Elastic potential energy calculator
In the elastic potential energy calculator, a 400 N/m spring stretched 10 cm stores ½ × 400 × 0.1² = 2 J and pulls back with 40 N.
Elastic potential energy U
- Elastic potential energy U U = ½kx² = ½ × 400 N/m × (0.1 m)² = 2 J.
- 2 J
- The unit algebra Stiffness times a length squared is a force times a length: work, in joules.
- N/m × m² = N·m = J
- Force at that extension F = kx. The energy is half of F × x, ½ × 40 N × 0.1 m, because the force grows from zero as the spring moves.
- 40 N
- Twice the extension would store Four times the energy, because the extension is squared.
- 8 J
- Launch speed for the mass above √(2U / m) with m = 0.05 kg: the speed if all the stored energy went into that mass, with no friction and a massless spring.
- 8.944 m/s
- Period with the mass above on it 2π√(m/k) with m = 0.05 kg: how fast that mass would bounce on this spring.
- 0.07025 s
- Hanging stretch per kg g / k at g = 9.807 m/s²: how far each kilogram hung on it pulls it down.
- 2.452 cm (0.02452 m)
| Extension | Force | Energy stored |
|---|---|---|
| 2.5 cm (0.025 m) | 10 N | 0.125 J |
| 5 cm (0.05 m) | 20 N | 0.5 J |
| 10 cm (0.1 m) | 40 N | 2 J |
| 15 cm (0.15 m) | 60 N | 4.5 J |
| 20 cm (0.2 m) | 80 N | 8 J |
- The force here is the one you apply to the spring. The spring pulls back with an equal and opposite force, which is why the law is often written F = −kx.
- Hooke's law holds only within the elastic limit. Stretched further, a spring deforms permanently and the line bends.
- Results are shown to four significant figures. Round to the precision of your least precise measurement.
How the Simple Harmonic Motion Calculator works
The elastic potential energy formula is U = ½kx², where k is the spring constant in N/m and x is the stretch or compression from the spring's natural length in meters. The answer is in joules. It is the spring potential energy formula, and it holds for anything that obeys Hooke's law, F = kx: a coil spring, a rubber band, a bending beam, each within its elastic limit.
Worked example: a spring with k = 400 N/m stretched 10 cm. In meters, x = 0.1, so U = ½ × 400 × 0.1² = ½ × 400 × 0.01 = 2 J. At that stretch the spring pulls back with F = kx = 400 × 0.1 = 40 N.
The ½ is there because the force is not constant. It starts at zero and grows in step with the stretch up to kx, so the work done is the average force, ½kx, times the distance x. On a graph of force against extension, that work is the area of the triangle under the line.
Because x is squared, doubling the stretch stores four times the energy: 20 cm on the same spring holds 8 J. Compression counts the same as stretch, so the spring pushed in by 10 cm also holds 2 J.
Released against a mass, the stored energy becomes kinetic energy: ½kx² = ½mv². A 50 g ball launched by the 2 J spring leaves at v = √(2 × 2 ÷ 0.05) = 8.944 m/s, as long as friction is negligible and the spring's own mass is small next to the ball's.
Simple Harmonic Motion Calculator FAQ
How do I find the spring constant from the energy?
Rearrange to k = 2U ÷ x². A spring that stores 2 J when stretched 0.1 m has k = 4 ÷ 0.01 = 400 N/m.
How far does a spring stretch to store a given energy?
x = √(2U ÷ k). To store 8 J, the 400 N/m spring stretches √(16 ÷ 400) = 0.2 m, or 20 cm.
What is elastic potential energy in US units?
With k in pounds-force per inch and x in inches, ½kx² comes out in inch-pounds. A 20 lbf/in spring compressed 2 in stores ½ × 20 × 2² = 40 in·lbf, which is 3.333 ft·lbf or 4.519 J.
More from the Simple Harmonic Motion Calculator
Sources
- NASA NSSDCA Moon Fact Sheet, bulk parameters table, row "Surface gravity (m/s2)", Moon column (page last updated January 11, 2024) (2026)
- NASA NSSDCA Mars Fact Sheet, bulk parameters table, row "Surface gravity (mean) (m/s2)", Mars column (page last updated May 19, 2025) (2026)