Hooke’s law calculator (F = kx)

Force of a stretched spring, stiffness k from a force-extension pair or from a hanging mass (k = m·g / x), and extension under a force. The elastic energy ½·k·x² is computed in every mode.

The stiffness k, in newtons per metre, is the slope of the force-extension line. A motionless hanging mass gives k without a dynamometer: at equilibrium, the spring carries its weight. The energy is the area of the triangle under that line, hence the ½.

Stiffness k

the force per unit extension, in N/m. 100 N/m: one newton per centimetre.

Extension x

the offset from the natural length, not the length of the spring.

Restoring force

the force exerted by the spring, directed towards the natural length; hence the sign in F = −k·x in equations of motion.

Elastic limit

the extension beyond which the spring no longer returns to its natural length and Hooke’s law no longer applies.

The law holds while the extension stays below the elastic limit, typically a few tens of percent of the natural length for a helical spring; beyond it, the deformation becomes permanent. Compression follows the same law as long as the coils do not touch. The spring is assumed massless and without internal friction.

The equivalent mass is the mass whose weight equals the computed force. In series, the inverse stiffnesses add; in parallel, the stiffnesses add.