Hooke's Law Spring Force & Stiffness Calculator
Hooke's Law states that the restoring force F exerted by a linear elastic spring below its yield limit is directly proportional to its axial displacement x: F = k × x.
Axial tension or compression force in Newtons
Spring rate in Newtons per meter (N/m)
Extension or compression from free length in meters
Step-by-Step Calculation
Hooke's Law states that the restoring force F exerted by a linear elastic spring below its yield limit is directly proportional to its axial displacement x: F = k × x.
Real-World Reference Values
Precision tactile actuation coil
35 to 90 N/mm suspension wheel rate
Energy stored in the compressed spring
20-Row Comparison Table
Spring Restoring Force (F) · Spring Stiffness Constant (k (N/m))
| # | Spring Stiffness Constant (k (N/m)) | Spring Restoring Force (N) | Δ % |
|---|---|---|---|
| 01 | 15,000.00 Current | 750 N | 0.00% |
| 02 | 36,974.62 | 1,848.73 N | +146.50% |
| 03 | 73,948.25 | 3,697.41 N | +392.99% |
| 04 | 110,921.87 | 5,546.09 N | +639.48% |
| 05 | 147,895.50 | 7,394.77 N | +885.97% |
| 06 | 184,869.12 | 9,243.46 N | +1132.46% |
| 07 | 221,842.75 | 11,092.14 N | +1378.95% |
| 08 | 258,816.37 | 12,940.82 N | +1625.44% |
| 09 | 295,789.99 | 14,789.50 N | +1871.93% |
| 10 | 332,763.62 | 16,638.18 N | +2118.42% |
| 11 | 369,737.24 | 18,486.86 N | +2364.91% |
| 12 | 406,710.87 | 20,335.54 N | +2611.41% |
| 13 | 443,684.49 | 22,184.22 N | +2857.90% |
| 14 | 480,658.12 | 24,032.91 N | +3104.39% |
| 15 | 517,631.74 | 25,881.59 N | +3350.88% |
| 16 | 554,605.37 | 27,730.27 N | +3597.37% |
| 17 | 591,578.99 | 29,578.95 N | +3843.86% |
| 18 | 628,552.61 | 31,427.63 N | +4090.35% |
| 19 | 665,526.24 | 33,276.31 N | +4336.84% |
| 20 | 702,499.86 | 35,124.99 N | +4583.33% |
Frequently Asked Questions
Can I enter spring compression x in millimeters (mm) or inches (in)?+
Yes. Switch the inline unit selector on x to Millimeters (mm) or Inches (in) for direct engineering readouts.
How do springs combine in parallel vs in series?+
In parallel, stiffnesses add directly (k_eq = k₁ + k₂); in series, reciprocals add (1/k_eq = 1/k₁ + 1/k₂).
How much potential energy is stored at deflection x?+
Elastic potential energy equals the area under the force-displacement curve: U = 0.5 × F × x = 0.5 × k × x².