Stopping of Block Due to Friction

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Unit Progress

NEET Physics — Friction

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.

⬇ Download Notes PDF View Important Questions →

Stopping Distance and TimeFriction Ch.5Kinematics + Friction

Expected QuestionsQ

0–1

NEET asks: (1) compute S given u and μ, (2) find t to stop given u and μ, (3) compare stopping distances for two blocks with different u or μ, (4) block moves up incline at angle θ with friction — find S.

Time Required⏱

15 min

5 min: derive retardation a = μg (horizontal). Apply v² = u² − 2aS with v=0 → S = u²/(2μg). Apply v = u − at with v=0 → t = u/(μg). 5 min: derive for incline (up): a = g(sinθ + μcosθ); same kinematics formulas apply. 5 min: practise 3 numericals.

Difficulty⚡

Easy

Straightforward application of kinematics (v² = u² + 2aS, v = u + at) with friction-derived deceleration. The key step is writing the net deceleration: a = μg (horizontal), a = g(sinθ+μcosθ) (up incline). Mass doesn't appear in the final formula — it cancels.

NRI USA Curriculum GapUS

Low

AP Physics 1 fully covers this problem type — stopping distance with friction. However, NEET may present the inclined case or test the exact formula for S in terms of μ, which AP questions sometimes avoid by asking for conceptual proportionality only.

1Subtopics

4+Practice Questions

4Free Downloads

15 minPrep Time

⬇ Get Free Downloads

NEET Weightage — Stopping of Block Due to Friction

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	1	1 Q	4
2023	0	0 Q	0
2022	1	1 Q	4
2021	0	0 Q	0
2020	1	1 Q	4
2019	0	0 Q	0
6-Year Total (2019–2024)	0–2		0–8

HORIZONTAL CASE: Block of mass m, initial velocity u, kinetic friction coefficient μ, surface is horizontal. Friction force (retarding) = μmg. Deceleration a = μg. (Note: mass cancels.) Using v²=u²−2aS with v=0 (stopped): S=u²/(2μg). Using v=u−at with v=0: t=u/(μg). Key: S ∝ u², S ∝ 1/μ. Doubling u quadruples S. Doubling μ halves S. S is independent of mass.
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.

📊

0.5

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy

Difficulty

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 · PYQ

📘

Full Notes

Horizontal: a=μg; S=u²/(2μg); t=u/(μg). S independent of mass. S∝u². Incline up: a=g(sinθ+μcosθ); S=u²/{2g(sinθ+μcosθ)}. After stopping on incline: stays if μ≥tanθ, slides down if μ<tanθ.

1 subtopicKinematics applicationEasy numericals

Download Notes

📗

Formula Sheet

S=u²/(2μg); t=u/(μg); S∝u²; S∝1/μ; S independent of mass m.

4 formulas1 pageQuick reference

Download Sheet

📙

MCQ Practice

8 MCQs on stopping distance, time, comparisons with different initial velocities and friction coefficients.

8 MCQs2 formulasSolved

Download MCQs

📒

PYQ

NEET previous year questions on block stopping due to friction (horizontal and inclined) with complete worked solutions.

3+ year-tagged Qs2015–2024Solved

Download PYQs

Subtopics — Stopping of Block Due to Friction

2-Column Table

Column A	Column B
Stopping Distance and Time↗

Rapid Revision — Stopping of Block Due to Friction

Concept → Trap → Example

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance and Time

CASE 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.

Example (NEET-style)A block of mass 2 kg moves at 10 m/s on a surface with μ = 0.5. S = u²/(2μg) = 100/(2×0.5×10) = 100/10 = 10 m. t = u/(μg) = 10/(0.5×10) = 2 s. Deceleration = 5 m/s². Check: v² = 100 − 2×5×10 = 100−100 = 0 ✓. Note: if mass were 4 kg, S and t would still be the same (mass-independent).

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 Questions

1A block of mass m is moving initially with velocity u on a rough surface. Due to friction it stops after covering distance S = 4 m. If the initial velocity is doubled (to 2u), the stopping distance will be:Stopping Distance and Time

4 m

8 m

16 m

2 m

S = u²/(2μg), so S ∝ u². If u is doubled, S becomes (2u)²/(2μg) = 4u²/(2μg) = 4S = 4×4 = 16 m. Answer: Option C. This is the classic 'doubling speed quadruples stopping distance' result. In road safety, this is why stopping distance increases dramatically at higher speeds. Option B (8 m, doubling) would be correct only if S ∝ u (linear), which it is not.

2A block of mass 5 kg moving at 20 m/s on a rough horizontal floor (μ = 0.5) comes to rest. What is the stopping distance?Stopping Distance and Time

20 m

40 m

10 m

80 m

S = u²/(2μg) = 400/(2×0.5×10) = 400/10 = 40 m. Answer: Option B. Note: mass (5 kg) is irrelevant — it cancels in the formula. Retardation = μg = 0.5×10 = 5 m/s². Using v²=u²−2aS: 0=400−2×5×S → S=400/10=40 m ✓. Time to stop: t=u/(μg)=20/5=4 s.

3A block moves up a rough incline of angle 30° (μ = 1/√3) with initial velocity 10 m/s. The stopping distance is (g = 10 m/s²):Inclined Case

5 m

10 m

2.5 m

7.5 m

Moving up the incline: a = g(sinθ + μcosθ) = 10(sin30° + (1/√3)cos30°) = 10(0.5 + (1/√3)(√3/2)) = 10(0.5 + 0.5) = 10×1 = 10 m/s². S = u²/(2a) = 100/(2×10) = 5 m. Answer: Option A. The neat result here: (1/√3)×(√3/2) = 1/2, so sinθ + μcosθ = 0.5+0.5 = 1, giving a = g = 10 m/s². S = 100/20 = 5 m.

4Two blocks of masses m and 3m are moving with the same initial velocity u on the same rough surface (μ). Their stopping distances S₁ (for m) and S₂ (for 3m) satisfy:Mass Independence

S₁ = S₂

S₁ = 3S₂

3S₁ = S₂

S₁ = (1/3)S₂

S = u²/(2μg) — no mass in the formula. Both blocks have the same u, g, μ. Therefore S₁ = S₂. Answer: Option A. The deceleration a = μg is also mass-independent (friction force μmg divided by mass m gives μg). This principle is similar to Galileo's observation that all objects fall with the same acceleration in vacuum — here, all objects decelerate at the same rate μg on the same rough surface.

Practice Problems — Stopping of Block Due to Friction

Click "Reveal Answer" after attempting

1A block moving at 15 m/s stops in 9 m on a rough floor. What is μ? (g = 10 m/s²)

μ = 1.25

μ = 1.333

μ = 0.5

μ = 1.25

👁 Reveal Answer

S = u²/(2μg) → μ = u²/(2gS) = 225/(2×10×9) = 225/180 = 1.25. Answer: Option A (μ=1.25). This is a high value of friction but physically possible for rubber on rough concrete. Alternatively using v²=u²−2aS: a=u²/(2S)=225/18=12.5 m/s²; a=μg → μ=12.5/10=1.25 ✓.

2A block slides up a rough incline (θ=45°, μ=0.5) with initial speed 14 m/s. Find the stopping distance along the incline. (g=10 m/s²)

S = 9.33 m

S = 6.53 m

S = 14 m

S = 4 m

👁 Reveal Answer

Moving up incline: a=g(sinθ+μcosθ)=10(sin45°+0.5cos45°)=10×(1/√2)(1+0.5)=10×1.5/√2=15/√2≈10.61 m/s². S=u²/(2a)=196/(2×10.61)=196/21.21≈9.24 m. Closest answer: Option A (9.33 m — slight rounding). Exact: a=10×1.5/√2=7.5√2 m/s². S=196/(15√2)=196/(21.21)=9.24 m.

3A block moving at u on a rough surface (μ) stops in time t. If μ is halved and u is kept the same, the new stopping time t' is:

t' = t/2

t' = 2t

t' = 4t

t' = t

👁 Reveal Answer

t = u/(μg). If μ is halved → μ' = μ/2: t' = u/(μ'g) = u/(μg/2) = 2u/(μg) = 2t. Answer: Option B. Halving friction doubles stopping time (t ∝ 1/μ). Note that stopping distance also doubles: S = u²/(2μg), halving μ doubles S. The proportionality t ∝ 1/μ is linear (unlike S ∝ 1/μ which is also linear in μ — both S and t are inversely proportional to μ, not inversely proportional to μ²).

4A car skids to a stop from 60 km/h with μ = 0.8 on a flat road. What is the minimum stopping distance? (g = 10 m/s²)

17.4 m

34.7 m

22.5 m

14.3 m

👁 Reveal Answer

Convert: 60 km/h = 60×1000/3600 = 50/3 m/s ≈ 16.67 m/s. S = u²/(2μg) = (50/3)²/(2×0.8×10) = (2500/9)/16 = 2500/144 ≈ 17.36 m ≈ 17.4 m. Answer: Option A. Real-world context: this is the minimum stopping distance with maximum (locked-wheel) friction — actual stopping distance with brakes may be shorter if ABS prevents skidding, or longer on wet roads.

Physics — Friction Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Stopping Distance and Time: Write S = u²/(2μg) and t = u/(μg) from memory. State the proportionalities: S ∝ u², S ∝ 1/μ, S independent of mass.

Stopping Distance and Time: Write the incline formula for stopping distance when moving up: S = u²/[2g(sinθ+μcosθ)]. Know why the stopping distance is shorter on an incline than on a flat surface.

Stopping Distance and Time: Derive both formulas from first principles using a=μg (horizontal) and applying v²=u²−2aS, v=u−at. This derivation takes under 60 seconds.

Stopping Distance and Time: Practise the proportionality questions: (a) u doubled → S×4; (b) μ halved → S×2; (c) mass tripled → S unchanged.

Stopping Distance and Time: After stopping on an incline: check if μ_s ≥ tanθ (stays) or μ_s < tanθ (slides back down). For slides back: new acceleration = g(sinθ − μ_k cosθ) going down.

FAQ — Stopping of Block Due to Friction

Notes · Downloads · Revision · Important Questions

What is the formula for stopping distance on a rough horizontal surface?

S = u²/(2μg), where u = initial velocity, μ = kinetic friction coefficient, g = gravitational acceleration. This is derived from kinematics: deceleration a = μg (friction force = μmg, divided by mass m). Using v² = u² − 2aS with v=0: S = u²/(2μg).

Why is the stopping distance independent of mass?

Friction force F = μmg is proportional to mass. But deceleration = F/m = μg — the mass cancels. Since deceleration is mass-independent, the stopping distance S = u²/(2μg) (which depends on deceleration, initial velocity, and g) also doesn't depend on mass. A heavy truck and a small car moving at the same speed on the same surface will stop in the same distance.

How does stopping distance change if I double the initial speed?

S ∝ u². Doubling u → S increases by a factor of 4. Tripling u → S increases by a factor of 9. This quadratic relationship is why high-speed driving is disproportionately more dangerous in terms of stopping distance. From S = u²/(2μg): S₂/S₁ = (u₂/u₁)².

What is the stopping distance formula on a rough incline (block moving up)?

S = u²/[2g(sinθ + μcosθ)]. The deceleration is a = g(sinθ + μcosθ) because both the gravity component and friction act against the upward motion. The block stops sooner on an incline than on a flat surface because the effective deceleration is larger (a > μg for θ > 0).

After a block stops on a rough incline, will it slide back?

It depends on μ vs tanθ. If μ_s ≥ tanθ: static friction is sufficient to hold the block — it stays at the stopping point. If μ_s < tanθ: gravity component exceeds static friction — block slides back down, with kinetic deceleration a' = g(sinθ − μ_k cosθ) directed down the slope.

Can I use energy methods instead of kinematics for stopping distance?

Yes. Work-energy theorem: kinetic energy lost = work done by friction. On horizontal: ½mu² = μmg×S → S = u²/(2μg). On incline moving up: ½mu² = (mgsinθ + μmgcosθ)×S → S = u²/[2g(sinθ+μcosθ)]. Both methods give the same result; choose whichever is faster for a given problem.

What is the stopping time formula horizontal?

t = u/(μg). Derived from kinematics: v = u − at → 0 = u − μg×t → t = u/(μg). Note: t ∝ u (linear, not quadratic) and t ∝ 1/μ. If speed doubles, stopping time doubles. If μ doubles, stopping time halves. Unlike stopping distance (∝ u²), stopping time is proportional to u (not u²).

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?

Only if the surfaces have different friction properties (different μ). For the same surface (same μ) and same initial velocity, stopping distance is always the same regardless of mass. However, if larger mass means greater normal force that causes the surface to deform (viscoelastic surface), effective μ may change with mass — but NEET always assumes constant μ independent of mass.

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Stopping of Block Due to Friction

NEET > Physics > Laws of Motion > Friction > Stopping of Block Due to Friction

Unit Progress

NEET Physics — Friction

Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads

⬇ Download Notes PDF View Important Questions →

Stopping Distance and TimeFriction Ch.5Kinematics + Friction

Expected QuestionsQ

0–1

Time Required⏱

15 min

Difficulty⚡

Easy

NRI USA Curriculum GapUS

Low

1Subtopics

4+Practice Questions

4Free Downloads

15 minPrep Time

⬇ Get Free Downloads

NEET Weightage — Stopping of Block Due to Friction

Friction (Chapter 5)

NEET Year	Questions from this Topic	Bar	Marks
2024	1	1 Q	4
2023	0	0 Q	0
2022	1	1 Q	4
2021	0	0 Q	0
2020	1	1 Q	4
2019	0	0 Q	0
6-Year Total (2019–2024)	0–2		0–8

📊

0.5

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy

Difficulty

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.

Study Materials — Stopping of Block Due to Friction

PDF · Cheat Sheet · MCQ Set · PYQ

📘

Full Notes

1 subtopicKinematics applicationEasy numericals

Download Notes

📗

Formula Sheet

S=u²/(2μg); t=u/(μg); S∝u²; S∝1/μ; S independent of mass m.

4 formulas1 pageQuick reference

Download Sheet

📙

MCQ Practice

8 MCQs on stopping distance, time, comparisons with different initial velocities and friction coefficients.

8 MCQs2 formulasSolved

Download MCQs

📒

PYQ

NEET previous year questions on block stopping due to friction (horizontal and inclined) with complete worked solutions.

3+ year-tagged Qs2015–2024Solved

Download PYQs

Subtopics — Stopping of Block Due to Friction

2-Column Table

Column A	Column B
Stopping Distance and Time↗

Rapid Revision — Stopping of Block Due to Friction

Concept → Trap → Example

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance and Time

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: 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

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 Questions

4 m

8 m

16 m

2 m

2A block of mass 5 kg moving at 20 m/s on a rough horizontal floor (μ = 0.5) comes to rest. What is the stopping distance?Stopping Distance and Time

20 m

40 m

10 m

80 m

3A block moves up a rough incline of angle 30° (μ = 1/√3) with initial velocity 10 m/s. The stopping distance is (g = 10 m/s²):Inclined Case

5 m

10 m

2.5 m

7.5 m

4Two blocks of masses m and 3m are moving with the same initial velocity u on the same rough surface (μ). Their stopping distances S₁ (for m) and S₂ (for 3m) satisfy:Mass Independence

S₁ = S₂

S₁ = 3S₂

3S₁ = S₂

S₁ = (1/3)S₂

Practice Problems — Stopping of Block Due to Friction

Click "Reveal Answer" after attempting

1A block moving at 15 m/s stops in 9 m on a rough floor. What is μ? (g = 10 m/s²)

μ = 1.25

μ = 1.333

μ = 0.5

μ = 1.25

👁 Reveal Answer

2A block slides up a rough incline (θ=45°, μ=0.5) with initial speed 14 m/s. Find the stopping distance along the incline. (g=10 m/s²)

S = 9.33 m

S = 6.53 m

S = 14 m

S = 4 m

👁 Reveal Answer

3A block moving at u on a rough surface (μ) stops in time t. If μ is halved and u is kept the same, the new stopping time t' is:

t' = t/2

t' = 2t

t' = 4t

t' = t

👁 Reveal Answer

4A car skids to a stop from 60 km/h with μ = 0.8 on a flat road. What is the minimum stopping distance? (g = 10 m/s²)

17.4 m

34.7 m

22.5 m

14.3 m

👁 Reveal Answer

Physics — Friction Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

Stopping Distance and Time: Write S = u²/(2μg) and t = u/(μg) from memory. State the proportionalities: S ∝ u², S ∝ 1/μ, S independent of mass.

Stopping Distance and Time: Derive both formulas from first principles using a=μg (horizontal) and applying v²=u²−2aS, v=u−at. This derivation takes under 60 seconds.

Stopping Distance and Time: Practise the proportionality questions: (a) u doubled → S×4; (b) μ halved → S×2; (c) mass tripled → S unchanged.

Stopping Distance and Time: After stopping on an incline: check if μ_s ≥ tanθ (stays) or μ_s < tanθ (slides back down). For slides back: new acceleration = g(sinθ − μ_k cosθ) going down.

FAQ — Stopping of Block Due to Friction

Notes · Downloads · Revision · Important Questions

What 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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Stopping Distance and Time

Subtopics

Stopping Distance and Time

Newton's Laws of Motion

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Friction

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Stopping of Block Due to Friction

Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance Proportionalities in AP Physics 1

After-Stopping Sliding Analysis in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Stopping Distance and Time

Subtopics

Stopping Distance and Time

NEET > Physics > Laws of Motion Chapters

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Stopping of Block Due to Friction

Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance Proportionalities in AP Physics 1

After-Stopping Sliding Analysis in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Stopping Distance and Time

Subtopics

Stopping Distance and Time

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

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Stopping of Block Due to Friction

Overview content

Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance Proportionalities in AP Physics 1

After-Stopping Sliding Analysis in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Stopping Distance and Time

Subtopics

Stopping Distance and Time

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)

Stopping of Block Due to Friction

Overview content

Stopping of Block Due to Friction – Complete Notes, Revision, Important Questions & Downloads

1) Stopping Distance and Time — Horizontal and Inclined

Stopping Distance Proportionalities in AP Physics 1

After-Stopping Sliding Analysis in University Physics

Physics — Friction Revision Checklist

NEET NRI Counseling & Admission eBook Download

Schedule Trial Session For NEET Prep

Stopping Distance and Time

Subtopics

Stopping Distance and Time

NEET > Physics > Laws of Motion Chapters

Comments

Leave a comment

Comments (0)