Acceleration of Block Against Friction – Complete Notes, Revision, Important Questions & Downloads
When a block is placed on an inclined plane, the TOC subtopic Inclined Plane covers three core scenarios. NEET tests: (1) a = g(sinθ−μcosθ) for sliding down, (2) deceleration = g(sinθ+μcosθ) for moving up, (3) a = (P−μmg)/m for horizontal surface. If angle > angle of repose: block slides down with a = g(sinθ − μ_k cosθ). If given initial push up: friction and gravity both oppose motion, deceleration = g(sinθ + μ_k cosθ). Horizontal surface with force P: a = (P − μmg)/m. Core NEET applications of Newton's second law with friction.
NEET Weightage — Acceleration of Block Against Friction
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 0 | 0 | |
| 2022 | 1 | 4 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 3–5 | 12–20 |
MOVING UP THE INCLINE: If a block is given an initial push up the incline: (1) gravity component mg sinθ acts DOWN the incline (opposing upward motion), (2) kinetic friction μ_k mg cosθ also acts DOWN the incline (always opposing motion, so now opposing upward motion). Both forces decelerate the block. Net deceleration: a = g(sinθ + μ_k cosθ). Block decelerates at a higher rate than it would accelerate going down (because friction reverses direction). The block may not return unless gravity overcomes friction at rest (depends on tan θ vs μ_s). Note: acceleration going down < deceleration going up because a_down = g(sinθ − μ), a_up = g(sinθ + μ) and a_up > a_down.
HORIZONTAL SURFACE WITH APPLIED FORCE: When a block on a horizontal surface has applied force P (horizontal): kinetic friction f_k = μ_k mg acts backward. Net force = P − μ_k mg. Acceleration a = (P − μ_k mg)/m = P/m − μ_k g. Block accelerates only if P > μ_k mg. If P ≤ μ_s mg: block doesn't move (static friction prevents motion). If μ_k mg < P ≤ μ_s mg: block doesn't start moving (not relevant since this gives no sliding). If P > μ_s mg: block slides with a = (P − μ_k mg)/m. NCERT: 'When body is moving under application of force P, then kinetic friction opposes its motion.'
How to Prepare Inclined Plane Friction Acceleration for NEET
Step 1 — Master the FBD on inclined plane Always draw forces on the inclined plane: (1) Weight mg vertically downward. Resolve: mg sinθ along incline (down), mg cosθ perpendicular to incline. (2) Normal reaction N = mg cosθ (perpendicular to incline, away from surface). (3) Kinetic friction f_k = μ_k N = μ_k mg cosθ (along incline, OPPOSING motion). Apply Newton's 2nd law along incline. For sliding down: a = g(sinθ − μ_k cosθ). For moving up: a = g(sinθ + μ_k cosθ) (deceleration).
Step 2 — Memorise both formulas and the difference DOWN: a = g(sinθ − μ_k cosθ). UP (decelerating): a = g(sinθ + μ_k cosθ). Key insight: a_down < a_up because friction helps gravity stop the block going up, but opposes gravity when block goes down. The two formulas differ only in the sign of the μ_k cosθ term. Remember: FRICTION ALWAYS OPPOSES MOTION — so its component along the incline direction changes sign based on direction of motion.
Step 3 — Practise time/velocity questions using kinematics NEET often combines friction acceleration with kinematics. Example: block slides distance L down incline starting from rest. First find a = g(sinθ − μcosθ), then use s = ½at² or v² = 2as. Another type: block given initial velocity v₀ up incline — find distance before stopping: v² = v₀² − 2a_up × s → s = v₀²/(2g(sinθ+μcosθ)). Practise these formulas until the substitution is automatic.
Study Materials — Inclined Plane Friction
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Rapid Revision — Inclined Plane Acceleration
Concept → Trap → Example1) Inclined Plane — Acceleration Formulas (Down and Up)
Inclined PlaneA block of mass m rests on an inclined plane of angle θ. Coefficient of static friction μ_s, kinetic friction μ_k. FORCES ON BLOCK: (1) Weight mg: vertically downward. Components: mg sinθ along incline (directed down the slope), mg cosθ perpendicular to incline (into the surface). (2) Normal reaction N = mg cosθ (perpendicular to incline, away from surface — balancing the perpendicular weight component). (3) Kinetic friction f_k = μ_k N = μ_k mg cosθ (along incline, OPPOSING motion). CASE 1 — BLOCK SLIDING DOWN (θ > angle of repose): Motion: downward along incline. Friction: upward along incline (opposes downward motion). Net force along incline: F_net = mg sinθ − μ_k mg cosθ = mg(sinθ − μ_k cosθ). Newton's 2nd law: ma = mg(sinθ − μ_k cosθ). Acceleration: a = g(sinθ − μ_k cosθ) [down the incline]. This is the acceleration of the block as it slides DOWN. Condition for this case: θ > angle of repose, i.e., tan θ > μ_s (initial condition). CASE 2 — BLOCK PUSHED UP THE INCLINE: Block has initial velocity upward along incline. Motion: upward along incline. Friction: downward along incline (opposes upward motion). Weight component: also downward along incline (mg sinθ). Both friction AND weight component are now in the SAME direction (both opposing upward motion). Net decelerating force: F = mg sinθ + μ_k mg cosθ = mg(sinθ + μ_k cosθ). Deceleration: a = g(sinθ + μ_k cosθ) [opposing upward motion, i.e., directed down the incline]. Note: since both gravity component and friction now point down-incline, the deceleration going UP is greater than the acceleration going DOWN. a_up_deceleration = g(sinθ + μcosθ) > a_down = g(sinθ − μcosθ). CASE 3 — HORIZONTAL SURFACE (applied force P): Normal reaction N = mg. Kinetic friction f_k = μ_k mg (opposing motion). Net force = P − μ_k mg. Acceleration a = (P − μ_k mg)/m = P/m − μ_k g. Block starts moving only if P > μ_s mg.
- CONDITION TO SLIDE DOWN: Static friction can hold the block if tanθ ≤ μ_s (angle ≤ angle of repose). If tanθ > μ_s: block slides. Once sliding: kinetic friction μ_k applies. Note: μ_k < μ_s, so once sliding starts the friction decreases from μ_s mg cosθ to μ_k mg cosθ — the block accelerates faster than the minimum needed to start. Condition: tanθ > μ_s is the requirement for a block to START sliding; during sliding, μ_k < μ_s is used.
- COMPARISON — a_down vs a_up: a_down = g(sinθ − μcosθ), a_up = g(sinθ + μcosθ). Since μcosθ > 0, clearly a_up > a_down. The block decelerates FASTER going up than it accelerates going down. Important implication: if a block is pushed up an incline and comes to rest at the top, it will not necessarily slide back down — it depends on whether θ > angle of repose. If it does slide back: the return speed at the bottom is less than the initial push speed (because a_return < a_decelerate). Ratio: a_down/a_up = (sinθ−μcosθ)/(sinθ+μcosθ).
- SPECIAL CASES: (1) Smooth surface (μ=0): a_down = a_up = g sinθ (same in both directions). Symmetry restored — no friction means identical acceleration magnitude for up and down. (2) θ = 90° (vertical surface): a = g (free fall). f_k = μ_k g cos90° = 0 (friction vanishes on vertical surface because N=0). a = g(sin90° − 0) = g. (3) Block on horizontal surface (θ=0°): a = g(0 − μcosθ) = −μg (deceleration) for a block sliding with no driving force — just sliding friction slowing it down. (4) For block on horizontal with applied force P: a = (P − μmg)/m = P/m − μg.
US Curriculum Gaps — Inclined Plane Friction
Topics in this section are in NEET but may be framed differently in US physics courses.Angle of Repose as Starting Condition in AP Physics 1
AP Physics 1 students work extensively with inclined plane + friction problems and know the formulas a = g(sinθ − μcosθ) and a = g(sinθ + μcosθ). However, NEET specifically uses the 'angle of repose' concept as the critical angle — if θ > angle of repose, the block slides. US AP Physics typically frames this as μ_s = tanθ at the critical angle without naming it 'angle of repose'. NRI students should explicitly connect: angle of repose α = tan⁻¹(μ_s) and: 'block slides if θ > α' is an NCERT-specific named condition.
- AP Physics 1: works with μ_s = tanθ for critical angle; does not name it 'angle of repose'
- NEET: 'angle of repose' is a named, tested concept directly tied to the sliding condition
- NCERT: sliding condition stated as 'angle of inclined plane more than angle of repose'
Asymmetric Acceleration for Up vs Down in AP Physics 1
AP Physics 1 students solve both 'block sliding down' and 'block pushed up incline' as separate problems using FBD, correctly applying friction direction in each case. The insight that a_up > a_down (deceleration going up is greater than acceleration going down) is tested in AP but not always stated as an explicit rule. NEET may ask this as a direct comparison, so the ratio a_down/a_up = (sinθ−μcosθ)/(sinθ+μcosθ) should be memorised for quick MCQ answering.
- AP Physics 1: solves both cases correctly by FBD; may not state asymmetry as a standalone rule
- NEET: may ask: 'if decelerating going up is a₁ and accelerating going down is a₂, which is greater?' → a₁ > a₂
- Memorise ratio: a_down/a_up = (sinθ−μcosθ)/(sinθ+μcosθ) for comparison MCQs
NEET-Style Practice Questions — Inclined Plane Acceleration
4 QuestionsPractice Problems — Inclined Plane Acceleration
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Inclined Plane Acceleration
Notes · Downloads · Revision · Important QuestionsWhat is the formula for acceleration of a block sliding DOWN an inclined plane with friction?
What is the formula for deceleration when a block is moving UP an incline?
Why is the deceleration going up an incline greater than the acceleration going down?
When does a block on an inclined plane NOT slide?
How do I set up the FBD on an inclined plane?
If a block slides down and comes back up after bouncing off a wall, will its speed be different?
What is the acceleration of a block on a horizontal surface under an applied force?
How does the result change for a block on a smooth inclined plane (μ = 0)?
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