Compute force (F), spring constant (k), or displacement (x) based on Hooke's Law F = k·x. Enter any two values (with sign: positive for tension, negative for compression) and get the third instantly, plus the elastic potential energy E = ½kx². The interactive graph plots the linear relationship and highlights your current point.
In 1676, the English physicist Robert Hooke discovered that, within the elastic limit, the extension (or compression) of a spring is directly proportional to the applied force. This fundamental relation is expressed as F = k·x, where F is the force in newtons (N), k is the spring constant (stiffness) in N/m, and x is the displacement from the natural length in meters (m). The negative sign often seen in textbooks (F = -kx) indicates that the restoring force opposes the displacement; here we treat direction through the sign of x and F.
F = k·x (magnitude form)
E = ½ k x² (elastic potential energy)
For continuous materials, the equivalent form is σ = E·ε (stress = Young's modulus × strain).
Hooke first published his law as a Latin anagram "ceiiinosssttuv" in 1676, and two years later revealed the solution: "Ut tensio, sic vis" — "As the extension, so the force." This was one of the first quantitative laws in physics and laid the foundation for elasticity theory, material science, and even modern seismology. Hooke's work influenced Newton, Euler, and the entire development of classical mechanics. Today, Hooke's law remains a cornerstone in engineering design, from micro‑scale MEMS devices to large‑scale bridge suspensions.
Given any two of the three variables {F, k, x}, the third is uniquely determined (provided k>0 and, when solving for k, x ≠ 0). The calculator uses simple algebraic rearrangements:
Elastic potential energy is always computed as E = ½ k x², which is positive for any non‑zero displacement (since x² ≥ 0 and k > 0). The graph displays the line F = k·x through the origin, with slope k. The current operating point (x, F) is highlighted in red.
| Application | Known values | Computed result | Notes |
|---|---|---|---|
| Automotive suspension | k = 25,000 N/m, x = -0.08 m (compression) | F = -2000 N, E = 80 J | Each spring supports about 200 kg under compression |
| Lab spring calibration | F = 6 N, x = 0.12 m | k = 50 N/m | Soft spring suitable for small force measurements |
| Toy spring (stretch) | k = 80 N/m, desired F = 4 N | x = 0.05 m (5 cm) | Linear relationship verified |
| Shock absorber design | k = 5000 N/m, max x = 0.2 m | Max F = 1000 N, stored energy = 100 J | Used in impact analysis |
| Seismic isolation | F = 200 N, x = 0.02 m | k = 10,000 N/m | Stiff spring for minimal motion |
A resistance band behaves like a spring with k = 120 N/m. When stretched by 0.6 m, the force exerted is F = 120 × 0.6 = 72 N. The elastic energy stored is E = 0.5 × 120 × (0.6)² = 21.6 J. This energy is released when the band contracts. Using the calculator, a trainer can verify if the band meets the desired resistance profile.
Often springs are used together. Their effective spring constant can be calculated as:
While this tool does not directly combine springs, you can use the computed k values to manually apply these formulas. A future version may include a combination mode.
Hang a known mass m (in kg) on the spring and measure the static extension x (in m). Then k = mg / x, where g ≈ 9.81 m/s². For example, m = 0.5 kg, x = 0.1 m → k = (0.5×9.81)/0.1 = 49.05 N/m. Enter F = 4.905 N and x = 0.1 m to verify.
The graph's slope equals the spring constant k. A steeper line means a stiffer spring. The area under the line from 0 to x (a right triangle) represents the work done to stretch/compress the spring, which equals the elastic potential energy stored: Area = ½ × base × height = ½ × x × F = ½ k x². When x is negative, the area is still positive because both base and height are negative in the third quadrant, but the physical energy remains positive.