MotionLab

Hooke's Law Calculator

The linear relationship between how far a spring stretches and how hard it pulls back.

Spring constant k (N/m)

Displacement (m)

Restoring force (N)

F = k · x

What you entered

Spring constant k (N/m): 200Displacement (m): 0.05

1
F = k × x
200 × 0.05= 10 N

Result

Restoring force (N): 10

A spring with k=200 N/m stretched 0.05m exerts 10 N.

Spring force from stiffness and displacement

Hooke's law: force equals the spring constant times the displacement from rest. The relationship is linear within the elastic limit.

Spring constant (N/m)	Displacement (m)	Force (N)
50	0.3	15.00
100	0.1	10.00
200	0.15	30.00
500	0.05	25.00
1000	0.02	20.00

Reading down, the spring constant rises twentyfold while the force barely changes, because the displacement falls to compensate - a stiff spring needs far less stretch for the same force. Stretch a spring twice as far and the force exactly doubles, which is what makes spring scales linear and easy to calibrate. The law holds only up to the elastic limit: beyond it the material deforms permanently and the linear relationship fails entirely, which is why an over-stretched spring never returns to its original length.

Only valid within the elastic limit

Hooke's law holds as long as the spring isn't stretched past its elastic limit — push a spring too far and it deforms permanently, at which point this linear relationship no longer applies and the spring won't return to its original shape.

Spring constant is a stiffness rating

A higher spring constant k means a stiffer spring that resists stretching more — this single number characterizes a spring's behavior completely within its elastic range, which is why it's the key spec on any spring's datasheet.

Frequently asked questions

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

F = kx = 200 × 0.15 = 30 N. That's about the force of holding a 3 kg weight. The spring pulls back toward its natural length with exactly this force — double the stretch, double the force.

I hang a 2 kg mass on a spring and it stretches 4 cm. What's the spring constant?

F = mg = 2 × 9.81 = 19.62 N. k = F/x = 19.62/0.04 = 490.5 N/m. This is a moderately stiff spring — car suspension springs have k ≈ 20,000-80,000 N/m, while a Slinky has k ≈ 1 N/m.

What happens if I stretch a spring past its elastic limit?

The spring deforms permanently — Hooke's law no longer applies, and the spring won't return to its original length when released. The force-displacement relationship becomes nonlinear and the spring is effectively ruined for precise applications.

How is Hooke's law related to spring energy?

The energy stored in a stretched spring is PE = ½kx². This comes from integrating F = kx over displacement. A spring stretched twice as far stores four times the energy. See the spring energy calculator for stored energy problems.

Does Hooke's law apply to things that aren't metal springs?

Yes — rubber bands, bungee cords, bones, and even atomic bonds follow Hooke's law for small deformations. The 'spring constant' varies enormously: a steel beam has k in millions of N/m, while a hair elastic might be 10 N/m.