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Hooke's Law — Definition, Formula & Examples

Hooke's Law is the rule that the force needed to stretch or compress a spring is proportional to how far you stretch or compress it. If you double the stretch, you double the force.

For an ideal elastic object within its elastic limit, the restoring force FF is a linear function of the displacement xx from the equilibrium (natural) length: F=−kxF = -kx, where k>0k > 0 is the spring constant. The negative sign shows that the restoring force acts opposite to the displacement. The magnitude of the applied force is ∣F∣=k∣x∣|F| = k|x|.

Key Formula

F=kxF = kx
Where:
  • FF = Magnitude of the force applied to the spring, in newtons (N)
  • kk = Spring constant (stiffness), in newtons per meter (N/m)
  • xx = Extension or compression: the change from the spring's natural length, in meters (m)

How It Works

Hooke's Law is a direct variation model, y=kxy = kx, applied to springs. The spring constant kk is the slope of the force-versus-extension graph, so a stiffer spring has a steeper line. To use the law, first find kk from one known force and stretch. Then use F=kxF = kx to predict any other force or stretch. The model works only up to the elastic limit. Past that point the spring deforms permanently and the graph stops being a straight line.

Worked Example

Problem: A force of 20 N stretches a spring 0.10 m beyond its natural length. How far will a 50 N force stretch the same spring?
Step 1: Find the spring constant from the known force and extension.
k=Fx=200.10=200 N/mk = \frac{F}{x} = \frac{20}{0.10} = 200 \text{ N/m}
Step 2: Set up Hooke's Law with the new force and solve the equation for xx.
50=200x⇒x=50200=0.25 m50 = 200x \quad\Rightarrow\quad x = \frac{50}{200} = 0.25 \text{ m}
Step 3: Convert to centimeters to check that the answer is reasonable. The force is 2.5 times larger, so the stretch should also be 2.5 times larger.
0.25 m=25 cm=2.5×10 cm0.25 \text{ m} = 25 \text{ cm} = 2.5 \times 10 \text{ cm}
Answer: The spring stretches 0.25 m (25 cm).

Another Example

Problem: A car suspension spring has k=5000k = 5000 N/m. What force compresses it by 0.04 m?
Step 1: Substitute the spring constant and the compression into the formula.
F=kx=5000×0.04F = kx = 5000 \times 0.04
Step 2: Multiply.
F=200 NF = 200 \text{ N}
Answer: A force of 200 N compresses the spring by 0.04 m.

Visualization

Why It Matters

Hooke's Law is often the first real-world example of a linear model in Algebra 1 and physics, because the slope has a physical meaning: stiffness. Engineers use it to design car suspensions, mattresses, and bathroom scales. In AP Physics it leads directly to spring potential energy, 12kx2\tfrac{1}{2}kx^2, and to simple harmonic motion.

Common Mistakes

Mistake: Using the spring's total length for xx instead of the change in length.
Correction: xx is the extension or compression only. A 0.30 m spring stretched to 0.40 m has x=0.10x = 0.10 m, not 0.40 m. Subtract the natural length first.
Mistake: Mixing units, such as using centimeters with a spring constant in N/m.
Correction: If kk is in N/m, convert xx to meters before you multiply. Using 5 cm as the number 5 instead of 0.05 m makes the force 100 times too large.

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