How to use this calculator
- Enter or choose solve for.
- Enter or choose force unit.
- Enter or choose extension unit.
- Enter or choose spring constant unit.
- The result updates as you type. Use Copy result or Share to keep it.
Method
F = k x; k = F ÷ x; x = F ÷ k; E = ½ k x²
- F
- force in N
- k
- spring constant in N/m (1 lbf/in = 175.127 N/m)
- x
- extension or compression from the free length in m
- E
- elastic potential energy in J
Worked example
- x = 25 mm = 0.025 m.
- k = 50 ÷ 0.025 = 2,000 N/m.
- E = ½ × 2,000 × 0.025² = 0.625 J.
A spring that stretches 25 mm under 50 N has a constant of 2,000 N/m (about 11.4 lbf/in) and stores 0.625 J.
What the result means
Measure k by hanging known weights and recording the stretch; the slope of force against extension is k. If the points stop lying on a straight line, the spring has passed its elastic limit.
Assumptions and limits
- Linear elastic behaviour, measured from the spring’s free (unloaded) length.
- Not valid beyond the elastic limit, for progressive-rate springs or for coils that touch (coil bind).
Questions
What happens to k for springs in series or parallel?
In parallel the constants add (k = k₁ + k₂). In series the reciprocals add (1/k = 1/k₁ + 1/k₂), so the combination is softer.
Why is the stored energy ½ k x² and not k x²?
The force rises from zero to k x as the spring stretches, so the average force is half the final force.