Spring forces and Hooke’s law
You will be able to: Relate spring force to signed deformation and infer stiffness from data.
What does a spring’s restoring force restore?
Stretch a light spring by 2 cm, then 4 cm. Within its elastic linear range, the second stretch requires twice the holding force. The spring resists deformation in either direction.
A useful starting point: Net force and acceleration →
Words and symbols before equations
- Relaxed length L₀
- Length when the spring is neither stretched nor compressed.
- Deformation x
- Signed change from relaxed length, in meters.
- Spring constant k
- Stiffness in N/m; a larger value means more force for the same deformation.
What this picture assumes
Horizontal ideal spring anchored on the left. Right is positive. Force is on the endpoint; lengths are schematic. Deformation stays in the assumed linear range.
Connect the picture to the physics
For a horizontal spring fixed on the left, let right be positive and x be the endpoint displacement from its relaxed position. The spring’s force on the endpoint is F_s=−kx: stretch right produces a leftward force; compression produces a rightward force.
The minus sign expresses direction, not a negative stiffness. A graph of signed spring force versus x has slope −k. A graph of holding-force magnitude versus extension has slope +k.
A hanging spring can be stretched while the mass is in equilibrium: spring force balances gravity. There the equilibrium position is not the relaxed length. The net restoring force about equilibrium and the spring force measured from relaxed length must not be confused.
A worked example, step by step
A spring has k=100 N/m and is stretched 0.04 m to the right from its relaxed position. Find its force. Then hang a 0.5 kg mass at rest from that spring using g=10 m/s².
- Horizontal case: F_s=−100×0.04=−4 N, meaning 4 N left.
- The extension 0.04 m is 4 cm; converting cm to m matters.
- Hanging equilibrium: spring force must balance weight, so kx=mg=5 N.
- Extension x=5/100=0.05 m, or 5 cm. Net force is zero although spring force is not zero.
Use change from relaxed length in Hooke’s law, not the spring’s full length.
Does a stretched hanging spring have zero spring force at equilibrium?
Compare with an explanation
No. It exerts an upward force equal to weight; the net force is zero.
Predict. Change one thing. Explain.
Hold k fixed. Move x through negative, zero and positive values. Compare the sign of the force and the slope of the force–deformation graph. Then double k.
Spring force -4 N. Grey line marks the relaxed endpoint. Deformation scale 1200 drawing units/m; force-arrow scale 4 drawing units/N.
Horizontal ideal spring anchored on the left. Right is positive. Force is on the endpoint; lengths are schematic. Deformation stays in the assumed linear range.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use a force or motion 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 questionMeasured extension and holding-force pairs are (0.02 m,2 N), (0.04 m,4 N), (0.06 m,6 N). (a) Specify graph axes, (b) find k, (c) predict extension under 3 N, and (d) describe one quality check in the experiment.
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Compare with the answer and four-point rubric
- 1 point: Extension on horizontal axis; holding force on vertical axis.
- 1 point: Slope 100 N/m.
- 1 point: 0.03 m.
- 1 point: Measure the relaxed length, repeat readings, and check that the graph stays linear without permanently deforming the spring.
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 length goes into Hooke’s law?
Signed deformation from relaxed length.
RECALL 2What does k measure?
Stiffness in N/m.
RECALL 3Why the minus sign?
Force opposes the endpoint’s displacement from the relaxed position in the stated convention.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Spring forces and Hooke’s law
- F_s=−kx for the signed horizontal convention described.
- At vertical static equilibrium: k(extension)=mg.
- Ideal spring; negligible spring mass and linear elastic range.
Remember: Use change from relaxed length in Hooke’s law, not the spring’s full length.
Conditions: Horizontal ideal spring anchored on the left. Right is positive. Force is on the endpoint; lengths are schematic. Deformation stays in the assumed linear range.
Refresh Kid · 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. Fall-2026 corrections also checked. 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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