Spring force Hooke law:
formula and worked example

What Hooke's law says about springs

Hooke's law describes a simple but powerful relationship: the force a spring exerts is proportional to how far it is stretched or compressed from its resting length. Twice the stretch, twice the pull. It holds for coil springs, many rubber-like elements in a limited range, and even for small bends in a metal beam.

The idea underlies spring scales, suspensions, shock absorbers and weighing mechanisms. By measuring how far a spring moves, you get a direct reading of the force on it.

The relationship and its two inputs

• Spring constant: stiffness in newtons per metre. A value of 250 N/m means each metre of stretch would give 250 N, so each centimetre gives 2.5 N.
• Extension: change from the natural, unloaded length, in metres.

The result is the magnitude of the restoring force. In full vector form there is a minus sign, F = -kx, because the spring pulls back opposite to the displacement. For sizing problems the magnitude is what you want.

Rearranged, the same law gives k = F ÷ x. This is how stiffness is found experimentally: hang a known load, measure the stretch, and divide.

A 250 N/m spring stretched by 4 cm

1. 1Convert extension to metres: 4 cm = 0.04 m.
2. 2Multiply by the spring constant: 250 × 0.04 = 10.
3. 3State the unit: 10 N.
4. 4Sense check by hanging: 10 N supports a mass of about 1.02 kg, since 10 ÷ 9.81 ≈ 1.02.

Because 10 N is about the weight of a one-kilogram bag of rice, the answer is comfortably in the range a small spring scale could report.

How to interpret the force and the limit

The force is the pull a spring applies to whatever is attached to it. In a balance at rest, this equals the weight hanging from it, which is how a spring scale reads.

• Linear zone: force rises in step with extension, so a graph of force against extension is a straight line with slope k.
• Elastic limit: past this the spring stays permanently stretched.
• Safe working range: designers usually keep operation well inside the linear region.

If measured points curve away from a straight line, the spring is out of range, or the load is not truly static. The formula is then only a first approximation.

What stiffness changes

A 10% stiffer spring, 275 N/m instead of 250, exerts 11 N at the same 0.04 m extension, a straight 10% increase. Force responds one for one to both k and x.

Look at it the other way: to carry the same 10 N with a stiffer spring, the stretch shrinks. At 275 N/m, 10 N needs only about 0.036 m. That is why stiff springs are used where small movement is wanted.

Where Hooke's law calculations go wrong

• Using total spring length instead of extension from the natural length.
• Entering centimetres directly, which gives a force 100 times too large.
• Applying the law beyond the elastic limit, where permanent deformation has set in.
• Combining two springs without adding constants for a side-by-side pair or using the reciprocal rule for a chain.

A routine for finding k from a test

1. 1Hang the spring and record its unloaded length.
2. 2Add a known mass and note the new length.
3. 3Compute force as mass times 9.81 and extension as the length difference.
4. 4Divide force by extension to get k, then repeat with a second mass to check linearity.

If the second value of k disagrees noticeably with the first, the spring is either too close to its limit or your length readings are off. Using a few loads and averaging gives a more trustworthy stiffness.

For combinations, two identical springs side by side share the load and their constants add, so two 250 N/m springs behave like one 500 N/m spring. Chained end to end, the stretches add and the effective constant halves, giving 125 N/m. Work out the combined constant before applying the law.