Connected objects and the system boundary
You will be able to: Use a combined-system equation for acceleration and a single-object equation for tension.
How can the same setup need two different force equations?
A 3 kg cart on a frictionless table connects over an ideal pulley to a hanging 2 kg mass. Both speed up with the same acceleration magnitude because the string stays taut and does not stretch.
A useful starting point: From net force to a motion function →
Words and symbols before equations
- Constraint
- A geometric condition connecting the allowed motions.
- Tension T
- Pull exerted by the ideal string.
- Combined system
- Both moving masses considered together for the acceleration calculation.
What this picture assumes
Frictionless table, taut massless inextensible string, negligible-inertia frictionless pulley. g = 10 m/s². Arrows on each dot are actual forces on that object.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Shared acceleration magnitude 4 m/s²; tension 12 N. Cart net = 12 N toward pulley; hanger net = 8 N downward.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Choose +x toward the pulley for the cart and positive downward for the hanger. These axes give the same scalar a for the string constraint. For the cart, T = m₁a. For the hanger, m₂g − T = m₂a.
Add the equations: the equal tension magnitudes cancel, leaving m₂g = (m₁ + m₂)a. The hanger’s weight accelerates both masses. Then return to one object to find tension.
Tension is smaller than m₂g while the hanger accelerates downward. It would equal the hanger’s weight only if its vertical acceleration were zero. A massive pulley, stretching string or friction requires a changed model.
A worked example, step by step
Find acceleration and tension for m₁ = 3 kg on the table and m₂ = 2 kg hanging, with g = 10 m/s².
- Cart: T = 3a. Hanger: 20 − T = 2a.
- Add: 20 = 5a, so a = 4 m/s².
- Use the cart: T = 3(4) = 12 N.
- Check hanger: 20 − 12 = 8 N = 2(4), consistent with downward acceleration.
The hanging weight is not the tension when the hanging mass accelerates.
Why cancel tension in the sum but still calculate it afterward?
Compare with an explanation
It is internal to the combined moving system but external to each individual mass.
Predict. Change one thing. Explain.
Increase only the tabletop mass. Predict what happens to acceleration and tension, then check each object’s force balance.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Shared acceleration magnitude 4 m/s²; tension 12 N. Cart net = 12 N toward pulley; hanger net = 8 N downward.
Frictionless table, taut massless inextensible string, negligible-inertia frictionless pulley. g = 10 m/s². Arrows on each dot are actual forces on that object.
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 questionA 6 kg cart connects to a 2 kg hanging mass in the stated ideal setup. Find acceleration, tension and the hanger’s net force.
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Compare with the answer and four-point rubric
- 1 point: Combined equation: 20 = 8a.
- 1 point: a = 2.5 m/s².
- 1 point: T = 6(2.5) = 15 N.
- 1 point: Hanger net force is 20 − 15 = 5 N downward, equal to 2(2.5).
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 do acceleration magnitudes match?
A taut inextensible string constrains their displacements.
RECALL 2Which equation finds the internal tension?
A force equation for an individual object.
RECALL 3What assumptions make tensions equal?
Massless string and an ideal frictionless, negligible-inertia pulley.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Connected objects and the system boundary
- T = m₁a; m₂g − T = m₂a for this setup.
- a = m₂g/(m₁ + m₂).
- Assume a taut massless string, ideal pulley and frictionless table.
Remember: The hanging weight is not the tension when the hanging mass accelerates.
Conditions: Frictionless table, taut massless inextensible string, negligible-inertia frictionless pulley. g = 10 m/s². Arrows on each dot are actual forces on that object.
Refresh Kid · AP Physics C: Mechanics Unit 2 (official Unit 2) · Objectives 2.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 2.5, objectives 2.5.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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