Elastic Potential Energy Calculator
Calculate the elastic potential energy stored in a spring from its spring constant and displacement, or solve for k or x when the energy is known. Supports N/m, N/cm, m, cm and mm units.
Calculate Stored Spring Energy
Elastic Potential Energy Formula
Where:
- \(PE\) — elastic potential energy (J)
- \(k\) — spring constant (N/m)
- \(x\) — displacement from the equilibrium position (m)
This formula, derived from Hooke's law, gives the energy stored in a spring that is stretched or compressed by a distance \(x\) from its natural length. The same relation can be rearranged to solve for the other two variables:
\[ k = \frac{2 \cdot PE}{x^2} \qquad x = \sqrt{\frac{2 \cdot PE}{k}} \]
Elastic Potential Energy – Calculation Example
Example: a spring with a spring constant of \(k = 100\ \text{N/m}\) is stretched by \(x = 0.1\ \text{m}\) from its equilibrium position. How much elastic potential energy is stored?
\[ PE = \frac{1}{2} \times 100 \times (0.1)^2 \]
\[ PE = 50 \times 0.01 = 0.5\ \text{J} \]
Result: the spring stores 0.5 joules of elastic potential energy.
How to Use the Calculator
- In the "I want to calculate" dropdown, choose whether you want the elastic potential energy (PE), the spring constant (k), or the displacement (x). The field you are solving for is filled in automatically.
- Enter the Spring Constant (k) and select its unit — N/m or N/cm (values in N/cm are converted to N/m internally).
- Enter the Displacement (x) and pick its unit — m, cm, or mm (converted to meters automatically).
- If you are solving for k or x, enter the known Elastic Potential Energy (PE) in joules instead of the corresponding input.
- Read the result under Result: the energy is given in joules (J), the spring constant in N/m, and the displacement in meters.
What Is Elastic Potential Energy?
Elastic potential energy is the energy stored in an object when it is stretched or compressed — think of a compressed spring, a drawn bow, or a stretched rubber band. As long as the deformation stays within the elastic limit, the object returns to its original shape when the force is removed, releasing the stored energy. This behavior is described by Hooke's law, which states that the restoring force is proportional to the displacement: \(F = kx\).
The spring constant \(k\) measures stiffness: a stiff spring might have \(k = 10{,}000\ \text{N/m}\), while a soft one might be \(k = 50\ \text{N/m}\). The larger the spring constant, the more force is needed for the same displacement, and the more energy is stored. Notice that energy grows with the square of the displacement — doubling the stretch stores four times as much energy, which is why over-stretched springs can be dangerous.
Because the spring force varies with displacement, the stored energy equals the work done against it: the average force \(\frac{kx}{2}\) acting over the distance \(x\) gives \(PE = \frac{1}{2}kx^2\). This is the linear-elastic analogue of rotational kinetic energy and other squared-variable energy formulas in physics.
The result tells you how much energy is available for release. When a compressed spring pushes a block, that elastic energy converts into kinetic energy; in a dart gun or a car's suspension, the same principle smooths impacts. Comparing the stored energy with other energies in a system (for example, in projectile motion problems where a spring launches a projectile) is a standard way to solve mechanics problems.
Assumptions and Limitations
- Linear elasticity (Hooke's law): the formula is only valid while force and displacement stay proportional. Beyond the elastic limit the material deforms permanently and the calculation no longer applies.
- Ideal spring: the spring's own mass, internal friction, and damping are neglected; real springs lose some energy as heat when they compress or recoil.
- Displacement from equilibrium: \(x\) must be measured from the spring's natural (unstressed) length — not from an arbitrary position. For a vertically hung spring carrying a weight, measure the additional stretch, not the total length.
- Unit consistency: the calculator converts N/cm to N/m and cm/mm to m automatically, but if you use the formula by hand, keep \(k\) in N/m and \(x\) in meters so that \(PE\) comes out in joules.
- Sign doesn't matter: since \(x\) is squared, compression and stretching by the same distance store the same energy; you can enter either as a positive value.
Frequently Asked Questions
How do I calculate elastic potential energy by hand?
Multiply the spring constant by the square of the displacement, then take half of that: \(PE = \frac{1}{2}kx^2\). For example, a spring with \(k = 200\ \text{N/m}\) compressed by \(0.05\ \text{m}\) stores \(\frac{1}{2} \times 200 \times 0.0025 = 0.25\ \text{J}\). Just make sure \(k\) is in N/m and \(x\) in meters so the result comes out in joules.
Can elastic potential energy be negative?
No. Because the displacement is squared, the result is always zero or positive — compressing a spring by 5 cm stores exactly as much energy as stretching it by 5 cm. If you get a negative value by hand, you likely entered compression as a negative number; use its absolute value instead. A negative result would only appear if you subtracted energies of different springs in a larger problem.
What is a typical spring constant value?
It depends on stiffness. Soft springs, like those in pens or toys, are around 20–100 N/m. Car suspension springs are roughly 20,000–50,000 N/m, and heavy industrial springs can exceed 100,000 N/m. If you don't know \(k\), you can measure it: hang a known mass and divide the weight by the extra stretch, \(k = F/x\), since Hooke's law links force, stiffness, and displacement.
How do I convert units when using the formula manually?
Convert everything to SI units before plugging in. Since \(1\ \text{N/cm} = 100\ \text{N/m}\), multiply N/cm values by 100. Divide centimeters by 100 and millimeters by 1000 to get meters. Mixing units is the most common source of wrong answers — a spring of 5 N/cm stretched 10 cm is \(k = 500\ \text{N/m}\) and \(x = 0.1\ \text{m}\), giving \(PE = 2.5\ \text{J}\).
Why is there a ½ in the formula and not just kx²?
The force needed to stretch a spring isn't constant — it grows from 0 up to \(kx\). The work done equals the average force times distance, and the average of 0 and \(kx\) is \(\frac{kx}{2}\). Multiplying by \(x\) gives \(\frac{1}{2}kx^2\). Geometrically, it's the area of a triangle under the force–displacement graph.
Does the formula work for rubber bands and other elastic objects?
Only approximately, and only for small deformations. The formula assumes ideal linear elasticity, where force is exactly proportional to stretch. Rubber bands are non-linear: their stiffness changes as they stretch, so \(\frac{1}{2}kx^2\) becomes increasingly inaccurate at large extensions. Metal springs follow Hooke's law much more closely, but even they deviate once you approach their elastic limit.
How does elastic potential energy relate to kinetic energy?
When a compressed or stretched spring is released, its stored energy converts into kinetic energy of the moving object: \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\) in an ideal, frictionless case. This lets you predict launch speed — a spring storing 4 J that releases into a 0.5 kg block gives it a speed of \(\sqrt{2 \times 4 / 0.5} = 4\ \text{m/s}\). Real systems lose some energy to friction and heat.