Stretch a spring and it pulls back; squeeze it and it pushes back. For most springs, as long as you don't overstretch them, the force is proportional to how far you deform them.
Key idea
Hooke's law: $F = kx$, where $x$ is the stretch or compression from the spring's natural (relaxed) length and $k$ is the spring constant in N/m. The spring's force points back toward the natural length.
Why it works
A spring is a long coil of metal, and each little piece of the wire twists slightly when you stretch it. Twice the stretch twists each piece twice as much, and the metal pushes back twice as hard. So the force grows in proportion, and $k$ tells how stiff the spring is: the newtons needed for each meter of stretch.
Worked examples
Worked example: Force from stretch
A spring with $k = 300\ \text{N/m}$ is stretched $8\ \text{cm}$. How hard does it pull?
- $x = 8\ \text{cm} = 0.08\ \text{m}$.
- $F = (300)(0.08) = 24\ \text{N}$.
Answer: $24\ \text{N}$
Worked example: Stretch, not length
A spring is $20\ \text{cm}$ long relaxed and $26\ \text{cm}$ long when stretched; $k = 150\ \text{N/m}$. Find the force.
- Stretch: $x = 26 - 20 = 6\ \text{cm} = 0.06\ \text{m}$.
- $F = (150)(0.06) = 9\ \text{N}$.
Answer: $9\ \text{N}$
Worked example: A hanging mass
A $0.5\ \text{kg}$ mass hangs at rest from a spring with $k = 20\ \text{N/m}$. How far is the spring stretched?
- At rest the forces balance: $kx = mg$.
- $$x = \dfrac{(0.5)(9.8)}{20} = 0.245\ \text{m} = 24.5\ \text{cm}.$$
Answer: $24.5\ \text{cm}$
Hooke's law holds only up to a spring's elastic limit. Stretched too far, a spring deforms permanently and no longer follows $F = kx$.
Common mistake
Using the spring's total length instead of its stretch from the natural length. A $26\ \text{cm}$ spring that is $20\ \text{cm}$ long relaxed is stretched only $6\ \text{cm}$. Check: when the spring is at its natural length, your $x$ (and so the force) must be zero.