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AP® Physics 1: Algebra-Based · Unit 2 · Topic 2.8

Spring Forces: Revision Notes (Physics 1 2.8)

One-page recap of ideal springs and Hooke's law for the algebra-based course: change in length, the spring constant, force direction, graphs, and the mistakes that cost marks.

Short on time? This page is the recap. For explanations, diagrams and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.

Recap

  • An ideal spring has negligible mass and exerts a force proportional to its change in length from its relaxed length.
  • A stretched spring pulls on what is attached to it; a compressed spring pushes.
  • The spring force always points back toward the position where the spring is relaxed. For a block on a horizontal spring, that is the equilibrium position.
  • The spring constant k (N/m) measures stiffness. Large k = stiff spring.
  • The force does not depend on which way the object is moving, only on where it is.

Key relationships

Relationship Symbols and units Use it to
F_s = −kΔx k in N/m, Δx in m find the spring force with direction
|F_s| = k|Δx| N find the size of the force
k = slope of |F_s| against |Δx| N/m find k from data
kΔx = mg hanging object at rest find k or the stretch
kΔx = mg sin θ object at rest on a smooth slope find the stretch on a slope

Assumptions behind the numbers

  • The spring is ideal: massless, and it obeys Hooke’s law over the range used.
  • The spring is not stretched so far that it stays permanently longer.
  • A hanging object is at rest when you read the stretch, so spring force equals weight.
  • g = 9.8 m/s².

Mistakes to avoid

  1. Using total length for Δx. Subtract the relaxed length first.
  2. Centimetres with k in N/m. Convert to metres.
  3. Spring force drawn in the direction of motion. It points toward the relaxed position.
  4. Treating −kΔx as a negative size. The sign is direction only.
  5. Joining the dots on a graph. Draw one best-fit straight line and take its slope from two points on the line.
  6. Thinking k changes with the load. The stretch changes; k is fixed for the spring.

Quick self-check

  1. A spring with k = 200 N/m is compressed by 4.0 cm. What is the size of the force it exerts? (200 × 0.040 = 8.0 N)
  2. A 0.30 kg mass hangs at rest and stretches a spring by 6.0 cm. Find k. (0.30 × 9.8 ÷ 0.060 = 49 N/m)
  3. A force of 1.2 N stretches a spring by 3.0 cm. What force stretches it by 7.5 cm? (k = 40 N/m, so 40 × 0.075 = 3.0 N)

Next: practice questions.

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