Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads
Stopping of Block Due to Friction covers the TOC subtopic Stopping Distance and Time. A block moving with initial velocity u on a rough horizontal surface (μ) decelerates due to kinetic friction and comes to rest. Stopping distance: S = u²/(2μg). Stopping time: t = u/(μg). On an inclined rough surface (angle θ, moving up): S = u²/[2g(sinθ + μcosθ)]. NEET tests these as direct numericals and conceptual questions about how S and t depend on mass, velocity, and μ.
NEET Weightage — Stopping of Block Due to Friction
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 0 | 0 | |
| 2022 | 1 | 4 | |
| 2021 | 0 | 0 | |
| 2020 | 1 | 4 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 0–2 | 0–8 |
INCLINED CASE (moving up): On incline at angle θ, moving up. Friction acts DOWN the incline (opposing upward motion). Gravity also acts down the incline. Net retarding force = mgsinθ + μmgcosθ. Deceleration a=g(sinθ+μcosθ). Stopping distance S=u²/[2g(sinθ+μcosθ)]. Stopping time t=u/[g(sinθ+μcosθ)]. Incline gives SHORTER stopping distance than horizontal (a is larger). Reversing: after stopping, friction must exceed gravity component for block to stay (μcosθ ≥ sinθ → μ ≥ tanθ).
INCLINED CASE (moving down with friction): Retarding friction acts UP, gravity component DOWN. Net force = mgsinθ − μmgcosθ (if sinθ > μcosθ, block accelerates; if sinθ < μcosθ, block decelerates). For a block moving down that must be stopped: a (retardation) = g(μcosθ − sinθ) — valid only if μ > tanθ.
Solving Stopping Distance Problems
Step 1 — Write the deceleration Horizontal: a = μg. Incline (moving up): a = g(sinθ + μcosθ). Incline (moving down while decelerating): a = g(μcosθ − sinθ).
Step 2 — Apply kinematics (v=0 at stop) Distance: v² = u² − 2aS → 0 = u² − 2aS → S = u²/(2a). Time: v = u − at → 0 = u − at → t = u/a.
Step 3 — Substitute a Horizontal: S = u²/(2μg); t = u/(μg). Incline up: S = u²/[2g(sinθ+μcosθ)]; t = u/[g(sinθ+μcosθ)]. Key: S ∝ u² and inversely proportional to the effective deceleration.
Study Materials — Stopping of Block Due to Friction
PDF · Cheat Sheet · MCQ Set · PYQSubtopics — Stopping of Block Due to Friction
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Rapid Revision — Stopping of Block Due to Friction
Concept → Trap → Example1) Stopping Distance and Time — Horizontal and Inclined
Stopping Distance and TimeCASE 1 — HORIZONTAL SURFACE: Block of mass m moving at velocity u on rough surface (kinetic friction μ). Kinetic friction force = μmg (acting backward, opposing motion). Deceleration: a = F/m = μmg/m = μg. (Mass cancels — deceleration is mass-independent.) KINEMATICS: Starting at u, decelerating at a until v=0. Distance: S = v²−u²/(−2a) = u²/(2a) = u²/(2μg). Time: t = (v−u)/(−a) = u/(μg). KEY PROPORTIONALITIES: S ∝ u² (doubling speed quadruples stopping distance); S ∝ 1/μ; S independent of mass — two blocks of different masses on same surface with same u stop in same distance. CASE 2 — INCLINED SURFACE (BLOCK MOVING UP): On incline of angle θ, block moving up. Both gravity (component down the incline = mgsinθ) and friction (opposing motion, down the incline = μmgcosθ) decelerate the block. Total retarding force = mg(sinθ + μcosθ). Deceleration: a' = g(sinθ + μcosθ). Stopping distance on incline: S' = u²/[2g(sinθ + μcosθ)]. Stopping time: t' = u/[g(sinθ + μcosθ)]. Since a' > a (for any θ > 0 and μ > 0), S' < S — block stops sooner on an incline than on a flat surface with the same initial speed.
- AFTER-STOPPING ANALYSIS: Once block stops on incline, check if it remains stationary or slides back. It stays if static friction ≥ gravity component: μ_s × mgcosθ ≥ mgsinθ → μ_s ≥ tanθ. If μ_s < tanθ, the block slides back down after stopping. Kinetic friction during sliding down = μ_k mgcosθ (acting UP, opposing downward motion). Net downward acceleration = g(sinθ − μ_k cosθ). This is a common NEET extension: compute where the block ends up after bouncing (it may slide back and stop before the original position).
- ALTERNATIVE PROOF USING ENERGY: Work done by friction = friction force × distance = μmgS (horizontal). This equals kinetic energy lost (since surface is horizontal and final KE=0): μmgS = ½mu² → S = u²/(2μg). Energy method gives the same result as kinematics and can be faster in NEET if you recall it. Similarly: work by friction + work by gravity = change in KE on incline moving up: (μmgcosθ + mgsinθ)S' = ½mu² → S' = u²/[2g(sinθ+μcosθ)].
- NEET COMMON QUESTION TYPE: 'A block is moving with initial velocity u. If μ is doubled, by what factor does the stopping distance change?' From S = u²/(2μg): doubling μ halves S (S₂ = S₁/2). Another type: 'Two blocks of masses m and 2m are moving with the same u on the same surface. Compare their stopping distances.' Answer: same stopping distance (S independent of mass). These conceptual proportionality questions are very common.
US Curriculum Gaps — Stopping of Block Due to Friction
Topics in this section are in NEET but may be framed differently in US physics courses.Stopping Distance Proportionalities in AP Physics 1
AP Physics 1 covers friction-induced deceleration and stopping distance. The key NEET extension is a heavy emphasis on proportionality reasoning (S ∝ u², S ∝ 1/μ, S independent of mass) and the inclined-plane stopping case with simultaneous friction and gravity components. NEET may ask numerical computations directly, while AP Physics 1 more often asks qualitative comparison responses.
- AP Physics 1: stopping distance covered; emphasises understanding relationships, not always exact formula
- NEET: direct formula application S=u²/(2μg) in 30 seconds — know the formula exactly
- Inclined version a=g(sinθ+μcosθ) for moving up: practise with θ=30°, θ=45°, θ=60°
After-Stopping Sliding Analysis in University Physics
Halliday & Resnick covers static vs kinetic friction transitions but doesn't emphasise the after-stopping-on-incline back-sliding scenario as a multi-step NEET problem. This requires: (1) compute stopping distance moving up, (2) check if μ_s ≥ tanθ for staying, (3) if not, compute back-sliding acceleration g(sinθ−μ_k cosθ) and where the block ultimately stops. This 3-step chain is NEET-characteristic but not typically tested as a unit in US courses.
- US university: individual concepts (friction, kinematics on inclines) covered separately
- NEET: multi-step problems combining stopping distance → checking static condition → back-sliding distance
- Key: distinguish μ_s (for staying condition) from μ_k (for kinetic friction during back-slide)
NEET-Style Practice — Stopping of Block Due to Friction
4 QuestionsPractice Problems — Stopping of Block Due to Friction
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Stopping of Block Due to Friction
Notes · Downloads · Revision · Important QuestionsWhat is the formula for stopping distance on a rough horizontal surface?
Why is the stopping distance independent of mass?
How does stopping distance change if I double the initial speed?
What is the stopping distance formula on a rough incline (block moving up)?
After a block stops on a rough incline, will it slide back?
Can I use energy methods instead of kinematics for stopping distance?
What is the stopping time formula horizontal?
A block of mass m moving at velocity u on a rough surface (μ) — is there any situation where two blocks of different masses stop in different distances?
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