater than zero.

k Unit

Displacement (x)

Enter a valid displacement greater than zero.

x Unit

💡 E = ½ × k × x²
Doubling displacement quadruples stored energy. Doubling k doubles it.

Spring Force

--

Sources & Methodology

✓Hooke’s Law formulas verified against Engineering Toolbox spring stiffness data and Shigley’s Mechanical Engineering Design. Unit conversions verified against NIST standard reference data.

📚

Engineering Toolbox — Hooke’s Law and Spring Stiffness

Spring constant calculation, elastic potential energy, series and parallel spring combinations, and unit conversions including N/m, N/mm, and lbf/in.

🔬

Engineering Toolbox — Typical Spring Constants

Reference values for spring constants across common applications from precision instruments (10–50 N/m) to automotive suspension (15,000–30,000 N/m) and seismic isolators (1,000,000+ N/m).

F = k × x k = F / x x = F / k E = ½kx²

Units: F in Newtons (N), k in N/m, x in meters (m), E in Joules (J)
Unit conversions: 1 N/mm = 1000 N/m · 1 lbf/in = 175.127 N/m · 1 lbf = 4.44822 N
Oscillation period: T = 2π√(m/k) · Frequency: f = (1/2π)√(k/m)

Last reviewed: April 2026

Hooke’s Law Calculator — F=kx Calculator, Spring Force & Force Constant Calculator

This is a Hooke’s Law calculator, spring force calculator, force constant calculator, and elastic force calculator in one tool. Whether you need to find spring force using F = kx, find the spring constant k from a measured force and extension, calculate extension in spring from a known load, or compute elastic potential energy — all four are here. It also works as a force vs extension calculator for plotting spring behavior across a range of displacements, and as a Hooke's Law formula calculator for any F = kx calculator problem you encounter in physics or engineering.

What is Hooke’s Law? What Does It Mean?

What is Hooke’s Law? It’s a statement about how springs and elastic materials behave: the restoring force they produce is directly proportional to how much you’ve stretched or compressed them. Double the extension, double the force. Halve the compression, halve the force. That linear relationship is the entire law — elegant in its simplicity.

What does Hooke's Law mean physically? It means a spring is a linear elastic system. Unlike, say, a rubber band (which gets progressively stiffer as you stretch it), an ideal spring produces a perfectly proportional force for any displacement within its elastic range. Plot force against extension and you get a straight line. The slope of that line is k — the spring constant, which defines the linear force displacement relationship, also called spring stiffness or force constant.

Robert Hooke discovered this in 1678 and concealed it as a Latin anagram — “ceiiinosssttuv” — which decodes as “ut tensio sic vis”: “as the extension, so the force.” He was protecting his priority before he could publish. Three hundred and forty-seven years later, every physics student knows his name because of it.

Hooke’s Law Formula — F = kx Explained

The Hooke’s Law formula is F = kx. This is the F = kx formula in its standard form. Here’s what each variable means and the unit of spring constant:

F = k × x [Spring Force Calculator] k = F ÷ x [Spring Constant Calculator — unit: N/m] x = F ÷ k [Spring Stretch / Extension Calculator — unit: m] E = ½ × k × x² [Elastic Potential Energy — unit: J]

F = spring force (Newtons, N) — the restoring elastic force. Force = spring constant × extension: this is the complete statement.
k = spring constant / force constant / spring stiffness (N/m) — what is k in Hooke's Law? k is the proportionality constant that defines spring stiffness. The slope of the force-extension graph. A stiffer spring has a higher k. Force = spring constant (k) times extension (x).
x = extension or compression (meters, m) — displacement from natural length
E = elastic potential energy stored in the deformed spring (Joules, J)

How to Calculate Spring Force — Hooke’s Law Solved Examples

The easiest way to understand how to calculate spring force, how to find spring constant k, and how to calculate extension in spring is through concrete worked problems. Below are the four most common exam and engineering scenarios — the same types you’ll find in physics exam spring force questions and numerical problems on Hooke’s Law.

How to Calculate Force Using Hooke's Law — F = kx Calculator Step by Step

Problem: A spring with k = 1500 N/m is stretched 10 cm. What is the spring force? This answers the classic question: how much force to stretch a spring by a given amount?

How to solve Hooke’s Law step by step: Convert x to meters first (10 cm = 0.10 m). Apply F = k × x. F = 1500 × 0.10 = 150 N. This is the elastic force the spring exerts, opposing the stretch. To calculate force of a spring stretched 10 cm with any k, just multiply k (in N/m) by 0.10.

Spring Constant Formula with Example — How to Find k

Problem: A spring stretches 8 cm when a 400 g mass hangs from it. What is the spring constant?

Step 1: Find the force. Weight = mass × g = 0.4 × 9.81 = 3.924 N.
Step 2: Convert extension: 8 cm = 0.08 m.
Step 3: Apply k = F / x = 3.924 / 0.08 = 49.05 N/m.

This is a classic spring constant physics problem. The key insight: always convert grams to kg first, then multiply by 9.81 to get Newtons. The unit of spring constant is always N/m in SI units (or N/mm in engineering).

Hooke's Law Solved Examples — Spring Constant Physics Problems & Spring Force Examples

Problem Type	Given	Find	Formula	Answer
Spring force example	k = 800 N/m, x = 5 cm	Force F	F = kx	40 N
Force vs extension example	k = 2000 N/m, x = 3 cm	Force F	F = kx	60 N
Spring constant problem	F = 50 N, x = 5 cm	k	k = F/x	1000 N/m
Extension calculation	F = 120 N, k = 3000 N/m	x	x = F/k	4 cm
Spring compression force	k = 5000 N/m, x = 2 cm	F (compression)	F = kx	100 N
Elastic energy	k = 1000 N/m, x = 5 cm	E	E = ½kx²	1.25 J

💡 Easy way to understand Hooke's Law for exams — how to memorize Hooke's Law formula: Think of k as “stiffness” — how hard the spring fights back per metre of stretch. A k of 1000 N/m means the spring pushes back with 1000 N for every 1 metre stretched. At 10 cm (0.1 m), it pushes back with 100 N. At 5 cm (0.05 m), 50 N. The force-displacement relationship is always a straight line through zero. That’s the quick mental model for how to memorize Hooke’s Law formula: stiffer spring, steeper line, higher k.

What is Spring Constant? Spring Stiffness, Real-Life Examples & Applications

What is the spring constant? The spring constant k (also called the force constant or spring stiffness) is a measure of how stiff a spring is. It tells you how much force is needed to stretch or compress the spring by one unit of length. A higher k means a stiffer spring — more force required for the same extension. The unit of spring constant is N/m (newtons per metre) in SI units, or N/mm in engineering applications.

Real Life Examples of Hooke's Law — Elastic Force Real Life Applications

Hooke's Law governs elastic force in real life across an enormous range of applications — from the spring in a ballpoint pen to the suspension system of a car. Every time an elastic system is designed to return to its original shape after deformation, F = kx is the underlying model.

Real-Life Application	k (N/m)	How Hooke’s Law Applies
Bathroom scale spring	1,000–5,000	Your weight compresses spring; display reads force via calibrated extension
Pen click mechanism	100–500	Spring compression force clicks cartridge in/out; snap feel is spring behavior
Trampoline mat spring	5,000–10,000	Each coil spring stores elastic potential energy on impact, releases on rebound
Car suspension spring	15,000–30,000	Force vs extension relationship absorbs road bumps; spring stiffness tuned to vehicle weight
Watch mainspring	10–50	Wound spring releases elastic energy slowly via escapement mechanism
Archery bow limb	500–2,000	Mechanical elasticity stores energy on draw; elastic system calculation determines arrow speed
Building seismic isolator	1,000,000+	High-stiffness spring behavior isolates structure from ground deformation physics

Spring Stiffness — How k Affects Spring Behavior Physics

Spring stiffness (k) determines the entire force-displacement relationship of an elastic system. This is the core of mechanical elasticity: materials and springs that behave linearly within their elastic range. High k = stiff spring = large force for small displacement. Low k = soft spring = small force for large displacement. In mechanical elasticity design, engineers choose k based on the loads and deflections required. A car suspension spring needs high enough k to support the vehicle weight without bottoming out, but low enough to absorb road irregularities — that balance is the core of spring behavior physics.

Elastic vs Plastic Deformation — When Hooke’s Law Fails

Elastic deformation is temporary — remove the force and the material returns to its original shape. This is the regime where Hooke’s Law applies and F = kx holds. Plastic deformation is permanent — the material yields and doesn’t recover. When does Hooke’s Law fail? At the proportional limit — the point beyond which the force-extension graph stops being linear. Just past that is the elastic limit (last point of full recovery), and beyond that is the yield point where plastic deformation begins.

Limitations of Hooke's Law — When Does It Fail?

1. Proportional limit: F = kx holds only up to this point. Beyond it, the graph curves — linear elasticity is lost.

2. Elastic limit: Beyond this, permanent deformation occurs even if the relationship was still approximately linear.

3. Yield point: The spring (or material) begins permanent plastic deformation. Hooke’s Law completely breaks down here.

4. Temperature dependence: k changes with temperature — springs are stiffer at low temperatures.

5. Non-ideal materials: Rubber, foam, and biological tissue don’t follow linear Hooke’s Law even in their “elastic” range. They need non-linear elasticity models.

Elastic Potential Energy — The x² Relationship

The elastic potential energy formula E = ½kx² captures something important that the force formula alone doesn’t: energy scales with the square of displacement. Double the stretch, quadruple the stored energy. This is why springs are excellent energy storage devices — a little extra compression at high deformation stores dramatically more energy. It’s the same principle behind why car crumple zones and spring-powered mechanisms are so effective at absorbing impact.

Spring k = 1000 N/m	Extension x	Force F	Elastic Energy E
2 cm	0.02 m	20 N	0.20 J
4 cm	0.04 m	40 N	0.80 J (4×)
6 cm	0.06 m	60 N	1.80 J (9×)
10 cm	0.10 m	100 N	5.00 J (25×)

Note: Force (F) scales linearly with x. Elastic energy (E) scales with x² — doubling extension doubles force but quadruples stored energy.

Frequently Asked Questions

Hooke’s Law formula is F = kx, where F is the spring force in Newtons, k is the spring constant (force constant) in N/m, and x is the extension or compression in metres. Rearranged: k = F/x to find spring stiffness; x = F/k to calculate extension in spring. Elastic potential energy: E = ½kx² in Joules. This is the complete Hooke’s Law formula calculator set — all four versions are in the tabs above.

To calculate spring force using Hooke’s Law: multiply the spring constant k (N/m) by the displacement x (m). F = k × x. Example: k = 2000 N/m, spring stretched 10 cm (0.10 m). Spring force = 2000 × 0.10 = 200 N. For spring compression force, use the same formula — direction reverses but magnitude calculation is identical. Enter your values in the “Find Force” tab above for instant results.

What is spring constant k? It’s the stiffness of a spring — how many Newtons of force the spring exerts per metre of deformation. A higher k means a stiffer spring. The unit of spring constant is N/m (Newtons per metre) in SI units. Engineering catalogs often use N/mm (where 1 N/mm = 1000 N/m) or lbf/in (imperial). Soft springs (k = 10–100 N/m) are used in precision instruments; stiff springs (k = 10,000–50,000 N/m) in automotive suspension and industrial machinery.

How does Hooke’s Law work: hang a spring vertically, add known weights, measure extension each time. Plot force (y-axis) vs extension (x-axis). The slope of the straight line = spring constant k. How to find spring constant k from experiment: k = F/x. Example: 200g mass (1.962 N) causes 3 cm (0.03 m) extension: k = 1.962/0.03 = 65.4 N/m. Take 5–6 readings and average for accuracy.

Limitations of Hooke’s Law: it only applies within the proportional limit. Beyond this, the force-extension graph curves and F ≠ kx. Beyond the elastic limit, the spring permanently deforms and won’t return to its original length. Beyond the yield point, plastic deformation begins. Practically: overstretch a spring (past its elastic limit) and it stays stretched. Hooke’s Law also fails for non-linear materials like rubber, biological tissue, and foam even in their elastic range.

Elastic deformation is reversible — remove the force and the spring or material returns to its original shape. Hooke’s Law applies here. Plastic deformation is permanent — the material yields and doesn’t recover. The boundary between elastic and plastic behaviour is the elastic limit (also called the yield point for engineering materials). Engineering springs are designed to operate at 60–70% of yield stress to stay safely in the elastic zone over millions of cycles.

Elastic force (or spring restoring force) is the force a deformed elastic object exerts to return to its natural length. It’s always directed opposite to the displacement — a stretched spring pulls back; a compressed spring pushes back. The applied force stretches or compresses the spring; the elastic force is the spring’s reaction. At equilibrium (no acceleration), elastic force equals applied force in magnitude. F = kx gives the magnitude of this elastic restoring force.

Rearrange F = kx to get x = F/k. Extension = Force ÷ Spring Constant. Example: force 75 N, spring constant 2500 N/m. Extension = 75/2500 = 0.03 m = 3 cm. This is the spring stretch calculator version of Hooke’s Law. Use the “Find Displacement” tab above — it also converts the result to mm, cm, and inches simultaneously.

Stress = force ÷ cross-sectional area (Pa = N/m²). Strain = extension ÷ original length (dimensionless). Hooke’s Law at the material level: stress = E × strain, where E is Young’s modulus. For a spring: F = kx is the macroscopic version; σ = Eε is the material-level version. Both express the same linear force-displacement relationship — one for the whole spring, one for the material it’s made from.

Real life examples of Hooke’s Law: bathroom scales (spring extension reads weight), car suspension (spring stiffness absorbs road bumps), ballpoint pens (click mechanism), trampoline coils (store and release elastic energy), archery bows (limb elasticity), atomic force microscopes (cantilever spring measures nano-scale forces), seismic building isolators, and any weighing or force-measuring device. Essentially anywhere an elastic element stores and releases mechanical energy predictably, Hooke’s Law is the governing equation.

How to solve Hooke’s Law questions fast: (1) Identify what’s given and what to find. (2) Convert all units to SI: cm to m, grams to kg, then multiply by 9.81 for Newtons. (3) Apply the correct rearrangement of F = kx. (4) Check the answer against the reference table — spring force for a stretched 10 cm spring with k = 1000 N/m should be 100 N. (5) For elastic energy problems, don’t forget the ½ in E = ½kx² and square x in metres, not cm. Unit conversion is the most common source of errors in physics exam spring force questions.

Linear elasticity means force and displacement are proportional — the force-displacement relationship is a straight line through the origin. The gradient (slope) of that line is the spring constant k. Linear force model: F = kx. This is a linear force model because doubling x doubles F. Above the proportional limit, the relationship becomes non-linear — this is exactly when does Hooke's Law fail. The deformation physics changes as the material structure begins to yield. Hooke’s Law is the fundamental expression of linear elasticity in both springs and elastic materials.

Springs in series (end to end, same force): 1/k_total = 1/k1 + 1/k2. Result is softer than either alone. Two equal springs: k_total = k/2. Springs in parallel (side by side, same displacement): k_total = k1 + k2. Result is stiffer. Two equal springs: k_total = 2k. Car suspension: both springs on an axle act in parallel for a central load, giving double the stiffness of one spring alone. This is an elastic system calculation and elastic force real life application used in every multi-spring engineering design.

Related Calculators