2.5
J
50
N
2.2361
m/s
3.5588
Hz
22.3607
rad/s
0.281
s
2.5
J/kg
2.5
J
50
N
2.2361
m/s
3.5588
Hz
22.3607
rad/s
0.281
s
2.5
J/kg
Elastic potential energy is the energy stored in a deformable object — such as a spring, rubber band, or bow — when it is stretched or compressed from its equilibrium position. This stored energy can be released to do work, powering everything from mechanical watches to car suspension systems. The formula $$PE_{elastic} = \frac{1}{2}kx^2$$ relates the energy to the spring constant and the displacement from equilibrium.
This Elastic Potential Energy Calculator computes the stored energy, the restoring force, and — if a mass is attached — the maximum velocity the mass would achieve when released and the natural oscillation frequency. These calculations are fundamental in mechanical engineering (spring design, vibration analysis), materials science (elastic deformation), and physics (simple harmonic motion). Hooke's Law ($$F = -kx$$) governs the behavior, and the quadratic dependence on displacement means even small additional compression stores significantly more energy.
The elastic potential energy stored in a spring obeying Hooke's Law is:
$$PE = \frac{1}{2}kx^2$$
where $$k$$ is the spring constant in N/m and $$x$$ is the displacement from equilibrium in meters. The restoring force is $$F = kx$$ (magnitude). If a mass $$m$$ is attached and released from displacement $$x$$, energy conservation gives $$\frac{1}{2}kx^2 = \frac{1}{2}mv_{max}^2$$, so $$v_{max} = x\sqrt{k/m}$$. The natural frequency of oscillation is $$f = \frac{1}{2\pi}\sqrt{k/m}$$, determining how quickly the spring-mass system vibrates.
A stiff spring (large $$k$$) stores more energy for the same displacement. Doubling the displacement quadruples the stored energy due to the $$x^2$$ relationship. The max velocity shows how fast the attached mass would move at the equilibrium point when released. The natural frequency helps identify resonance conditions in engineering systems, which must be avoided to prevent catastrophic vibrations.
Inputs
Results
A 25,000 N/m suspension spring compressed 5 cm stores 31.25 J and exerts 1,250 N restoring force.
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Results
A toy spring (k=200 N/m) compressed 8 cm launches a 50 g ball at about 5.1 m/s.
The spring constant $$k$$ (in N/m) measures a spring's stiffness — the force required per unit of displacement. A large $$k$$ means a stiff spring. It is determined by the spring material, wire diameter, coil diameter, and number of coils.
Yes. The formula $$PE = \frac{1}{2}kx^2$$ applies to both compression and extension, as long as the spring obeys Hooke's Law (stays within its elastic limit). The displacement $$x$$ is the absolute distance from equilibrium.
Hooke's Law is valid only within the elastic limit of the material. Beyond this point, permanent deformation occurs and the force-displacement relationship becomes nonlinear. For metals, this typically happens at strains above 0.1–1%.
Simple harmonic motion (SHM) is oscillatory motion where the restoring force is proportional to displacement. A mass on a spring is the classic example. The motion is sinusoidal with frequency $$f = \frac{1}{2\pi}\sqrt{k/m}$$ and the system continuously exchanges kinetic and elastic potential energy.
Both are forms of potential energy. Gravitational PE ($$mgh$$) is linear in height, while elastic PE ($$\frac{1}{2}kx^2$$) is quadratic in displacement. In a vertical spring-mass system, both types interact — the equilibrium position shifts downward by $$mg/k$$.
Approximately, yes, for small deformations. However, rubber and bungee cords have nonlinear stress-strain curves and exhibit hysteresis (energy loss due to internal friction). The linear Hooke's Law model is a rough approximation for these materials.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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