Same endpoints, different routes
You will be able to: Compare work along two routes and identify conservative interactions.
Why is gravitational work path independent while sliding friction is not?
Raise the same backpack 1 m by a direct lift or by a longer ramp. Gravity does the same negative work either way. Dragging a block over a rough surface is different: a longer sliding route generally dissipates more energy.
A useful starting point: Net work changes kinetic energy →
Words and symbols before equations
- Conservative force
- A force whose work between configurations is independent of route; its work around a closed path is zero.
- Configuration
- Relative positions of the interacting objects.
- Path length L
- Total distance traveled along the route, not endpoint displacement.
- Nonconservative force
- A force for which a position-only potential-energy description does not capture its work in general.
What this picture assumes
Geometric paths in three spatial coordinates with uniform downward gravity, g = 10 m/s². Start (0,0,0), end (4,h,0); detour midpoint (2,h/2,z), all in metres. These are prescribed paths, not free-flight trajectories. Work shown is gravity alone.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Direct path 4.123 m; detour 5.745 m. Both rise 1 m and have W_g = -20 J. The shape is a geometric path, not a trajectory computed from gravity alone.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Near Earth, gravity is F = (0, −mg, 0). The dot product gives dW_g = −mg dy, so W_g = −mg(y_f − y_i). Horizontal detours contribute no gravitational work. For a closed path, the endpoint height returns to its initial value and W_g = 0.
A kinetic-friction force opposing sliding on a fixed surface does W_f = −∫f_k ds. If its magnitude is constant, W_f = −f_k L. The length L grows during a round trip even when the displacement returns to zero; friction work remains negative.
The 3D investigation compares a straight rising route and a sideways detour between the same endpoints. Only gravitational work is calculated there. The separate friction comparison assumes a constant resisting magnitude; a real ramp’s normal force and friction may differ, so equal coefficients alone do not imply equal friction forces.
| Property | Gravity | Sliding friction |
|---|---|---|
| Depends on route? | No, for fixed endpoints | Yes, generally |
| Work on a round trip | Zero | Negative when sliding against a fixed surface |
| Potential energy function? | Yes | No single position-only friction potential |
A worked example, step by step
A 2 kg object rises by 1 m along either a 2 m route or a 5 m route. Use g = 10 m/s². Separately, a constant 3 N resistance opposes its motion on each route. Compare the works.
- The gravitational interaction depends only on height: W_g = −(2)(10)(1) = −20 J on either route.
- Resistance along the 2 m path does −(3)(2) = −6 J.
- Resistance along the 5 m path does −(3)(5) = −15 J.
- Same endpoints guarantee equal gravity work, not equal total work when nonconservative resistance is also present.
A longer path does not change conservative work between the same endpoints; it can change dissipative work.
Does a force being constant automatically mean it does positive work?
Compare with an explanation
No. A constant force is conservative, but its work is positive, negative or zero depending on displacement.
Predict. Change one thing. Explain.
Increase only the sideways detour. Watch route length change while endpoints and gravity work remain fixed. Rotate the 3D view to inspect the geometry; compare the flat top view and work bars.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Direct path 4.123 m; detour 5.745 m. Both rise 1 m and have W_g = -20 J. The shape is a geometric path, not a trajectory computed from gravity alone.
Geometric paths in three spatial coordinates with uniform downward gravity, g = 10 m/s². Start (0,0,0), end (4,h,0); detour midpoint (2,h/2,z), all in metres. These are prescribed paths, not free-flight trajectories. Work shown is gravity alone.
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 questionA 1 kg object moves from height 0 to 2 m and back, with g = 10 m/s². Total path length is 10 m and a constant 1 N resistance acts throughout. Compute gravity work on each leg, total gravity work, and resistance work.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Upward gravity work is −20 J.
- 1 point: Return gravity work is +20 J.
- 1 point: Total gravity work is zero because the endpoints coincide.
- 1 point: Resistance work is −10 J because it depends on total sliding distance.
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 test characterizes a conservative force?
Zero work around every closed path in the relevant domain.
RECALL 2Which distance belongs in −fL?
The path length over which constant opposing friction acts.
RECALL 3Why does a sideways detour leave gravity’s work unchanged?
Gravity is vertical, so only the endpoint height difference contributes to its work.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Same endpoints, different routes
- Conservative force: W on a closed path = 0.
- Uniform gravity: W_g = −mgΔy.
- Constant sliding resistance on a fixed surface: W_f = −fL.
Remember: A longer path does not change conservative work between the same endpoints; it can change dissipative work.
Conditions: Geometric paths in three spatial coordinates with uniform downward gravity, g = 10 m/s². Start (0,0,0), end (4,h,0); detour midpoint (2,h/2,z), all in metres. These are prescribed paths, not free-flight trajectories. Work shown is gravity alone.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.2, objectives 3.2.A. 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.
Want to work through this with a tutor?
Bring your question about Same endpoints, different routes. Your explanation and answers remain free to access.