Elastic Potential Energy Calculator
Compute the energy stored in a stretched or compressed spring, or work backward to the spring constant or the extension, with a chart for each mode.
🔩 What is Elastic Potential Energy?
Elastic potential energy is the energy stored in an elastic object, most often a spring, when it is stretched or compressed away from its relaxed length. The work you do pulling the spring is not lost. It is stored in the deformed material, and it comes back as motion, sound or heat when the spring is released. For a spring that follows Hooke's law, the stored energy is U = ½ k x², where k is the spring constant that measures stiffness and x is how far the spring has moved from its relaxed length.
This calculator has three modes. In Find energy you enter k and the displacement x and get the energy, the force at that stretch, the work needed to go from one displacement to another, and the speed an attached mass would have when the spring snaps back to its relaxed length. In Find k you work backward from a measurement: a force and an extension give k = F / x, an energy and an extension give k = 2U / x², and an optional mass gives the oscillation period of the mass-spring system. In Find extension you know k and the energy you want to store and solve for the stretch, x = √(2U / k).
Spring energy turns up everywhere: the dart in a toy launcher, the suspension of a car, the bounce of a trampoline, the return spring in a pen, the lab spring in a physics class, and the safety spring in a door latch. It is also the first example of a potential energy function that students meet, and the same quadratic shape appears in the quantum harmonic oscillator, where the potential is ½ mω²x² with k = mω². A common mistake is to treat the energy as proportional to the stretch. It is not. Because U depends on x², doubling the stretch quadruples the energy, while the force only doubles. Another mistake is to forget that the formula applies only within the elastic, linear range of the spring.
The results show the energy in joules, foot-pounds and calories, the spring constant in N/m, N/cm and lbf/in, and a chart for each mode. For related ideas see the kinetic energy calculator for what the released energy becomes, the work, energy and power calculator for the work-energy theorem, the force calculator for F = ma, the mechanical advantage calculator for simple machines, and the quantum harmonic oscillator calculator for the quantum version of the same potential.
📐 Formula
📖 How to Use This Calculator
Steps
🔗 Squared Growth, Series and Parallel Springs
Why energy scales with x². The force on a Hooke's law spring rises in a straight line from zero to kx. The work to stretch it is the area under that line, a triangle with base x and height kx, so U = ½ x (kx) = ½ k x². The second chart in Find energy mode shades that triangle. Stretching twice as far doubles both the base and the height, so the area is four times larger.
Springs in parallel share the stretch and add their forces, so the combined stiffness is k = k₁ + k₂. Two springs of 25 N/m and 75 N/m side by side behave like one 100 N/m spring. Springs in series share the force and add their stretches, so 1/k = 1/k₁ + 1/k₂, and the pair is softer than either spring. For the same two springs the series stiffness is 25 × 75 / 100 = 18.75 N/m. Check it by force: pulling with 10 N stretches the 25 N/m spring 0.4 m and the 75 N/m spring 0.1333 m, a total of 0.5333 m, and 10 / 0.5333 = 18.75 N/m. Enter the combined k in the calculator to get the energy of the whole arrangement.
Limits. Every real spring has an elastic limit. Past it the spring is permanently deformed and part of the work is lost as heat, so U = ½ k x² no longer applies. Even below that limit, Hooke's law is only a linear approximation. Coil springs are close to linear over much of their range, but a coil that is nearly fully closed, a spring pushed to its maximum length, or a rubber band follows a curved force-stretch relation instead. For those, the measured force-distance area gives the energy, and this calculator gives a first estimate only.
💡 Example Calculations
Example 1 - Lab spring stretched 0.2 m
A lab spring with k = 25 N/m is stretched 0.2 m with a 0.5 kg mass attached
Example 2 - Toy dart launcher
A toy launcher spring of 1.2 N/cm is compressed 3 cm behind a 10 g dart
Example 3 - Car suspension spring (illustrative rate)
An illustrative 400 lbf/in coil spring compressed 2 in
Example 4 - Trampoline-like value, work from 0.2 m to 0.5 m
An illustrative effective stiffness of 4000 N/m, deflected from 0.2 m to 0.5 m
Example 5 - Find k from a measurement
A spring stretches 8 cm under a 12 N pull and stores 0.48 J, with a 0.3 kg mass
Example 6 - Find the extension for a given energy
How far does an 800 N/m spring stretch when it stores 10 J, with a 2 kg mass?
❓ Frequently Asked Questions
🔗 Related Calculators
What is the formula for elastic potential energy?
For a spring that obeys Hooke's law, U = 1/2 k x^2, where k is the spring constant in N/m and x is the stretch or compression from the relaxed length in metres. The result is in joules. A spring with k = 25 N/m stretched 0.2 m stores 0.5 x 25 x 0.04 = 0.5 J.
Why is spring energy proportional to x squared?
The force grows linearly with stretch, F = kx, and the energy is the area under the force-distance line, which is a triangle with area 1/2 x (kx). Because the area depends on both the base and the height, and both grow with x, the energy grows with x squared. Doubling the stretch quadruples the energy.
How do you find the spring constant?
Divide the force by the extension, k = F / x. A 5 N pull that stretches a spring 0.2 m gives k = 25 N/m. If you know the stored energy instead, use k = 2U / x^2, so 0.5 J at 0.2 m also gives 25 N/m.
How do you find the extension from energy?
Solve U = 1/2 k x^2 for x to get x = sqrt(2U / k). With k = 800 N/m and U = 10 J, x = sqrt(20 / 800) = 0.158 m. Four times the energy only doubles the extension.
How much work does it take to stretch a spring from x1 to x2?
The work is the change in stored energy, W = 1/2 k (x2^2 - x1^2). For k = 4000 N/m going from 0.2 m to 0.5 m, W = 0.5 x 4000 x (0.25 - 0.04) = 420 J. It is not the same as stretching from rest to the difference.
How fast does a mass leave a spring?
If all the stored energy becomes kinetic energy when the spring passes its relaxed length, 1/2 m v^2 = 1/2 k x^2, so v = x sqrt(k / m). A 0.5 kg mass on a 25 N/m spring released from 0.2 m moves at 1.414 m/s. Friction and the spring's own mass make real launches slower.
What is the period of a mass on a spring?
For an ideal mass-spring system T = 2 pi sqrt(m / k). With m = 0.3 kg and k = 150 N/m, T = 0.281 s, which is a frequency of about 3.56 Hz. The period does not depend on the amplitude as long as the spring stays linear.
What happens to the total spring constant of springs in series and parallel?
In parallel the stiffnesses add, k = k1 + k2. In series the reciprocals add, 1/k = 1/k1 + 1/k2. For 25 N/m and 75 N/m, parallel gives 100 N/m and series gives 18.75 N/m.
What is the elastic limit of a spring?
The elastic limit is the largest stretch from which the spring still returns to its original length. Beyond it the spring is permanently deformed, and the formula U = 1/2 k x^2 no longer describes it. Stay well inside the range given by the manufacturer.
Does Hooke's law work for a rubber band?
Only for small stretches. A rubber band is non-linear, so its force is not proportional to its stretch over a large range, and it also loses energy as heat on each cycle. Treat the 1/2 k x^2 result as a rough estimate for rubber.