Hooke's Law Practice Problems and Solutions

What Hooke's Law Actually Is

Hooke's Law describes how springs behave when you stretch or compress them. The formula is F = -kx.

Breaking that down:

• F = restoring force (the pushback from the spring)
• k = spring constant (how stiff the spring is, measured in N/m)
• x = displacement from equilibrium (how far you stretched/compressed it)
• The negative sign means the force pushes opposite to the displacement direction

That's it. No magic, no complexity. Just a linear relationship between force and displacement for ideal springs.

The Formula You Need to Memorize

F = -kx

Or rearranged depending on what you're solving for:

• k = F/x — find the spring constant
• x = F/k — find the displacement
• F = kx — find the force (ignoring the negative sign for magnitude problems)

The negative sign only matters when you're dealing with direction. Most textbook problems ask for magnitude, so you can drop it.

Units Matter — Get This Right

Spring constant k is measured in Newtons per meter (N/m).

Displacement x is in meters (not centimeters).

Force F is in Newtons.

Most students lose points because they forget to convert centimeters to meters. Don't be that person.

Spring Constant Comparison

Real springs don't always behave perfectly linearly, but Hooke's Law works fine for small displacements.

Practice Problem 1: Finding the Force

Problem: A spring with k = 250 N/m is compressed by 0.08 m. What force is required?

Solution:

Use F = kx

F = 250 × 0.08

F = 20 N

Simple multiplication. That's the whole problem.

Practice Problem 2: Finding the Spring Constant

Problem: A spring stretches 0.15 m when you hang a 6 kg mass from it. What is the spring constant?

Solution:

First, find the force from the mass:

F = mg = 6 × 9.8 = 58.8 N

Now solve for k:

k = F/x = 58.8 / 0.15

k = 392 N/m

Two steps. Find force, then divide by displacement.

Practice Problem 3: Finding the Displacement

Problem: A spring with k = 1,200 N/m supports a weight of 360 N. How far does it stretch?

Solution:

x = F/k = 360 / 1,200

x = 0.3 m (or 30 cm)

This is useful for figuring out how much a real suspension system compresses under load.

Practice Problem 4: Energy Stored in a Spring

Problem: A spring with k = 800 N/m is compressed 0.12 m. What potential energy is stored?

Solution:

Elastic potential energy uses:

PE = ½kx²

PE = ½ × 800 × (0.12)²

PE = 400 × 0.0144

PE = 5.76 J

Remember: it's ½kx², not kx². Forgetting the ½ is a common mistake.

Practice Problem 5: Two Springs in Series

Problem: Two springs with k₁ = 300 N/m and k₂ = 600 N/m are connected end-to-end. What is the effective spring constant?

Solution:

For springs in series:

1/k_total = 1/k₁ + 1/k₂

1/k_total = 1/300 + 1/600

1/k_total = 0.00333 + 0.00167 = 0.005

k_total = 1/0.005 = 200 N/m

Adding springs in series always makes the system weaker. Two identical springs in series give you half the spring constant.

Practice Problem 6: Two Springs in Parallel

Problem: Two identical springs with k = 400 N/m are mounted side-by-side. What is the effective spring constant?

Solution:

For springs in parallel:

k_total = k₁ + k₂

k_total = 400 + 400 = 800 N/m

Parallel springs double the stiffness. This is how vehicle suspension works — multiple springs share the load.

Common Mistakes That Cost You Points

• Forgetting the negative sign — only matters for direction, but some professors deduct for it
• Using centimeters instead of meters — k = 50 N/cm is wrong; convert to N/m
• Confusing force with energy — F = kx, PE = ½kx²
• Mixing up series and parallel formulas — series: 1/k = 1/k₁ + 1/k₂; parallel: k = k₁ + k₂
• Using g = 10 m/s² when the problem specifies 9.8 — match what the problem gives you

How to Solve Any Hooke's Law Problem

1. Identify what the problem asks for — force, displacement, spring constant, or energy
2. Write down what you know — circle the given values
3. Pick the right formula — F = kx for force problems, PE = ½kx² for energy
4. Convert units — everything must be in meters, Newtons, N/m
5. Solve algebraically — isolate the unknown variable
6. Plug in numbers — calculate carefully
7. Check your answer — does the magnitude make sense?

When Hooke's Law Breaks Down

Hooke's Law only works up to the elastic limit of a material. Past that point:

• The material deforms permanently
• The relationship stops being linear
• You enter plastic deformation territory

The yield strength is where permanent deformation starts. The ultimate strength is where it breaks. Hooke's Law doesn't apply past the yield point.

For most homework problems, you can ignore this — they assume ideal conditions. But in real engineering, it matters.