Spring force and the reference length
You will be able to: Relate signed spring deformation to force and distinguish relaxed length from loaded equilibrium.
What does the minus sign in Hooke’s law mean?
Stretch a horizontal spring 0.10 m to the right. If k = 100 N/m, it pulls the attached block 10 N to the left. Compress it the same amount and the force reverses.
A useful starting point: Static friction adjusts up to a limit →
Words and symbols before equations
- Spring constant k
- Stiffness, measured in N/m.
- Deformation x
- Signed change from the spring’s relaxed length along the chosen axis.
- Restoring force
- A force opposing the spring’s deformation.
- Equilibrium
- A position where all forces on the object sum to zero.
What this picture assumes
Ideal horizontal spring, deformation from relaxed length; spring mass neglected. The force is on the attached block and opposes deformation.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- x = 0.1 m from relaxed length; F_s = -10 N. Force points left.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For an ideal horizontal spring, F_s = −kx. A positive extension gives a negative force on the attached block; compression gives the opposite sign. The equation uses deformation, not the total spring length.
The slope of a spring-force versus deformation graph is −k. A graph of the balancing applied force against extension instead has slope +k. Identify whose force is plotted before interpreting the sign.
For a vertical hanging mass, equilibrium extension is mg/k because spring force balances weight. The spring force there is not zero. Only the net force is zero; a small displacement from this loaded equilibrium produces a net restoring force. Ideal springs have negligible mass and a linear force law within the assumed elastic range.
A worked example, step by step
A 2 kg mass hangs at rest from a vertical spring with k = 100 N/m. Find its extension and spring force.
- The downward weight is 20 N.
- At equilibrium the spring pulls upward with 20 N.
- Use kx = mg: extension x = 20/100 = 0.20 m.
- The spring is stretched at equilibrium; its 20 N force balances gravity rather than vanishing.
Loaded equilibrium and relaxed spring length are different when another force, such as gravity, is present.
Is the spring force zero at the equilibrium of a hanging mass?
Compare with an explanation
No. It is upward with magnitude mg; the total force is zero.
Predict. Change one thing. Explain.
Move deformation from positive through zero to negative while keeping k fixed. Predict force direction. Then double k at the same deformation.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
x = 0.1 m from relaxed length; F_s = -10 N. Force points left.
Ideal horizontal spring, deformation from relaxed length; spring mass neglected. The force is on the attached block and opposes deformation.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, system boundary, acceleration or calculus relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAn ideal horizontal spring has k = 80 N/m. Find its force at x = +0.15 m and x = −0.10 m, describe directions and identify the force-graph slope.
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Compare with the answer and four-point rubric
- 1 point: At +0.15 m, F_s = −12 N.
- 1 point: At −0.10 m, F_s = +8 N.
- 1 point: The first force points left; the second points right.
- 1 point: The F_s-versus-x slope is −80 N/m.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is x in Hooke’s law?
Signed deformation from relaxed length.
RECALL 2What does the minus sign express?
The spring force opposes that deformation.
RECALL 3How does doubling k change force at fixed x?
It doubles the force magnitude.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Spring force and the reference length
- F_s = −kx for signed deformation from relaxed length.
- k has units N/m.
- Vertical equilibrium: kx_eq = mg.
Remember: Loaded equilibrium and relaxed spring length are different when another force, such as gravity, is present.
Conditions: Ideal horizontal spring, deformation from relaxed length; spring mass neglected. The force is on the attached block and opposes deformation.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.8.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.8, objectives 2.8.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 2: Force and Translational Dynamics. The unit covers Topics 2.1–2.10. Calculus is introduced where it is needed for continuous mass and velocity-dependent forces. Shell theorem is applied without requiring a proof; spring combinations are purely series or purely parallel. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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