AP Physics 1 · Unit 2: Forces & Translational Dynamics · Lesson 2.8

Deep Dive: Spring Forces

🔬 Deep Dive

This is your textbook for this topic. Take your time. Read it more than once.

2.8.A.1Concept

The Ideal Spring Model

Like the object model in kinematics, physics uses an idealized version of springs to keep problems clean. An ideal spring has three defining properties:

Negligible mass

The spring itself has no mass. Only the object attached to it matters for dynamics calculations.

Perfectly linear

The force is exactly proportional to the displacement — always, no matter how far it's stretched or compressed.

No energy loss

Ideal springs don't heat up, vibrate at odd frequencies, or wear out. All energy stored is fully returned.

Works in both directions

Extension and compression both obey the same law. Stretch it or squeeze it — same relationship.

💡Real springs approximate these properties well for small displacements. AP Physics 1 always uses the ideal model — don't worry about springs that stretch past their elastic limit or have significant mass.

2.8.A.2Math

Hooke's Law

The force a spring exerts is directly proportional to how far it's been displaced from its natural (equilibrium) position:

F_s = −kΔx

Three quantities, each with a specific meaning:

F_sSpring force (N) — the force the spring exerts on the object attached to it.

kSpring constant (N/m) — how stiff the spring is. Larger k = harder to stretch.

ΔxDisplacement from equilibrium (m). Positive = stretched, negative = compressed.

Drag the sliders to stretch or compress the spring. The force arrow and graph update live — watch force always point back toward equilibrium.

Spring constant k200 N/m

Displacement Δx+0.15 m

Stretched — force pulls LEFT (toward equilibrium)

F vs. Δx — slope = k

Spring constant

200 N/m

k — stiffness

Displacement

+0.15 m

Δx from equilibrium

Spring force

-30.0 N

F_s = −kΔx

ExampleWorked Example — Finding Spring Force

A spring with k = 300 N/m is stretched 0.08 m from equilibrium. What force does it exert? In which direction?

2.8.A.3Concept⚠ Watch Out

The Restoring Force

The negative sign in Hooke's Law is the most important character in the equation. It tells you that spring force and displacement always point in opposite directions. Displace the object to the right — force points left. Displace left — force points right. The spring always pushes or pulls back toward its equilibrium position. That's what "restoring" means.

⚠️Don't drop the negative sign. Students often write F_s = kΔx (positive) and then manually assign direction. That works if you're careful, but the AP exam expects you to handle signs algebraically. Keep the negative sign in the equation and let the algebra tell you the direction.

🔑A restoring force is the physical mechanism behind all oscillatory motion. Any time something oscillates — a mass on a spring, a pendulum, a vibrating string — it's because a restoring force keeps pulling it back toward equilibrium. Hooke's Law is the simplest and most important example you'll encounter.

2.8.A.4Math

The Force vs. Displacement Graph

Because Hooke's Law is linear — force is directly proportional to displacement — the F vs. Δx graph is a straight line through the origin. The slope of that line is the spring constant k.

🔑The slope equals k. This is the most tested graph interpretation for springs on the AP exam. If you're given a force-displacement graph and asked for the spring constant, calculate the slope: k = ΔF / ΔΔx = rise / run.

Your infographic shows this clearly: doubling Δx doubles F. Triple Δx triples F. The relationship is exact and linear. A steeper line on the graph means a stiffer spring — the same displacement produces more force.

ExampleGuided Example — Reading k from a Graph

A force vs. displacement graph shows a straight line through the origin passing through the point (0.10 m, 45 N). What is the spring constant?

Step 1 — Identify slope

k = slope = ΔF / ΔΔx = 45 N / 0.10 m = 450 N/m

2.8.A.5Math

Applying Hooke's Law

Hooke's Law connects directly to Newton's Second Law. When a mass on a spring is in equilibrium, the spring force balances gravity. When it's not in equilibrium, the net spring force drives an acceleration — leading to the oscillatory motion of Unit 7.

ExampleWorked Example — Mass on a Vertical Spring

A 0.5 kg mass hangs at rest from a vertical spring with k = 100 N/m. How far is the spring stretched from its natural length? Use g = 10 m/s².

💡Looking ahead to Unit 7: The equilibrium position for a vertical spring-mass system is at Δx = mg/k, not at the spring's natural length. When you analyze oscillations in Unit 7, measure displacement from this equilibrium position, and gravity effectively drops out of the equation.

← Back to Lesson 2.8 Next: Lesson 2.9 →Circular Motion — the season finale. Everything in Unit 2 comes together.

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