Hooke’s Law Calculator

Hooke's law says spring force is proportional to extension: F = k x. At k = 200 N/m and x = 0.1 m you get 20 N. At 500 N/m and 0.1 m you get 50 N: a stiffer spring, a larger force. At x = 0.05 m and k = 200 you get 10 N. At 500 N/m and 0.02 m that is 10 N another way. At 50 N/m and 0.05 m that is 2.5 N. At 150 N/m and 0.2 m that is 30 N. At 300 N/m and 0.15 m that is 45 N.

At 20,000 N/m and 0.08 m, F = 1600 N. At 4000 N/m and 0.25 m, F = 1000 N. At 8000 N/m and 0.06 m, F = 480 N. That same 200 N/m and 0.1 m pair gives 1 J of elastic energy next door, because Eₛ = ½ k x². Force grows in a straight line with x; energy grows with the square.

Stiffness k is in N/m, extension x in meters. The result is in newtons. 200 N/m is 2 N/cm, not 200 in centimeters. The model is linear and holds in the elastic range. Beyond that F is not kx. A comma and a period are the same x: 0,1 and 0.1. Typed 0 in k or x gives F = 0. Both fields need a number before 20 can appear. The sign of x follows the problem: compression versus stretch.

Energy Eₛ = ½ k x² is on the neighbouring page. The mass-spring period is farther on. Here the result stays force, not energy and not frequency. Work of a constant force W = F·s does not replace ½kx² for a spring.

The unit switch does not change N/m into another stiffness scale. The header symbol does not pull the spring.

Type 200 and 0.1, click Calculate, and match 20 N. Then 500 and 0.1 for 50 N. Treat 20 N as a product inside the elastic range, not as a snapped spring at a large x.