Include friction and driving work on a rough slope
Calculate path-dependent friction work, height-dependent potential change and the reaction change caused by an angled pull.
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
- Resolve weight and an applied force relative to a plane.
- Use μR for sliding friction with its direction opposite motion.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For a constant sliding friction f over travel distance s, what work enters the particle’s energy equation?
What you will learn
- Use a mechanical-energy balance with supplied friction and drive work.
- Find reaction before friction for an angled pull and check contact and stopping.
Friction work uses path distance while gravity uses height
A 5 kg particle slides down a plane with sin θ = 3/5, cos θ = 4/5 and μ = 1/4. Reaction is 40 N and sliding friction is 10 N up the plane. Over distance 4 m, vertical drop is 2.4 m and potential-energy decrease is 120 J. Friction transfers 40 J out of mechanical energy, so a particle starting from rest ends with K = 80 J and speed 4√2 m s⁻¹.
Equivalently, total work on the particle is gravitational work +120 J plus friction work −40 J, giving ΔK = +80 J. Use either this form or K + U with non-gravitational work, not both gravity contributions together. On the upward leg friction and gravity both remove kinetic energy: for that particle projected up at speed 8, initial K is 160 J and the energy cost per metre is 30 + 10 = 40 J, giving stopping distance 4 m.
An angled applied force can change the loss
For a 5 kg particle pulled up a 30° slope by 40 N at angle φ above the plane, with sin φ = 3/5 and cos φ = 4/5, the normal equation is R = 50 cos 30° − 40 sin φ = 25√3 − 24 N, about 19.30 N. At μ = 1/5 the friction magnitude is (25√3 − 24)/5, about 3.86 N. The along-plane drive is 32 N and downslope weight component is 25 N, so the net uphill work per metre is 32 − 25 − friction.
If this particle starts from rest and moves 3 m up, final K is 3[7 − (25√3 − 24)/5] = 177/5 − 15√3 J, about 9.42 J, giving speed about 1.94 m s⁻¹. The positive reaction permits contact and the uphill drive exceeds weight plus maximum friction, supporting the stated start. A pull large enough to make R negative would invalidate this contact calculation. At a stopping event, test static balance before assuming a return.
Pause and explain
An upward angled pull reduces a slope particle’s reaction. What must be done before calculating sliding friction?
Worked example
A 5 kg particle starts sliding down a rough slope from rest and travels 4 m. Given sin θ = 3/5, cos θ = 4/5 and μ = 1/4, find its final speed.
Show the worked solution
- R = 40 N and friction is 10 N, so friction work is −40 J.
- The vertical drop is 2.4 m, so potential-energy decrease is 120 J.
- Final K is 120 − 40 = 80 J; ½ × 5 × v² = 80 gives speed 4√2 m s⁻¹.
Answer Final K = 80 J; speed 4√2 m s⁻¹.
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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 5 kg particle starts from rest and slides 4 m down a rough slope with sin θ = 3/5, cos θ = 4/5 and μ = 1/4. Find final kinetic energy and speed.
Give me a hint
Subtract friction’s path loss from the height-based gravitational release.
Compare my reasoning
- Reaction is 40 N, friction is 10 N and friction loss over 4 m is 40 J.
- Vertical drop is 4 × 3/5 = 2.4 m, releasing 120 J gravitational energy.
- K = 80 J, so v² = 32 and speed is 4√2 m s⁻¹.
Final K = 80 J; speed 4√2 m s⁻¹.
Look for these in your work
- Use different geometric distances for gravity and friction.
- Retain the friction loss with the correct sign.
A 5 kg particle is projected up the same long rough slope at speed 8 m s⁻¹. Find the distance to first rest, and check whether it restarts downhill using the same coefficient for the static limit.
Give me a hint
During upward sliding, both gravity and friction remove kinetic energy.
Compare my reasoning
- Initial K is ½ × 5 × 64 = 160 J.
- The energy cost per metre is 50(3/5) + (1/4)50(4/5) = 40 J/m, so stopping distance is 4 m.
- At rest, the 30 N downslope weight component exceeds the 10 N static limit, so it restarts downhill under the stated model.
Stops after 4 m; downhill restart occurs because 30 N exceeds the 10 N static limit.
Look for these in your work
- Count uphill potential gain and friction loss together.
- Test the static state at first rest before asserting restart.
A 5 kg particle is pulled up a 30° plane by 40 N at angle φ above the plane, with sin φ = 3/5 and cos φ = 4/5. With μ = 1/5, it starts from rest and travels 3 m up. Find reaction, final kinetic energy and speed, checking contact.
Give me a hint
The pull’s normal component changes the friction before you form the along-path energy balance.
Compare my reasoning
- R = 25√3 − 24 N, approximately 19.30 N, so contact is admissible; friction is R/5.
- Driving work is 32 × 3 = 96 J, potential gain is 25 × 3 = 75 J and friction loss is 3R/5.
- Final K = 177/5 − 15√3 J, approximately 9.42 J, and v = √(2K/5), approximately 1.94 m s⁻¹.
R = 25√3 − 24 N; K = 177/5 − 15√3 J ≈ 9.42 J; speed ≈ 1.94 m s⁻¹.
Look for these in your work
- Check positive reaction and uphill drive before applying the sliding model.
- Include drive work, potential gain and reaction-dependent friction exactly once.
Common mistakes
- Friction uses path distance; mgΔh uses vertical height.
- An angled pull changes R and therefore μR.
- Do not claim mechanical-energy conservation when friction does work.
What can you explain now?
Does a rough slope generally conserve K + U for the moving particle?
Compare with the explanation
No; friction transfers mechanical energy to other forms.
Include its negative work and any supplied driving work in the balance.
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Make it stick
Resolve contact first, then balance drive work against kinetic change, vertical potential change and friction’s path loss.
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