AP Physics 1 · Unit 2 · Topic 2.8

Spring Forces

An ideal spring produces a restoring force proportional to its change in length. The force points along the spring toward the configuration that reduces the deformation, providing a direct connection between force diagrams, linear graphs, and Newton's laws.

Titration response · Spring ForcesThe equivalence region produces the steepest change; its location depends on stoichiometry and acid-base strength.

Learning Goals

After this lesson, you should be able to:

define deformation relative to a spring's relaxed length;
apply Hooke's law with correct vector direction and units;
distinguish spring force from the net force on an attached object;
calculate stiffness from force–deformation data;
analyze horizontal and vertical spring systems with Newton's laws;
interpret the slope and intercept of a spring-force graph; and
design an experiment that tests the range over which a spring is approximately ideal.

1. Ideal-Spring Model

An ideal spring has negligible mass and responds linearly over its modeled range. Choose the spring axis as \(x\), and define deformation from the relaxed length:

\[\Delta x=L-L_0\]

Here \(L_0\) is relaxed length. Positive \(\Delta x\) can represent extension and negative \(\Delta x\) compression, depending on the chosen axis.

2. Hooke's Law

The force exerted by an ideal spring on the attached object is

\[\vec F_s=-k\Delta\vec x\]

The spring constant \(k\), measured in \(\mathrm{N/m}\), describes stiffness. A larger \(k\) means more force is required for the same deformation. The minus sign communicates the restoring direction; it does not make the force magnitude negative.

3. Compression, Relaxation, and Extension

**Spring Forces**Topic-matched fine pixel-art model · use the lesson's equations and labeled diagrams for exact scientific meaning.

4. Magnitude and Direction Are Separate Steps

The magnitude is

\[F_s=k|\Delta x|\]

Then determine direction from the configuration: a stretched spring pulls its attachment toward the spring, while a compressed spring pushes it away. State the axis before assigning a signed component.

5. Relaxed Length Versus System Equilibrium

Relaxed length is the spring length at which the spring itself exerts zero force. The equilibrium position of an attached object is the location where all forces on that object sum to zero.

For a vertical spring, gravity shifts the object–spring equilibrium below the relaxed-length position. The spring force there is not zero; it balances weight. When displacement is measured from this shifted equilibrium, the net force is restoring.

6. Worked Example: Determine the Spring Constant

Question: A spring is stretched \(0.080\,\mathrm m\) from its relaxed length by a \(12.0\,\mathrm N\) force while held at rest. Find \(k\).

\[k=\frac{F_s}{|\Delta x|}=\frac{12.0}{0.080}=150\,\mathrm{N/m}\]

Answer: The spring constant is \(150\,\mathrm{N/m}\). The external holding force and the spring force have equal magnitudes only because the endpoint is in equilibrium.

7. Worked Example: Horizontal Dynamics

Question: A \(2.0\,\mathrm{kg}\) cart is attached to a \(120\,\mathrm{N/m}\) horizontal spring on a frictionless track. At an instant when the spring is stretched \(0.15\,\mathrm m\), find the cart's acceleration.

\[F_s=-k\Delta x=-(120)(0.15)=-18\,\mathrm N\]

\[a_x=\frac{F_s}{m}=\frac{-18}{2.0}=-9.0\,\mathrm{m\,s^{-2}}\]

Answer: The acceleration is \(9.0\,\mathrm{m\,s^{-2}}\) toward the relaxed position. This calculation describes one instant because the force changes with position.

8. Vertical Spring Equilibrium

For a mass hanging motionless from a vertical spring, the upward spring force balances downward weight:

\[k\Delta L=mg\]

\[\Delta L_{\mathrm{eq}}=\frac{mg}{k}\]

A \(0.50\,\mathrm{kg}\) mass on a \(98\,\mathrm{N/m}\) spring produces equilibrium extension

\[\Delta L_{\mathrm{eq}}=\frac{(0.50)(9.8)}{98}=0.050\,\mathrm m\]

The spring is extended \(5.0\,\mathrm{cm}\) at system equilibrium.

9. Force–Deformation Graph

A graph of spring-force component versus deformation is a straight line through the origin for an ideal spring:

\[F_{s,x}=-k\Delta x\]

Its slope is \(-k\). A graph of the magnitude of an applied equilibrium force versus \(|\Delta x|\) instead has slope \(+k\). Always read the axis labels before interpreting the sign.

Curvature or failure to return to the original relaxed length signals that the ideal linear model is no longer adequate.

10. Proportional Reasoning

Within the Hooke's-law range:

doubling deformation doubles force magnitude;
doubling \(k\) doubles force at the same deformation;
for the same force, doubling \(k\) halves deformation;
reversing deformation reverses force direction without changing magnitude.

\[\frac{F_2}{F_1}=\frac{k_2|\Delta x_2|}{k_1|\Delta x_1|}\]

11. Two Springs in Parallel

If two ideal springs connect the same object to a wall in parallel and undergo the same displacement, their forces add:

\[F_{\mathrm{net}}=-(k_1+k_2)x\]

\[k_{\mathrm{eq}}=k_1+k_2\]

The combination is stiffer than either spring alone because both oppose the deformation.

12. Two Springs in Series

For massless ideal springs in series at equilibrium, the same force magnitude passes through both while total extension adds:

\[x=x_1+x_2=\frac{F}{k_1}+\frac{F}{k_2}\]

\[\frac{1}{k_{\mathrm{eq}}}=\frac{1}{k_1}+\frac{1}{k_2}\]

The series combination is less stiff than either individual spring. This is an extension of Hooke's-law and system reasoning rather than a separate force law.

13. A Spring Force Is Not Always the Net Force

An attached object can also experience gravity, normal force, friction, tension, or an applied force. Write

\[\sum\vec F=\vec F_s+\vec F_g+\vec N+\vec f+\cdots=m\vec a\]

Do not set \(F_s=ma\) unless the free-body diagram shows that the spring is the only unbalanced force in that direction.

14. Investigation: Test Hooke's Law

Mount a spring vertically beside a meter scale and add known masses in small increments.

Measure relaxed length before loading.
After each mass comes to rest, record extension and applied load \(mg\).
Plot force magnitude versus extension and fit a straight line.
Use the slope to estimate \(k\), with units and uncertainty.
Unload in reverse order and check whether the spring returns to the same lengths.
Identify the data range that supports a linear model; do not overload the spring.

A nonzero intercept may indicate zero-position error, preload, spring mass effects, or an unaccounted attachment force.

15. Common Traps and Key Takeaways

Measure deformation from relaxed length, not automatically from the wall or floor.
The minus sign gives direction; force magnitude is \(k|\Delta x|\).
Relaxed length and system equilibrium can differ when other forces act.
Spring force is not necessarily net force; use a complete free-body diagram.
Keep centimeters and meters consistent when using \(k\) in \(\mathrm{N/m}\).
An ideal spring is a model with a limited linear range; permanent deformation invalidates it.
On a force–deformation graph, slope and sign depend on which force and axis are plotted.

Checkpoint

A \(200\,\mathrm{N/m}\) horizontal spring is compressed \(0.060\,\mathrm m\) to the left of its relaxed position while attached to a \(1.5\,\mathrm{kg}\) cart. Determine the spring-force vector and the cart's acceleration at that instant if friction is negligible. Then state how both answers change if the compression doubles.

Concept scope cross-checked with the College Board AP Physics 1 course framework and the open educational references OpenStax University Physics: Common Forces and OpenStax Physics: Hooke's Law and Simple Harmonic Motion. Explanations, diagram, examples, and investigation are independently written for Insomnia Studyroom.

Official curriculum reference: College Board AP Physics 1 course page. The explanation and worked example are independently written for this study site.