Elastic Potential Energy Calculator – Spring Energy Formula
Calculate elastic potential energy, spring constant, or displacement using Hooke's law. Solve for any variable in U = ½kx² instantly.
Select the variable you want to solve for, enter the two known values, and get the result with the applied formula.
Elastic Potential Energy Calculator – Spring Energy Formula
Calculate elastic potential energy, spring constant, or displacement using Hooke's law. Solve for any variable in U = ½kx² instantly.
About the Elastic Potential Energy Calculator
Elastic potential energy is the energy stored in a deformed elastic object — most commonly a spring — as a result of its deformation from the equilibrium position. When you compress or stretch a spring and then release it, this stored energy is converted into kinetic energy, driving the motion of whatever is attached to the spring. The formula is U = ½kx², where U is the elastic potential energy in joules, k is the spring constant in N/m (a measure of the spring's stiffness), and x is the displacement from the equilibrium position in metres.
This relationship follows directly from Hooke's Law, which states that the restoring force exerted by an ideal spring is proportional to its displacement: F = −kx. The negative sign indicates that the force opposes the displacement (a stretched spring pulls back; a compressed spring pushes back). The elastic potential energy is the integral of this force over the displacement from 0 to x: U = ∫₀ˣ kx dx = ½kx². This quadratic dependence on displacement means that doubling the stretch quadruples the stored energy — a fact with important implications for spring design, shock absorbers, and energy storage.
The spring constant k is a property of the material and geometry of the spring. A stiff spring (high k, e.g. a car suspension coil spring at 20 000–40 000 N/m) stores much more energy for a given displacement than a soft spring (low k, e.g. a pen spring at 1–5 N/m). Springs are manufactured from high-carbon steel, stainless steel, titanium, and beryllium copper alloys, each chosen for specific strength, fatigue life, and corrosion resistance requirements.
Elastic potential energy appears across a huge range of engineering applications. In mechanical watches and clockwork, a coiled mainspring is the energy reservoir that drives the gear train. In automotive suspension systems, coil springs and leaf springs store impact energy and release it smoothly to maintain tyre contact. Trampolines, bows and arrows, catapults, and rubber bands all rely on elastic potential energy storage and release. Even at the molecular scale, covalent bonds behave approximately like springs, and the elastic potential energy of stretched bonds determines vibrational spectra and reaction rates.
This calculator solves all three forms of the Hooke's Law energy equation. Given k and x, it computes U. Given U and x, it solves k = 2U/x². Given U and k, it solves x = √(2U/k). All three operations find frequent use in physics homework, spring selection for product design, energy budget calculations for robotics actuators, and analysis of elastic collisions and vibration damping.
A practical note on units: if spring constant is in N/m and displacement in metres, energy is in joules. If you work in N/cm or N/mm, convert to SI units first. The displacement x in the formula represents total deformation from the natural (unstressed) length, not the absolute position of the spring end.
Elastic Potential Energy Calculator Examples
Three worked examples showing how to find elastic potential energy, spring constant, and displacement.
| Known Values | Result | Application |
|---|---|---|
| k = 100 N/m, x = 0.5 m | U = 12.5 J | U = ½ × 100 × 0.5² = 12.5 J. A medium-stiffness spring (e.g. a small mechanical seal) compressed 50 cm stores 12.5 J. |
| U = 50 J, x = 0.2 m | k = 2500 N/m | k = 2 × 50 / 0.2² = 2500 N/m. A stiff spring (comparable to a car door latch spring) that stores 50 J at 20 cm deflection. |
| U = 8 J, k = 200 N/m | x = 0.283 m | x = √(2 × 8 / 200) = √0.08 ≈ 0.283 m. A spring-driven toy launcher that releases 8 J when compressed about 28 cm. |
| k = 40 000 N/m, x = 0.05 m | U = 50 J | U = ½ × 40 000 × 0.05² = 50 J. Typical car suspension coil spring absorbing a 5 cm bump impact; stores 50 J per corner. |
How to Use the Elastic Potential Energy Calculator
- Select the variable you want to calculate from the dropdown: Potential Energy (U), Spring Constant (k), or Displacement (x).
- Enter values for the two known quantities in the enabled input fields. All values must be positive numbers in SI units (N/m for k, metres for x, joules for U).
- Click Calculate to get the result along with the formula that was applied.
- To find the spring constant of an unknown spring, measure how far it deflects under a known force (F = kx → k = F/x), then use U = ½kx² to find the stored energy at any displacement.
- Click Reset to clear all fields and start a new calculation.
Elastic Potential Energy Calculator FAQ
What is elastic potential energy?
Elastic potential energy is the energy stored in a stretched or compressed elastic object, such as a spring, rubber band, or bow, as a result of its deformation from the natural equilibrium position. When the deforming force is removed, this stored energy is released and converted into kinetic energy or another form. The formula for an ideal spring is U = ½kx².
What is Hooke's Law and how does it relate to elastic potential energy?
Hooke's Law states that the force F required to stretch or compress a spring by a displacement x from its natural length is F = kx, where k is the spring constant. Elastic potential energy is the work done against this force: U = ∫₀ˣ kx dx = ½kx². The law is named after Robert Hooke who described it in 1678 and holds for small deformations; beyond the elastic limit the spring deforms permanently.
What are the units of the spring constant k?
The SI unit of spring constant is newtons per metre (N/m), also written as kg/s². It tells you how many newtons of force are required to stretch or compress the spring by one metre. Common springs range from about 1 N/m (soft pen spring) to 100 000 N/m or more (heavy industrial springs).
Why does elastic potential energy depend on x² rather than x?
Because the spring force itself increases linearly with displacement. The first bit of stretch requires very little force; subsequent stretching requires proportionally more force because the spring is already under tension. The total work (energy stored) is the area under the force-displacement graph, which is a triangle for a linear spring — giving ½ × k × x × x = ½kx².
What is the difference between elastic and gravitational potential energy?
Gravitational potential energy (U = mgh) depends linearly on height h and is due to the gravitational field. Elastic potential energy (U = ½kx²) depends on the square of deformation and is due to internal stress in the elastic material. Both are stored forms of mechanical energy that can be converted to kinetic energy: a stretched spring and a raised mass both release energy when released, but via different mechanisms.
Is U = ½kx² valid for all springs?
It is valid for ideal linear-elastic (Hookean) springs deformed within their elastic limit. Real springs deviate from this model at large deformations (nonlinear spring behaviour), after yielding (plastic deformation), or when used at temperatures near the material's glass transition. Rubber and elastomeric springs are inherently nonlinear and require more complex hyperelastic models for accurate energy calculations.