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Hooke's Law Calculator

Physics
The other two quantities become the required inputs.
Stiffness — must be greater than 0.
Positive stretches the spring, negative compresses it.
Optional — entering a measured force cross-checks it against k·x.
Load a realistic stiffness.
Decimal places, 0 to 10.
Flags results that fall outside the linear range.
Drag to sweep the displacement and watch every result update.

Force |F|

30 N

Direction
Spring is stretched — the restoring force pulls back toward equilibrium (negative, opposite to the displacement).
Restoring force F = −kx
-30 N
Force magnitude |F|
30 N
Elastic PE (½kx²)
1.8 J
Average force (½kx)
15 N
Spring constant in SI
250 N/m
Displacement in SI
0.12 m

Spring diagram

natural lengthapplied Frestoring −kxstretched by 12 cm

The applied force and the restoring force always point in opposite directions — that opposition is what the minus sign in F = −kx records.

Force and energy vs displacement

The shaded area under the force line is the work done, and it equals the stored elastic energy of 1.8 J.

Force rises as a straight line while energy curves upward as a parabola — double the displacement and the force doubles, but the stored energy quadruples.

Unit conversions

Force magnitude

UnitValue
N30
kN0.03
lbf6.7443
kgf3.0591
dyn3,000,000

Spring constant

UnitValue
N/m250
N/cm2.5
lbf/ft17.1304
lbf/in1.4275

Displacement

UnitValue
m0.12
cm12
mm120
ft0.3937
in4.7244

Elastic energy

UnitValue
J1.8
kJ0.0018
mJ1,800
cal0.4302
ft·lb1.3276
Leave empty to skip this panel.

About This Tool

Hooke's Law Calculator – Spring Force, Energy and Oscillation

Hooke's Law is the rule that makes springs predictable. It states that the restoring force an ideal elastic element exerts is proportional to how far it has been pushed or pulled from its natural length, and always points back toward that natural length. In symbols, F = −k·x, where k is the spring constant in newtons per metre and x is the displacement from equilibrium. This calculator solves that relationship in every direction — give it any two of force, stiffness and displacement, and it returns the third along with the energy, work and oscillation results that follow from it.

Reading the minus sign

The minus sign is the part students most often drop, and it carries real physical content. It says the force and the displacement point in opposite directions. Stretch a spring to the right and it pulls left; compress it to the left and it pushes right. When a problem asks for the force you must apply to hold the spring in place, the answer is +k·x— equal in size, opposite in direction to the spring's own restoring force. This tool reports both the signed restoring force and the plain magnitude so the distinction never gets lost.

Why the stored energy is ½kx²

A spring is not a constant-force device. As you stretch it, the force you must supply climbs steadily from zero up to k·x. The work done is therefore the area under the force–displacement line, which is a triangle of area ½ · x · kx = ½kx². That is the elastic potential energy stored in the spring. The average force over the stretch is ½kx, and using that average does make the familiar W = F̄ · x work out correctly.

The most common spring mistake
Multiplying the final force by the distance moved overstates the work, often badly. Stretching a 300 N/m spring from 10 cm to 25 cm takes ½ × 300 × (0.25² − 0.10²) = 7.875 J, but the naive 75 N × 0.15 m gives 11.25 J — about 43 percent too high.

Because x is squared, compression and extension of the same size store exactly the same energy, and doubling the displacement quadruples the energy. That quadratic growth is why the last centimetre of travel on a stiff spring feels so much more expensive than the first.

Springs in series and in parallel

Real assemblies rarely contain one spring. Connected in series, end to end, every spring carries the same force while their extensions add, so the combination is softer than any individual spring: 1/k_eq = Σ 1/kᵢ. Connected in parallel, side by side, they all share the same extension while their forces add, making the combination stiffer: k_eq = Σ kᵢ. A 200 N/m spring and a 300 N/m spring give 120 N/m in series but 500 N/m in parallel — a four-fold difference from the same two components.

From stiffness to natural frequency

Attach a mass and the same spring constant sets how fast the system oscillates. The angular frequency is ω = √(k/m), the period is T = 2π√(m/k), and the frequency is f = 1/T. This is the single most useful step in suspension and vibration work: a stiffness specification converts directly into a natural frequency for a given sprung mass. Notably, the period depends on neither the amplitude nor gravity — pull the mass twice as far and it simply covers twice the distance in the same time.

Hang a mass from a vertical spring instead and it settles where the spring force balances the weight, at x = mg/k. A useful check on intuition: the gravitational energy released in reaching that point is exactly twice the energy stored in the spring. The missing half is carried off by whatever lowers the mass gently. Let go of it instead and it overshoots to twice the static extension before springing back.

Where the linear model runs out

Hooke's Law is a small-deformation approximation, and it is only as good as the elastic range of the material. Past the elastic limit a spring deforms permanently and the force–extension curve bends away from a straight line. Progressive and conical springs are deliberately nonlinear from the very first millimetre. Fit a measured dataset in this calculator and the reported R² tells you how straight your real spring actually is; anything below about 0.98 is a signal to stop treating k as a constant.

Beyond springs
The same relationship governs any axially loaded bar in the small-strain regime through k = EA/L, which links spring stiffness to Young's modulus, cross-sectional area and length. It is also the working principle behind load cells, force gauges and mechanical scales, which all measure force by measuring a tiny, precisely calibrated deflection.

Getting reliable answers

Enter values in whatever units your problem uses — the calculator normalises everything to SI before applying the formula and shows each conversion step. Keep an eye on the sign of the displacement, since that is the one input where a negative number is meaningful rather than an error. If you are determining k from measurements, take several readings across the working range rather than one, and fit them: a single point can hide curvature that a fitted line and its R² will expose immediately.

Frequently Asked Questions

Is the Hooke's Law Calculator free?

Yes, Hooke's Law Calculator is totally free :)

Can I use the Hooke's Law Calculator offline?

Yes, you can install the webapp as PWA.

Is it safe to use Hooke's Law Calculator?

Yes, any data related to Hooke's Law Calculator only stored in your browser (if storage required). You can simply clear browser cache to clear all the stored data. We do not store any data on server.

How does this Hooke's Law calculator work?

Pick which quantity you want to find — force, spring constant, or displacement — then enter the other two in any supported units. The calculator converts everything to SI, applies F = −k·x, and returns the signed restoring force along with the elastic potential energy (½kx²), the average force over the stretch, a full unit-conversion strip, and a step-by-step derivation. Optional panels add spring networks, oscillation frequency, and a least-squares fit of k from measured data.

Why is there a minus sign in F = −k·x?

The minus sign encodes direction, not magnitude. It says the spring's restoring force always points opposite to the displacement: stretch the spring (x positive) and it pulls back (F negative); compress it (x negative) and it pushes out (F positive). When a textbook writes F = kx it is usually talking about the magnitude, or about the force you apply to the spring rather than the force the spring applies to you.

Why is the stored energy ½kx² and not kx?

Because the spring force is not constant — it grows linearly from zero to kx as you stretch it. The work done is the area under the force–displacement line, which is a triangle of area ½ × x × kx = ½kx². Using the final force for the whole distance would double-count the effort, which is why W = F·d gives the wrong answer for springs. The average force is ½kx, and that value does make W = F̄·x correct.

Why do springs in series get softer while springs in parallel get stiffer?

In series each spring carries the same force, so the extensions add and the combination stretches further for that force — the equivalent stiffness follows 1/k_eq = Σ1/kᵢ and is always smaller than the softest spring. In parallel each spring shares the same extension, so their forces add and k_eq = Σkᵢ, which is stiffer than any individual spring. Two 200 N/m and 300 N/m springs give 120 N/m in series but 500 N/m in parallel.

Does the mass on a spring oscillate faster if I pull it further?

No. For an ideal linear spring the period T = 2π√(m/k) depends only on the mass and the stiffness, not on the amplitude — pull it twice as far and it simply travels twice as far in the same time. The period is also independent of gravity, so a horizontal spring–mass system and a vertical one with the same k and m oscillate at exactly the same rate.

When does Hooke's Law stop being accurate?

Hooke's Law is a small-deformation approximation. Past the elastic limit a real spring deforms permanently and the force–extension curve bends away from a straight line, so the calculated values become extrapolation rather than prediction. Progressive and conical springs are deliberately nonlinear from the start. Enter an optional elastic limit and the calculator will flag results that fall outside the linear range, and the dataset-fitting mode reports R² so you can see how linear your actual spring is.