Where is the energy stored?
In the deformation of the spring material. It converts to kinetic energy when released, driving oscillation.
Elastic Potential Energy Calculator
PE = ½kx²
Calculate elastic potential energy stored in a spring. PE = ½kx² — energy from deformation.
Calculate
Enter values above to calculate
How it calculates
1
PE = ½ × k × x²
2
k in N/m, x (extension/compression) in metres
3
Result in joules
Formula
PE = ½kx²
Variable Table
| Symbol | Quantity | SI Unit |
|---|---|---|
| PE | Elastic potential energy | J |
| k | Spring constant | N/m |
| x | Extension or compression | m |
→ Simple Harmonic Motion article
How to Use This Calculator
This elastic potential energy calculator is built for quick physics checks and worked-problem review. Enter values in the units shown beside each input, then compare the result with the formula and variable table before using it in a longer solution. The calculator does the arithmetic, but the physics still depends on choosing a model that matches the situation.
Start by identifying the system, the known quantities, and the quantity you want to find. If a value is given in a non-SI unit, convert it before substitution. A correct numerical answer with mixed units can still be physically wrong, especially when squared units, inverse seconds, charges, temperatures, or distances are involved.
Assumptions and Limits
The formula PE = ½kx² is a model, not a universal description of every possible case. It assumes the quantities in the variable table are the relevant quantities for the problem and that hidden effects are either negligible or already included in the inputs. If friction, drag, relativistic speeds, changing fields, non-constant temperature, or geometry-specific effects matter, check whether a more complete model is needed.
Use the result as a magnitude and units check. Ask whether the answer has the right sign, whether it grows or shrinks when an input changes, and whether the limiting cases make sense. Setting an input to zero, doubling a quantity, or using a very large value is often enough to catch a formula choice or unit mistake before it reaches a final answer.
Worked Example
A spring (k = 200 N/m) compressed 0.1 m. Find stored energy.
Step 1: PE = ½kx²
Step 2: PE = ½ × 200 × (0.1)²
Step 3: PE = 100 × 0.01
Answer: PE = 1 J
Common Mistakes
- Note the ½ factor and that x is squared — doubling x quadruples energy
- Valid only within the elastic limit (Hooke's law region)
- Energy is the same for extension or compression of equal magnitude
Related
Frequently Asked Questions
Why x² and not x?
Because force increases linearly with x (F = kx), the work to stretch it (area under the F-x graph) is the triangle ½kx².