Physics & Mechanics

Elastic Potential Energy Calculator

Calculate the potential energy stored in a compressed or stretched spring based on the spring constant and displacement.

N/m
m
Elastic Potential Energy
10

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Energy in Springs and Elastic Materials

Elastic Potential Energy is the mechanical potential energy stored in the configuration of a material or physical system as it is subjected to elastic deformation by work performed upon it.

When you stretch a rubber band, compress a coil spring, or draw back a bowstring, you are doing work against the object's restorative intermolecular forces. This work is stored within the material as elastic potential energy.

Hooke's Law and Energy Storage

According to Hooke's Law, the force required to stretch or compress an ideal spring is directly proportional to the distance it is stretched. Because the force increases the further you stretch it, the energy stored increases with the square of the displacement.

This principle is used everywhere:

  • Vehicle Suspensions: Shock absorbers and springs store energy from bumps in the road, releasing it slowly to keep the ride smooth.
  • Clocks and Watches: Mechanical timepieces use tightly wound mainsprings to store energy that powers the ticking mechanism for days.
  • Archery: The massive potential energy stored in the bending limbs of a bow is instantly transferred to the arrow upon release.

The Formula

Ue=12kx2\begin{aligned} U_e = \frac{1}{2} k x^2 \end{aligned}

Where:
UeU_e=
Elastic Potential Energy (Joules, J)
k=
Spring Constant (N/m)
x=
Displacement or stretch distance (meters)

Example Calculation

Consider a heavy-duty vehicle suspension spring with a spring constant of $10,000 , ext{N/m}$. If a pothole compresses the spring by $0.05 , ext{meters}$ (5 cm):

  1. Square the displacement: $0.05^2 = 0.0025$
  2. Multiply by spring constant and 0.5: $0.5 cdot 10,000 cdot 0.0025 = 12.5 , ext{Joules}$.

The spring stores $12.5 , ext{J}$ of energy from that single bump.

Frequently Asked Questions

The spring constant ($k$) is a measure of a spring's stiffness. A higher value means the spring is stiffer and requires significantly more force to stretch or compress a given distance.

No. Because the displacement ($x$) is squared in the formula, stretching a spring by 5 cm stores the exact same amount of energy as compressing it by 5 cm.

If you exceed the material's 'elastic limit,' Hooke's Law no longer applies. The spring will undergo plastic deformation (permanent bending) and will not return to its original shape, meaning it can no longer store energy efficiently.