Measure spring stiffness with an energy graph
You will be able to: Design a spring-launch investigation, linearize the data and interpret slope and systematic loss.
How can motion data test an energy model?
A compressed spring launches a cart. Greater compression produces greater speed, but the raw speed graph is inconvenient. Plotting speed squared against compression squared turns the ideal energy prediction into a straight line.
A useful starting point: Mechanical energy can become thermal energy →
Words and symbols before equations
- Linearization
- Transforming plotted quantities so the model predicts a straight line.
- Best-fit slope
- Change of the vertical plotted variable divided by change of the horizontal variable, estimated from the trend.
- Control variable
- A quantity kept fixed so the comparison isolates one relationship.
- Systematic effect
- A repeated influence that biases results, such as friction reducing every measured launch speed.
What this picture assumes
Synthetic model data, not measured student results. Cart released from rest, speed sampled at relaxed spring length; no rolling energy. A fixed fraction of initial spring energy is removed in this illustration. Real losses can have different dependence on compression.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Synthetic slope = 320 s⁻²; inferred k = 160 N/m. True model k = 160 N/m; lost fraction 0%. Repeated measurements and systematic checks are still needed in a real experiment.
- 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 initially stationary cart and ideal spring on a smooth horizontal track, ½kx² = ½mv² when the spring reaches its relaxed length. Thus v² = (k/m)x². The slope of a v²-versus-x² plot is k/m, with units s⁻²; multiply by measured mass to obtain k.
Measure total moving mass, compression from the relaxed position, and speed at the same release location, for example with a photogate and a flag of measured length. Use several compressions within the spring’s linear range and repeat each run. Keep mass, track level, release point and speed-measurement method fixed.
Plot all repeated data and inspect scatter and the intercept rather than forcing a zero intercept automatically. A negative intercept or a trend below the ideal line can signal losses or offsets. If losses depend on compression, the slope may also be biased. Our idealized loss-fraction control shows one possible systematic effect; it is not a model of every laboratory friction mechanism.
A worked example, step by step
A 0.50 kg cart gives v² values 0.8, 3.2 and 7.2 m²/s² for x² values 0.0025, 0.0100 and 0.0225 m². Infer spring stiffness and state the ideal-model assumption.
- Plot x² horizontally in m² and v² vertically in m²/s².
- Use two well-separated points to illustrate the trend slope: (7.2 − 0.8)/(0.0225 − 0.0025) = 320 s⁻².
- k = m × slope = 0.50 × 320 = 160 N/m.
- This inference assumes spring energy transfers to translational cart energy with negligible losses and no significant rotational or spring kinetic energy. Real data should use a best-fit line over all measurements.
The slope is k/m, not k. Label both axes and their units before interpreting the graph.
If every speed measurement is too small, is the inferred k trustworthy?
Compare with an explanation
It can be biased low. Check the speed calibration and energy losses; a straight graph does not remove systematic error.
Predict. Change one thing. Explain.
With mass and stiffness fixed, increase the modeled lost fraction. Predict how the slope and inferred stiffness change. Explain why numerical agreement of a line alone does not validate the physical assumptions.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Synthetic slope = 320 s⁻²; inferred k = 160 N/m. True model k = 160 N/m; lost fraction 0%. Repeated measurements and systematic checks are still needed in a real experiment.
Synthetic model data, not measured student results. Cart released from rest, speed sampled at relaxed spring length; no rolling energy. A fixed fraction of initial spring energy is removed in this illustration. Real losses can have different dependence on compression.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant work, system boundary, energy 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 questionDesign a test of v² ∝ x² for one spring and cart. Specify measurements, controlled conditions, graph and slope interpretation, and one way to assess uncertainty or model limitations.
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Compare with the answer and four-point rubric
- 1 point: Measure cart mass, compression from relaxed length, and speed at a fixed position for several compressions.
- 1 point: Keep mass, track level and release method fixed; use a spring range where Hooke’s law applies.
- 1 point: Plot v² versus x²; fit a line, then calculate k = m × slope and examine the intercept.
- 1 point: Repeat trials and inspect scatter; check calibration, friction and rotational energy rather than assuming a straight line proves the model.
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 1Why square both plotted variables?
The energy equation predicts a linear relation between v² and x².
RECALL 2Why repeat each trial?
To estimate variability and avoid relying on a single possibly atypical measurement.
RECALL 3What does a good fit not establish?
That systematic errors are absent or all model assumptions hold.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Measure spring stiffness with an energy graph
- Ideal spring launch: ½kx² = ½mv².
- Plot v² versus x²: slope = k/m and ideal intercept = 0.
- Estimated k = measured mass × fitted slope.
Remember: The slope is k/m, not k. Label both axes and their units before interpreting the graph.
Conditions: Synthetic model data, not measured student results. Cart released from rest, speed sampled at relaxed spring length; no rolling energy. A fixed fraction of initial spring energy is removed in this illustration. Real losses can have different dependence on compression.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.4, objectives 3.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026 alongside the Fall 2026 clarifications. This is Mechanics Unit 3: Work, Energy, and Power. The unit covers Topics 3.1–3.5. Calculus connects work to force integrals, force to potential-energy derivatives, and power to the rate of energy transfer. Models distinguish object-only and multi-object systems; translational models exclude rotational energy unless explicitly noted. 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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