Hooke’s Law Calculator

Stiffness is the slope of a straight line

Hooke’s law says that the force a spring resists with grows in direct proportion to how far it has been moved from its natural length. Write it as F = kx and the spring constant k is nothing more than the slope of that line, measured in newtons per metre. A 500 N/m spring needs 5 N for every centimetre of stretch, and it needs the same 5 N for the tenth centimetre as it did for the first. Textbooks often write F = -kx, where the minus sign records that the spring pulls back towards its rest position rather than along the displacement. The calculator works in magnitudes, so the sign is dropped.

The proportionality is a property of the material, not a universal truth. It holds only up to the limit of proportionality. Past that the line curves over, and any figure calculated from F = kx will overstate what the spring can still give back. The elastic limit comes at or after the limit of proportionality: up to it the spring still returns to its original length when released, and past it the metal takes a permanent set.

Worked example

A spring of stiffness 500 N/m is pulled 5 cm past its natural length.

• Convert the extension: 5 cm = 0.05 m
• F = k x = 500 × 0.05
• F = 25 N
• E = 0.5 × k × x² = 0.5 × 500 × 0.05²
• E = 0.5 × 500 × 0.0025 = 0.625 J

So a 25 N pull is held by 0.625 J of stored elastic energy. Pull the same spring to 10 cm and the force doubles to 50 N while the energy quadruples to 2.5 J. The force per centimetre readout stays at 5 N/cm throughout, because that figure is just k in other units.

Why the energy is half k x squared

The force is not constant while you stretch a spring. It begins at zero and only reaches kx at full extension. Work is the area under the force against extension graph, and because that graph is a straight line through the origin, the area is a triangle of base x and height kx. Half base times height gives 0.5 k x². Treating the final force as though it applied for the whole stretch would give k x², exactly twice the truth, which is the single most common error in this calculation.

Because the extension is squared, energy climbs much faster than force. Double the stretch for four times the energy; triple it for nine times. Split a bow’s draw into two halves and the second half stores three times what the first half did.

Combining springs follows from the same reasoning. Two identical springs in series each carry the full load and each stretch fully, so the pair moves twice as far for the same force and the effective constant halves: two 500 N/m springs in series behave as 250 N/m. In parallel they share the load, so the pair is twice as stiff at 1000 N/m. Cutting one spring in half doubles its constant, since half the coils stretch half as far under the same pull.

Common mistakes

• Using the total length rather than the extension. A spring 20 cm long at rest and pulled out to 25 cm has an extension of 5 cm, not 25 cm. Everything in this formula is measured from the natural length.
• Mixing centimetres into a constant quoted in N/m. Feeding 5 into F = 500x gives 2500 N instead of 25 N, a hundredfold error. The unit selector on the extension field exists to prevent exactly this.
• Multiplying kx by x for the energy. That gives 1.25 J here instead of 0.625 J. The factor of one half is not decoration; it is the difference between a triangle and a rectangle.
• Extrapolating past the working range. A constant measured over a 5 cm stretch says little about the same spring at 50 cm, where it may already be permanently deformed.

Converting units first? Use the length, force and spring constant conversion tables.