Hooke's Law is the cornerstone of elasticity: the force a spring exerts is directly proportional to how far it is stretched or compressed, F = kx. The constant k is the spring's stiffness in newtons per metre, and x is the displacement from its natural length. Pull twice as far and the spring pulls back twice as hard. The energy you store doing so is ½kx² — the triangular area under the force–extension line — and because x is squared, that energy climbs much faster than the force itself.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: Hooke's Law F = kx and elastic energy ½kx².
The Hooke's Law equations
The spring constant k is the slope of the force–extension graph, so you can read it off any force–displacement pair as F/x. The stored energy is the area under that line, a triangle giving ½kx². To find how far a known force stretches a spring, use x = F/k. All of this applies only in the elastic region; once a spring is overstretched past its elastic limit, the proportionality no longer holds.
Worked example — stretching a spring
Scenario: A spring of stiffness 200 N/m is stretched 0.1 m. What force does it exert, and how much energy is stored?
The spring pulls back with 20 N and stores 1 joule. Stretch it to 0.2 m and the force doubles to 40 N, but the stored energy quadruples to 4 J — the square on x at work. Halve the stretch to 0.05 m and the force is 10 N with just 0.25 J stored. This square law is why winding a spring or bow further always feels disproportionately harder near the end.
Frequently Asked Questions
F = k × x. e.g. 200 N/m × 0.1 m = 20 N. Also k = F/x and x = F/k.
Stiffness k in N/m — force per metre of stretch. It's the slope of the force–extension graph.
E = ½kx². For k=200, x=0.1: ½×200×0.01 = 1 J. Double the stretch → 4× energy.
The max stretch where F = kx still holds and the spring returns to shape. Beyond it, permanent deformation.
The restoring force opposes the displacement. For magnitude use F = kx; the sign matters for oscillations.