CourseCourseQ BankQuestion BankDocsDocuments

2.8 Spring Forces

Syllabus
2024
Topic
2.8
Level
—

Learning objectives

2.8A—Describe the force exerted on an object by an ideal springDescribe the force exerted on an object by an ideal spring• An ideal spring has negligible mass and exerts a force that is proportional to the change in its length as measured from its relaxed length.• The magnitude of the force exerted by an ideal spring on an object is given by Hooke’s law:• The force exerted on an object by a spring is always directed toward the equilibrium position of the object–spring system. TOPIC 2.9 Circular Motion | AP Physics 1: Algebra-Based Course and Exam Description

An ideal spring pushes back toward equilibrium

Define the ideal model

An ideal spring has negligible mass. Its force is proportional to the change in length Δx⃗\Delta\vec x measured from its relaxed length.

Use Hooke's law

F⃗s=−k Δx⃗\vec F_s=-k\,\Delta\vec x

Read the restoring direction

Spring change Displacement sign Spring-force direction
Stretched to the right Δx>0\Delta x>0 Fs<0F_s<0: left, back toward equilibrium
Compressed to the left Δx<0\Delta x<0 Fs>0F_s>0: right, back toward equilibrium
Relaxed Δx=0\Delta x=0 Fs=0F_s=0

Calculate and interpret

Worked example: take right as positive. A spring with k=200 N/mk=200\,\text{N/m} is stretched Δx=+0.030 m\Delta x=+0.030\,\text{m}.

Fs=−kΔx=−(200 N/m)(+0.030 m)=−6.0 NF_s=-k\Delta x=-(200\,\text{N/m})(+0.030\,\text{m})=-6.0\,\text{N}.

The spring exerts a 6.0 N6.0\,\text{N} force to the left, toward equilibrium.

Separate sign and magnitude

The minus sign gives direction; force magnitude is ∣Fs∣=k∣Δx∣|F_s|=k|\Delta x|. The spring constant kk has units N/m\text{N/m} and measures stiffness. Hooke's law is the ideal linear model: doubling the displacement doubles the force magnitude within that model.