Conserve mechanical energy on smooth slopes and slides
Exchange kinetic and gravitational potential energy in a loss-free model, including a curved slide whose speed depends on height change.
Section 4.5 · Energy, work and power
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use kinetic energy and gravitational height changes.
- State the assumptions required for a smooth fixed path.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When may K + U be conserved on a fixed path?
What you will learn
- Use K + U conservation when no supplied force removes mechanical energy.
- Calculate endpoint speed on a smooth curved slide using only vertical height change.
A loss-free model exchanges kinetic and potential energy
For a smooth fixed slope with no supplied resistance or drive, gravity transfers energy between K and U while the normal reaction does no work. Thus ½mu² + mgh₁ = ½mv² + mgh₂. A particle released from rest and dropping vertically by 5 m reaches speed √(2 × 10 × 5) = 10 m s⁻¹. Positive mass cancels, so the predicted speed is independent of mass under these assumptions.
A particle entering a smooth descent at speed 4 m s⁻¹ and dropping 2 m has v² = 4² + 2 × 10 × 2 = 56, so v = 2√14 m s⁻¹. That is an endpoint speed; it does not by itself provide travel time or acceleration at every point. Mechanical energy need not be conserved when a drive, friction, resistance or impact changes the balance.
A curved path uses its overall height change
A particle sliding on a fixed smooth curved track, with contact stipulated throughout and no resistance, obeys the same endpoint energy relation. The curved shape changes local forces and directions, but the gravitational potential change depends only on vertical height. We do not need to compute a normal reaction or a curved-motion acceleration to answer the endpoint-speed question in this section.
If a particle starts at speed 8 m s⁻¹ and climbs a smooth path, its greatest possible height increase before speed becomes zero is 8²/(2 × 10) = 3.2 m. An endpoint 4 m higher would require v² = 64 − 80 < 0 and cannot be reached under that loss-free initial-energy model. This is an energy bound, subject to the stated contact and path assumptions, rather than a claim about a real track’s safety or full trajectory.
Pause and explain
For a smooth curved slide, which information determines endpoint speed in the loss-free model?
Worked example
A particle is released from rest on a fixed smooth curved slide with a 5 m vertical drop to its endpoint. Contact is maintained and resistance is neglected. Find endpoint speed.
Show the worked solution
- The potential-energy decrease is 5mg.
- By mechanical-energy conservation, ½mv² = 5mg.
- Cancel positive m: v² = 2 × 10 × 5 = 100, so endpoint speed is 10 m s⁻¹.
Answer Endpoint speed 10 m s⁻¹, independent of mass in this model.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A particle is released from rest on a smooth fixed slide with vertical drop 5 m. Contact is maintained and resistance is neglected. Find its endpoint speed.
Give me a hint
Equate lost potential energy to gained kinetic energy.
Compare my reasoning
- The decrease in potential energy is 5mg.
- The energy equation is ½mv² = 5mg.
- Thus v² = 100 and speed is 10 m s⁻¹.
Endpoint speed 10 m s⁻¹.
Look for these in your work
- Use the supplied vertical drop.
- State the loss-free smooth-contact assumptions.
A particle enters a smooth fixed descent at speed 4 m s⁻¹ and drops vertically by 2 m. With contact maintained and no resistance, find endpoint speed.
Give me a hint
Initial kinetic energy is non-zero and must be retained.
Compare my reasoning
- Energy gives ½mv² = ½m(4²) + mg(2).
- Cancel mass to obtain v² = 16 + 40 = 56.
- Endpoint speed is √56 = 2√14 m s⁻¹.
Endpoint speed 2√14 m s⁻¹.
Look for these in your work
- Include initial kinetic energy.
- Use the non-negative speed root.
A particle starts up a smooth fixed path at speed 8 m s⁻¹, with no driving force or resistance and contact maintained while it moves. Find the maximum possible vertical rise and decide whether it can reach a point 4 m higher.
Give me a hint
At the height limit, kinetic energy is zero; a higher point would require negative kinetic energy.
Compare my reasoning
- The limiting equation is ½m(8²) = mgΔh, so Δh = 64/20 = 3.2 m.
- At rise 4 m, the predicted squared speed would be 64 − 80 = −16.
- That is impossible, so the point cannot be reached under the stated energy model.
Maximum vertical rise 3.2 m; a point 4 m higher is unreachable in this model.
Look for these in your work
- Use zero final speed at the energy limit.
- Reject a negative squared-speed prediction rather than changing its sign.
Common mistakes
- Use vertical drop, not curved path length.
- Energy does not by itself supply the travel time.
- Contact and no-loss assumptions must be stated.
What can you explain now?
Does the curved path’s total length replace vertical drop in mgΔh?
Compare with the explanation
No; gravitational potential change depends on vertical height change.
Other path-dependent work must be supplied separately if friction or resistance acts.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Under a stated loss-free smooth model, exchange K and U using vertical height differences and check the available-energy bound.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.