Trade height for speed on a smooth track
You will be able to: Conserve mechanical energy between two points and check physical constraints.
How much can an energy equation predict without knowing travel time?
A small slider starts from rest 2 m above the lowest point of a smooth fixed track. At the bottom, g = 10 m/s² gives a speed √40 ≈ 6.32 m/s. The track shape changes the trip, but not that speed when the height change and assumptions match.
A useful starting point: Choose the system before writing energy terms →
Words and symbols before equations
- Ideal smooth track
- A fixed track with negligible friction; the normal is perpendicular to motion.
- Energy bar chart
- A same-unit comparison of the energy forms in a chosen state.
- Constraint
- An additional physical condition, such as staying in contact with a track.
- Speed versus time
- Energy connects states; it does not by itself give the time between them.
What this picture assumes
Point slider on an ideal fixed frictionless track, g = 10 m/s², U = 0 at height 0. Energy accessibility is checked; contact geometry is not modeled here. This is not a rolling-body model.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Total E = 64 J; U = 20 J. K = 44 J; speed = 6.633 m/s. Energy-allowed maximum height 3.2 m. Track contact must be checked separately.
- 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 slider + Earth and neglect rotation, air resistance and friction. The fixed track’s normal does no work, so ½mv_i² + mgh_i = ½mv_f² + mgh_f. Rearrange to v_f² = v_i² + 2g(h_i − h_f).
The mass cancels because both gravitational U and translational K scale with m. At the same height a particle has the same speed for the same total energy, regardless of whether it is moving upward or downward. Direction and travel time require more information.
Energy alone does not guarantee a proposed path is physically possible. For a slider on the inside of a vertical circular loop, top contact also needs N ≥ 0. At the top, mg + N = mv²/R, so v_top² ≥ gR. Combining with energy from rest at height h above the bottom yields h ≥ 5R/2.
A worked example, step by step
A particle starts from rest at height h above the bottom of a frictionless inside loop of radius R = 2 m. Find the minimum h for just maintaining top contact, using g = 10 m/s².
- At the limiting top contact, N = 0 and v_top² = gR = 20 m²/s².
- The top is at height 2R = 4 m above the bottom.
- Conservation gives mgh = mg(2R) + ½m(gR), hence h = 2.5R = 5 m.
- A height of 4 m only provides enough energy to reach the top with zero speed in an energy-only calculation; it cannot maintain the required circular contact.
An energy equation predicts a speed only for an accessible state. Verify contact, string tension and other constraints separately.
Would a rolling ball have the same speed as the point slider?
Compare with an explanation
Not generally. Rolling divides kinetic energy into translational and rotational parts, so the slider formula is incomplete for it.
Predict. Change one thing. Explain.
Change the selected height while keeping starting height and initial speed fixed. Compare K and U. Then request a height above the available energy and explain why the state is inaccessible.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Total E = 64 J; U = 20 J. K = 44 J; speed = 6.633 m/s. Energy-allowed maximum height 3.2 m. Track contact must be checked separately.
Point slider on an ideal fixed frictionless track, g = 10 m/s², U = 0 at height 0. Energy accessibility is checked; contact geometry is not modeled here. This is not a rolling-body model.
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 2 kg point slider starts at height 3 m with speed 2 m/s. It reaches height 1 m on a fixed smooth track. Use g = 10 m/s². Calculate initial energy, final K, final speed and name a limitation.
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Compare with the answer and four-point rubric
- 1 point: Initial E = ½(2)(2²) + (2)(10)(3) = 64 J.
- 1 point: At 1 m, U = 20 J and K = 44 J.
- 1 point: Speed = √(2×44/2) = √44 = 6.63 m/s.
- 1 point: This result assumes maintained contact and negligible friction and rotation; energy alone does not give travel time.
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 can mass cancel in a gravity-only speed calculation?
Both K and U are proportional to mass.
RECALL 2What extra condition is needed at the top of an inside loop?
The required normal force must be nonnegative.
RECALL 3What does energy conservation not determine by itself?
Direction, travel time, or whether a proposed contact constraint can hold.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Trade height for speed on a smooth track
- K_i + U_i = K_f + U_f for the stated conservative model.
- v_f² = v_i² + 2g(h_i − h_f).
- Inside-loop top contact: v_top² ≥ gR; from rest, h ≥ 2.5R.
Remember: An energy equation predicts a speed only for an accessible state. Verify contact, string tension and other constraints separately.
Conditions: Point slider on an ideal fixed frictionless track, g = 10 m/s², U = 0 at height 0. Energy accessibility is checked; contact geometry is not modeled here. This is not a rolling-body model.
Refresh Kid · AP Physics C: Mechanics Unit 3 (official Unit 3) · Objectives 3.4.A; 3.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 3.4, objectives 3.4.A; 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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