Hooke’s Law Calculator works out spring force, spring constant or displacement from F = −kΔx. Lengths or a hanging mass can be entered, and it adds stored energy and bounce period.
Enter Δx as positive for a stretched spring and negative for a compressed one.
Spring Length, Hanging Mass and Units
Step-by-Step Working
- Known values: k = 80 N/m, Δx = 0.15 m
- Formula: F = −k × Δx
- Substitute: −80 × 0.15
- Result: F = −12 N
Force and Energy at Other Displacements
| Displacement | Δx (m) | Force (N) | Stored energy (J) |
|---|---|---|---|
| ¼ Δx | 0.0375 | −3 | 0.05625 |
| ½ Δx | 0.075 | −6 | 0.225 |
| Δx (yours) | 0.15 | −12 | 0.9 |
| 1½ Δx | 0.225 | −18 | 2.025 |
| 2 Δx | 0.3 | −24 | 3.6 |
| 3 Δx | 0.45 | −36 | 8.1 |
Three Ways to Use the Hooke’s Law Calculator
The Hooke’s Law Calculator solves the spring equation for whichever value is missing. Solve For switches between spring force, spring constant and displacement, and the other two become the inputs.
It opens with a spring constant of 80 N/m and a displacement of 0.15 m. The spring pulls back with a restoring force of −12 N. The result cards add 0.9 J of stored energy and a bounce period of 0.7771 s.
Lab data rarely arrives as a neat Δx. The accordion accepts the spring’s initial and final length instead, or a hanging mass in place of a force. The mass becomes a weight through standard gravity, 9.80665 m/s², which is about 32.17 ft/s². The Gravitational Force Calculator shows where that 9.8 figure comes from.
Force Shown As sets the sign convention. Restoring force uses F = −kΔx and reports the spring’s own pull. Applied force uses F = kΔx and reports the push or pull of whatever is deforming the spring. Both have the same size and opposite directions.
The spring law is one of several routes to a force. The Force Calculator covers F = ma and pressure on an area, and the rest of the Dynamics and Statics calculators handle the other forces that act on a body.
Hooke’s Law Formula
Hooke’s law states that, for small deformations, the size of the deformation is directly proportional to the force that causes it. Britannica’s entry on Hooke’s law gives it as F = kx and notes that it is sometimes written F = −kx. For a spring the second form is the usual one.
$$F = -k \Delta x$$
F is the spring force in newtons and k is the spring constant in newtons per meter. Δx is the displacement from the rest length in meters, positive for a stretched spring and negative for a compressed one. Mathematically, Hooke’s law is a straight line through the origin with slope k.
What is the k in F = −kx?
The k is the spring constant, also called stiffness, force constant or spring rate. It is the force needed for each unit of deformation. A spring of 80 N/m needs 80 N to stretch one meter, or 0.8 N for each centimeter. It is always positive, and a larger k means a stiffer spring.
Why is F = −kx negative?
The minus sign says the spring force points opposite to the displacement. Pull the free end to the right and the spring pulls back to the left. Squeeze it and it pushes outward. That is why it is called a restoring force. When only the size matters, drop the sign and use the applied-force form.
Units in the Hooke’s Law Calculator
The SI units for Hooke’s law are newtons for force, meters for displacement and newtons per meter for the spring constant. One N/m is the same as one kg/s². The unit menus also take N/mm, kN/m, lbf/in and lbf/ft for stiffness, with inches, feet, pounds-force and kilograms-force for the rest.
The conversions follow the NIST SP 811 conversion tables. One pound-force is 4.448222 N, one lbf/in is 175.1268 N/m and one foot pound-force is 1.355818 J. Mixed entries, such as a rate in lbf/in with a deflection in millimeters, are converted to SI before the formula runs.
Hooke’s law formula triangle
The Hooke’s law formula triangle puts F at the top with k and Δx side by side underneath. Cover the quantity you want and the other two show the operation. Covering F leaves k times Δx. Covering k or Δx leaves F over the remaining letter, as below.
$$k = \frac{F}{\Delta x}$$
$$\Delta x = \frac{F}{k}$$
The triangle gives sizes only. The calculator adds the signs from the convention you picked, so a positive restoring force pairs with a compressed spring and a negative one with a stretched spring.
Hooke’s Law Calculation Example
To calculate Hooke’s law for spring force, put the spring constant and the displacement in matching units and multiply them. Then attach the sign. The Step-by-Step Working panel writes the lines out, which helps when physics homework asks for the working. For the opening values it reads as follows.
- Known values are k = 80 N/m and Δx = 0.15 m.
- The formula is F = −k × Δx.
- Substituting gives −80 × 0.15.
- The result is F = −12 N, a restoring force of 12 N toward the rest position.
A second example uses US units. A compression spring rated at 150 lbf/in is squeezed by 2 in, so Δx is −2 in. The force is 150 × 2 = 300 lbf, which is 1,334 N. The restoring force comes out positive because the spring pushes back in the positive direction.
The cards fill in the rest for that spring. It stores 33.9 J, exactly 25 ft·lbf, and the fourth card restates the rate as 26,270 N/m. A load of 136.1 kg resting on the spring would hold it at this compression.
The third mode of the Hooke’s Law Calculator finds displacement. A 30 N pull on a 1,200 N/m spring stretches it by 30 ÷ 1,200 = 0.025 m, or 25 mm. Set Force Shown As to Applied force for this kind of question. Left on Restoring force, a +30 N entry returns −0.025 m, because a positive restoring force belongs to a compressed spring.
Spring Constant Calculation Formula
The spring constant comes from one load and the deflection it causes. Divide the force by the displacement and ignore the signs. When the load is a hanging mass, the force is its weight, which gives the form used in most school experiments.
$$k = \frac{m g}{\Delta x}$$
Take a spring that is 0.20 m long and grows to 0.26 m with 0.5 kg hung from it. The displacement is 0.06 m and the weight is 0.5 × 9.80665 = 4.903 N. The spring constant is 4.903 ÷ 0.06 = 81.72 N/m. In the Hooke’s Law Calculator, choose Spring constant, then enter the two lengths and the hanging mass.
Spring makers work the other way and predict the rate from the coil’s dimensions. Acxess Spring’s rate formula for a helical compression spring is the one below, and the maker applies the same formula to extension springs.
$$k = \frac{G d^4}{8 D^3 N}$$
G is the shear modulus of the wire, d is the wire diameter, D is the mean coil diameter and N is the number of active coils. Wire diameter is raised to the fourth power. A wire only 10% thicker makes the spring about 46% stiffer.
Is k constant in Hooke’s law?
Yes, for one spring inside its working range. The value is fixed by the material and the geometry, not by the load. It changes if coils are cut off, and it stops being constant once the spring is pushed past its elastic limit.
Extension springs add one wrinkle. Many are wound with initial tension, and Acxess gives their force as k times extension plus that preload. The calculator assumes no preload, so add the maker’s initial tension to the result for such a spring.
How strong is a spring?
Spring strength is quoted as a rate, normally in lbf/in or N/mm. The fourth card of the Hooke’s Law Calculator converts it for you. The 80 N/m lab spring is only 0.4568 lbf/in, or 5.482 lbf/ft, and gives 12.5 mm for every newton. The 150 lbf/in spring from the example is about 330 times stiffer.
How heavy a load can a spring hold?
Hooke’s law treats the spring itself as weightless, so “how heavy” is a question about the load. The Hanging mass row on the first card answers it. The 12 N force equals the weight of 1.224 kg, found from kΔx ÷ g. A heavier load is fine as long as the deflection stays in the elastic range.
How do you weaken a spring?
Lowering k means changing the terms in the maker’s formula. Thinner wire, a larger coil diameter or more active coils all make a softer spring. Joining springs end to end also works. Two 80 N/m springs in series act as one 40 N/m spring, because each one stretches under the full load.
What are the four types of springs?
Coil spring catalogs such as Acxess Spring’s list four families, which are compression, extension, torsion and conical springs. Compression and extension springs follow F = kΔx directly. A torsion spring obeys the same law in rotation, with torque proportional to the angle of twist.
Hooke’s Law Energy Formula and Calculus
Deforming a spring takes work, and the force is not constant while you do it. It climbs from zero to kΔx. Calculus handles that by integrating the force over the displacement.
$$U = \int_0^{\Delta x} k x \, dx = \frac{1}{2} k \Delta x^2$$
This is the Hooke’s law formula for energy, treated in OpenStax University Physics, section 15.2. On a force against displacement graph it is the area of the triangle under the line, half of base Δx times height kΔx. For the opening values, ½ × 80 × 0.15² = 0.9 J.
The link also runs backward. Differentiate the stored energy with respect to position and the spring force appears as the negative slope of the energy curve, which returns the original law.
$$F = -\frac{dU}{dx} = -k x$$
The second card of the Hooke’s Law Calculator shows the square at work. At twice the displacement the stored energy is 3.6 J, four times as much. The Force and Energy at Other Displacements table carries this from ¼ Δx to 3 Δx. Force runs in a straight line from −3 N to −36 N, while energy climbs from 0.05625 J to 8.1 J.
Bounce Period of a Hanging Load
Put the spring force into Newton’s second law and the result is a differential equation. Its solution is simple harmonic motion, a steady bounce about the rest position. The Hooke’s Law Calculator reports the timing of that bounce on its third card.
$$m \frac{d^2 x}{dt^2} = -k x$$
OpenStax section 15.1 derives the period of that motion. It depends only on the mass and the spring constant, not on how far the load is pulled before release. A stiffer spring shortens the period and a heavier load lengthens it.
$$T = 2 \pi \sqrt{\frac{m}{k}}$$
A load hanging at rest satisfies mg = kΔx, so m ÷ k equals Δx ÷ g. The third card uses that shortcut, which needs only the static stretch and no separate entry for the mass or the spring constant.
$$T = 2 \pi \sqrt{\frac{\Delta x}{g}}$$
A 0.15 m stretch gives 0.7771 s, a frequency of 1.287 Hz and an angular frequency of 8.086 rad/s. The 0.5 kg load that stretched its spring by 0.06 m would bob every 0.4915 s. The card applies only when the force is a weight hanging on the spring.
The same force sets the acceleration at the moment of release, a = kΔx ÷ m. The Acceleration Calculator takes the force from here and the mass, and returns that figure. For the opening spring moving a 0.5 kg mass it is 12 ÷ 0.5 = 24 m/s².
Hooke’s Law in Mechanics of Materials
In mechanics of materials the Hooke’s law formula is written with stress and strain, so that it describes the material and not one particular part. Engineers use this version for rods, wires and beams.
$$\sigma = E \varepsilon$$
Stress σ is force per unit area, strain ε is the change in length divided by the original length, and E is Young’s modulus. Substituting F ÷ A and ΔL ÷ L turns a straight bar into a spring with the constant below.
$$k = \frac{E A}{L}$$
A steel wire 1 m long with a cross-section of 1 mm² and E = 2.0 × 10¹¹ Pa has k = 200,000 N/m. Enter 200 kN/m in the Hooke’s Law Calculator and a 100 N load stretches it by 0.5 mm.
What is Hooke’s law in class 11?
Class 11 physics states Hooke’s law in the stress form. The NCERT textbook says that for small deformations stress and strain are proportional, written as stress = k × strain, where k is the modulus of elasticity. The spring form F = −kx appears later with oscillations. Both are the same law.
When does Hooke’s law apply?
Use Hooke’s law when the deformation is small and the part returns to its original length. A quick test is to double the load. If the deflection doubles too, the law holds. Past the elastic limit the part stays bent, and the straight line no longer describes it.
Some materials never follow it closely. NCERT points to the elastic tissue of the aorta, which stretches a long way without obeying the law over most of its range. A car body in a collision is far past the limit as well, so the Car Crash Calculator works from stopping distance and not from a spring constant.
Who Discovered Hooke’s Law?
The English scientist Robert Hooke discovered the law, and Britannica dates it to 1660. He first printed it as a Latin anagram in 1676 and gave the solution in 1678 as “ut tensio, sic vis”, meaning as the extension, so the force. Because the law of elasticity carries his name, he is the usual answer to who the father of elasticity is.
What is Hooke’s law used for today?
Every spring scale reads weight from stretch by this law. Hooke applied it to the balance springs of watches, and the same rule sizes valve springs and suspension coils by their rate. In structural work its stress and strain form is the starting point for predicting how beams and wires deflect.
Its importance is that it turns stiffness into a single number. Once k is known, the force, the stored energy and the timing of any oscillation follow from a displacement alone.