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Hooke's Law — F = kx

Hooke's law F = kx explained: spring force, the spring constant k, elastic limits, and a worked example.

What it means

A spring pushes back in proportion to how far you displace it. Stretch it twice as far and it pulls back twice as hard — the minus sign often written as F = −kx simply records that the force opposes the displacement, always dragging the mass back toward equilibrium. The spring constant k is the spring's personality: a stiff car suspension has a huge k, a soft screen-door spring a small one. Because the force grows with displacement, springs naturally produce the sinusoidal back-and-forth of simple harmonic motion.

Variables

SymbolNameUnitMeaning
FFSpring forceNRestoring force exerted by the spring
kkSpring constantN/mSpring constant — stiffness of the particular spring
xxExtensionmDisplacement from the equilibrium (unstretched) position

Assumptions

  • Deformation stays within the spring's elastic limit
  • Spring mass is negligible
  • Force applied along the spring's axis (no bending)

Derivation

Hooke's law is empirical: within the elastic limit, measured stretch is directly proportional to applied force. The constant of proportionality k characterizes the particular spring. Beyond the elastic limit the linear relation breaks down and permanent deformation begins.

Worked example

Problem

A spring with stiffness k = 200 N/m is stretched 0.15 m from equilibrium. What restoring force does it exert?

Given:

  • k = 200 N/m
  • x = 0.15 m

  1. Write Hooke's law: F=kxF = kx

  2. Substitute: F=200×0.15F = 200 \times 0.15

  3. Multiply: F=30 NF = 30\ \mathrm{N}

30 N (directed back toward equilibrium)

Pulling the spring 15 cm develops 30 N of restoring force — about the weight of a 3 kg mass. Release it and that force accelerates the mass back, which is the engine of every spring oscillation.