Calculate spring force, displacement, or spring constant from Hooke's law, then check restoring direction, elastic potential energy. Use it to test different inputs quickly, compare outcomes, and understand the main factors behind the result before moving on to related tools or deeper guidance.
Last updated
Hooke's law calculator Solve spring force, stiffness, or displacement from the linear elastic relationship F = kx while also checking restoring-force direction, elastic potential energy, and the
equivalent hanging mass on Earth.
Solve for
Use this mode when the spring stiffness and the amount of stretch or compression are known and you need the restoring force magnitude.
Displacement direction
Enter positive magnitudes only. The direction buttons describe whether the spring is stretched or compressed so the restoring-force note stays physically clear.
Practical scope
Use this page for ideal linear springs. Once the spring is near coil bind, permanent deformation, or a non-linear material regime, the straight-line Hooke’s law model no longer holds.
Result
5 N
F = k × x
F = 100 N/m × 0.05 m = 5 N
Force
5 N
Spring constant
100 N/m
Displacement
0.05 m
Elastic energy
0.13 J
Equivalent hanging mass
0.51 kg
Stretch per 1 N
0.01 m/N
Restoring-force interpretation
The restoring force points back toward equilibrium against the extension.
Signed restoring force with the usual convention: -5 N
Hooke’s law only applies while the spring stays in its linear elastic range, where force remains proportional to displacement.
Hooke's law calculator: solve spring force, stiffness, displacement
A Hooke's law calculator helps when you know any two of spring force, spring constant, and displacement and need the third value quickly. This version also shows the restoring-force direction, the elastic potential energy stored in the spring, and the equivalent hanging mass that would create the same force under Earth gravity.
What this Hooke's law calculator solves
The core linear spring relationship is the magnitude form F = kx, where force is proportional to displacement from equilibrium. That means a spring twice as stiff needs twice the force for the same extension, and the same spring needs twice the force when it is stretched or compressed twice as far within its elastic range.
This page separates the workflow into three practical solve modes. You can solve force when stiffness and displacement are known, solve spring constant from a measured force and displacement pair, or solve displacement from a known force and spring constant. The result panel then extends the raw answer into spring energy and an Earth-gravity hanging-mass equivalent so the number is easier to interpret physically.
The formulas behind Hooke's law
In its common classroom form, Hooke's law is written as F = kx when you are working with force magnitude. In vector or sign-aware form, it is often written F = -kx because the restoring force always points opposite the direction of the displacement. The minus sign is a direction statement, not a claim that the spring constant itself can be negative.
Once the main relationship is known, two closely related formulas follow immediately. Rearranging gives k = F/x and x = F/k. The spring also stores elastic potential energy equal to one half of k times displacement squared, which is why spring energy rises quickly as displacement grows even though the force-distance relationship is linear.
For a vertical hanging-mass problem at static equilibrium, the weight force balances the spring force so mg = kx. That relationship is useful for turning a measured extension into an approximate supported mass, provided the spring remains within the linear region and the setup is genuinely static rather than oscillating.
F = k x
Magnitude form of Hooke's law, where F is force, k is spring constant, and x is displacement from equilibrium.
F = -k x
Signed restoring-force form showing that the spring force points opposite to the displacement.
U = 1/2 k x²
Elastic potential energy stored in an ideal linear spring.
m g = k x
Static-equilibrium relationship for a hanging mass on a vertical spring.
Worked examples for force, stiffness, and hanging mass
Suppose a spring has a stiffness of 100 N/m and you extend it by 0.05 m. Hooke's law gives F = 100 × 0.05 = 5 N. The same setup stores U = 1/2 × 100 × 0.05² = 0.125 J of elastic potential energy. That is a small but measurable amount of stored energy, and it shows why extension alone is not enough to describe the spring's behaviour without knowing k.
Now reverse the problem. If a lab measurement shows that a 12 N pull extends a spring by 0.03 m, then the spring constant is k = 12 / 0.03 = 400 N/m. A higher k means the spring is stiffer, so it takes more force per metre of stretch. In practice, engineers and students often infer k this way from measured data rather than from manufacturer labels alone.
For a vertical equilibrium example, if a hanging load stretches a spring by 0.08 m and the spring constant is 250 N/m, then the supporting spring force is 250 × 0.08 = 20 N. Dividing by standard gravity gives an equivalent hanging mass of about 2.04 kg. That conversion is useful for intuition, but it assumes the spring is at rest and not bouncing through a simple harmonic motion cycle.
Where Hooke's law stops being reliable
Hooke's law is a linear elastic model, not a universal spring rule for all loads and all materials. The straight-line force-displacement relationship holds only while the spring and material stay inside the elastic region. Once the load is high enough to approach yielding, coil bind, non-linear geometry, or permanent deformation, the proportional model starts to break down.
This matters because real springs can behave differently in compression and extension at large deflections, and some springs are intentionally designed with progressive rates rather than one constant k. The calculator is strongest for ideal spring problems, bench measurements in the linear region, and introductory engineering checks. It is not a substitute for a manufacturer spring curve, a finite-element model, or a test rig when the design is safety-critical or the spring is heavily loaded.
Another common mistake is mixing the restoring-force sign with force magnitude. If you only care about how much force is required to hold a spring at a displacement, use the positive magnitude form F = kx. If you care about motion direction, use the negative sign convention and remember that the spring force always points back toward equilibrium.
Further reading
Britannica — Hooke's law — Reference overview of the linear elasticity law, its proportional range, and the signed F = -kx form.
The negative sign in F = -kx shows that the spring's restoring force points opposite to the displacement from equilibrium. If you pull the spring outward, the spring pulls back inward. If you compress it inward, the spring pushes outward. When people use F = kx instead, they are usually talking about force magnitude only rather than direction.
Can Hooke's law be used for compression as well as extension?
Yes, as long as the spring stays within its linear elastic range. Hooke's law applies to both stretching and compressing because the key idea is proportional displacement from equilibrium. The important caution is that some real springs become non-linear near coil bind or at large deflections, so the same constant-k model may stop matching reality.
How do I measure the spring constant experimentally?
Measure the applied force and the resulting displacement in the linear region, then compute k = F/x. In a hanging-mass experiment, force comes from weight, so F = mg. Taking several measurements and checking whether force and displacement stay proportional is better than relying on one reading because it confirms that the spring is actually behaving linearly.
Why does Hooke's law stop working for large stretches?
Because the material and spring geometry eventually leave the simple proportional region. At larger loads, coils can approach contact, material behaviour can become non-linear, and permanent deformation can begin. Hooke's law is therefore a local linear model, not a guarantee that every extra millimetre of displacement always requires the same extra force.
