Springs, rubber bands, bow limbs and trampoline beds all store energy when you bend them out of shape, and they hand it back when released. This calculator finds that stored energy with U = ½kx², or solves for the spring constant or deflection when the energy is what you know. Add a projectile mass and it also estimates the fastest speed the spring could give it, a useful ceiling for toy and lab launchers.
How to use the elastic potential energy calculator
- Under Solve for, choose Energy, Spring constant or Stretch.
- Enter the Spring constant (k) in N/m, N/cm, N/mm, kN/m, lbf/in or lbf/ft. Spring catalogs often list it as the “rate”.
- Enter the Stretch or compression (x) in m, cm, mm, in or ft, measured from the spring’s relaxed length. Its sign does not matter.
- When solving for k or x, enter the Elastic potential energy (U) in J, mJ, kJ, cal, ft·lbf or erg instead.
- Optionally fill in Mass to launch (kg, g, lb or oz) to get an ideal launch speed.
- Read the energy or unknown with conversions, plus the Spring force at this stretch, the peak force |F| = k|x| the spring exerts at full deflection.
Elastic potential energy formula
An ideal spring pushes back with a force that grows in proportion to the deflection. The work needed to deform it is the area under that straight force line, a triangle with base x and height kx:
The calculator rearranges it for the other unknowns:
When you solve for x, the answer is shown as ± because a compression and an extension of the same size store the same energy. The optional launch speed assumes all of U becomes kinetic energy of the mass m:
k must be greater than zero and U cannot be negative.
Worked example
A 400 N/m spring pulled back 10 cm, launching a 50 g ball
U = ½ × 400 N/m × (0.1 m)² = 2 J (about 1.48 ft·lbf). At full stretch the spring pulls with 400 × 0.1 = 40 N.
With 50 g in the mass box: v = √(2 × 2 J ÷ 0.05 kg) = √80 ≈ 8.94 m/s, about 32 km/h. That is the best case; a real launcher will fall short of it.
In US units, a 20 lbf/in spring compressed 3 in stores ½ × 20 × 3² = 90 in·lbf, which the calculator reports as 7.5 ft·lbf, with a peak force of 60 lbf (shown as 266.9 N).
Why stretch matters more than stiffness
Energy grows with the square of the deflection but only in proportion to k. For the 400 N/m spring above:
| Stretch x | Stored energy U | Peak force k·x |
|---|---|---|
| 5 cm | 0.5 J | 20 N |
| 10 cm | 2 J | 40 N |
| 20 cm | 8 J | 80 N |
| 30 cm | 18 J | 120 N |
Tripling the pull from 10 cm to 30 cm stores nine times the energy. Put another way, at a fixed peak force F the stored energy is U = F² ÷ 2k, so a softer spring that travels farther stores more. Two springs each loaded to 40 N show it clearly: the 400 N/m spring stretches 10 cm and holds 2 J, while an 800 N/m spring stops at 5 cm with only 1 J. Long draw lengths, not just stiff springs, are what give slingshots and bows their punch.
Real springs, bows and trampolines
The formula describes an ideal linear spring, and the calculator’s note says so. Real elastic objects depart from it in a few predictable ways:
- Elastic limit. Push a coil spring past its rated travel and it takes a permanent set: the energy spent bending the wire is lost and the spring no longer returns to its original length.
- Non-linear materials. Rubber bands stiffen and soften as they stretch and give back noticeably less energy than you put in; the difference warms the rubber. For these, ½kx² is only a rough estimate.
- Bows. A bow’s draw-force curve is not a straight line, and compound bows deliberately flatten it. The stored energy is the area under the actual curve, which a straight-line model can over- or underestimate.
- Moving parts. Some energy always goes into accelerating the spring itself, the string or the bow limbs, plus friction, vibration and sound. For a uniform coil spring, roughly a third of its own mass behaves like extra load.
- Trampolines. Each bounce loses energy to air and to damping in the mat and springs, so jumpers must push with their legs to keep their height.
Related calculators
For the restoring force and its sign convention, see the Hooke’s law calculator. Energy stored by height is on the gravitational potential energy calculator, and a launched object’s motion on the kinetic energy calculator.
Frequently asked questions
What is the formula for elastic potential energy?
U = ½kx², where k is the spring constant and x is the distance the spring is stretched or compressed from its rest length. A 400 N/m spring stretched 10 cm (0.1 m) stores ½ × 400 × 0.1² = 2 J.
Why does doubling the stretch quadruple the energy?
Because the force grows along with the stretch, so the work you put in grows with the stretch squared. Stretching the same 400 N/m spring 20 cm instead of 10 cm stores 8 J instead of 2 J.
Does it matter whether the spring is stretched or compressed?
Not for the energy. x is squared, so a 10 cm compression and a 10 cm extension store the same 2 J in an ideal spring, and the calculator accepts either sign. The direction only matters for the force, which always points back toward the rest length.
How do I find the spring constant from stored energy?
Rearrange to k = 2U ÷ x². A spring that stores 0.5 J when compressed 2 cm has k = 2 × 0.5 ÷ 0.02² = 2,500 N/m, which is 25 N/cm or about 14.3 lbf/in. The calculator rejects a zero stretch or zero energy here because neither defines a stiffness.
Is the ideal launch speed what a real spring launcher achieves?
No, it is an upper limit. It assumes every joule becomes kinetic energy of the projectile, while real launchers also move their own spring and parts, rub, vibrate and make noise. Expect the measured speed to come in below the figure shown.