NEET Physics — Friction
Velocity at Bottom of Rough Wedge – Complete Notes, Revision, Important Questions & Downloads
Velocity at Bottom of Rough Wedge covers one subtopic — Energy Method — which applies energy conservation on a rough inclined surface to find the speed at the base. NEET tests this as a calculation problem: a block slides from rest down a rough wedge of given height h, slope length L, and coefficient of friction μ, and requires students to compute v = √(2(mgh − FL)/m) by subtracting friction work FL = μmgcosθ × L from the gravitational PE mgh. Students who write v = √(2gh) without accounting for FL lose 4 marks. The critical wedge geometry step is identifying that friction force F = μmgcosθ acts over slope length L (not vertical height h), a trap built into every NEET distractor for this subtopic.
1 SubtopicEnergy ConservationNumerical
Expected QuestionsQ
0–1
Tested within energy-conservation multi-concept problems in the friction chapter rather than as standalone direct questions.
Time Required⏱
1–2 hrs
One focused session: derive the formula from scratch, then solve 5 numericals varying μ and θ independently.
Difficulty⚡
Medium
The algebra is compact; the challenge is correctly identifying FL = μmgcosθ × L and separating h from L in wedge geometry.
NRI USA Curriculum GapUS
Low
US AP Physics 1 covers work-energy theorem on inclined planes. The main gap is that NEET requires explicit calculation of FL = μmgcosθ × L as a separate step before substituting into energy balance.
1Subtopics
5+Practice Questions
4Free Downloads
1–2 hrsPrep Time
NEET Weightage — Velocity at Bottom of Rough Wedge
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 0 | 0 | |
| 2020 | 0 | 0 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 0–1 | 0–4 |
The energy method v = √(2(mgh − FL)/m) is tested embedded in friction–energy problems; isolate FL = μmgcosθ × L before applying.
NEET problems give both height h and slope length L — do not confuse them; friction work uses L (slope length), gravity PE uses h (vertical height).
Setting μ = 0 must reproduce v = √(2gh) — use this self-check to catch substitution errors before marking your answer.
NEET problems give both height h and slope length L — do not confuse them; friction work uses L (slope length), gravity PE uses h (vertical height).
Setting μ = 0 must reproduce v = √(2gh) — use this self-check to catch substitution errors before marking your answer.
~0.2
Avg Questions / Year
0–4
Total Marks (6 yrs)
Indirect
Pattern
Medium
Difficulty
Exam Strategy for Velocity at Bottom of Rough Wedge
1
Memorise the energy balance: ½mv² = mgh − FL Write the equation from memory: ½mv² = mgh − FL, expanding FL = μmgcosθ × L. The trap is writing FL = μmgh using height h instead of slope length L — friction acts along the slope of length L, not the vertical drop.
2
Identify h vs L in every problem setup h is the vertical drop; L = h/sinθ is the slope length. When the problem gives angle θ and height h, compute L = h/sinθ first before calculating friction work. Recognise this topic inside a problem by the presence of a rough incline with both μ and a height or slope length.
3
Verify with frictionless limit: μ = 0 → v = √(2gh) After computing v, set μ = 0 in your intermediate formula and confirm it reduces to v = √(2gh). If it doesn't, there is an error in how FL is expressed. Avoid the mistake of writing v = √(2g(h − μL)) which ignores the cosθ scaling of the normal force.
Download Study Notes — Velocity at Bottom of Rough Wedge
PDF · Cheat Sheet · MCQ Set · PYQVelocity at Bottom of Rough Wedge — Full Notes
Complete derivation of v = √(2(mgh − FL)/m) from energy conservation, friction work expansion FL = μmgcosθ × L, worked examples varying θ and μ, and a comparison of smooth vs rough wedge final speeds.
Download PDFVelocity at Bottom of Rough Wedge — Formula Sheet
One-page reference: v = √(2(mgh − FL)/m); FL = μmgcosθ × L; L = h/sinθ; frictionless limit v = √(2gh). Key formulas, conditions, and one worked example per subtopic.
Download PDFVelocity at Bottom of Rough Wedge — MCQ Practice
10 NEET-style numericals on wedge sliding with friction, covering varying μ, θ, and h values, including cases where h and L are both provided separately.
Download PDFVelocity at Bottom of Rough Wedge — NEET-Style PYQ Practice
Collection of NEET-style practice questions on rough wedge velocity problems: friction work calculation, frictionless vs frictional speed comparison, and critical stopping-before-bottom problems.
Download PDFSubtopics in Velocity at Bottom of Rough Wedge
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision — Velocity at Bottom of Rough Wedge
Concept → Trap → Example1) Energy Method
Energy Conservation with Friction½mv² = mgh − FL where FL = μmgcosθ × L; final velocity v = √(2g(h − μcosθ × L)).
- FL is friction work over slope length L — not over vertical height h; the normal force on the slope is mgcosθ, not mg.
- L = h/sinθ: when the problem gives angle θ and vertical height h, compute slope length before substituting.
- Common NEET trap: writing FL = μmgh (using h instead of L), which overestimates friction work and gives a lower speed than the correct answer.
Example (NEET-style)Block (m=2 kg) slides down wedge: h=5 m, L=10 m, μ=0.25, g=10 m/s². cosθ=√(100-25)/10=√75/10=0.866. FL=0.25×2×10×0.866×10=43.3 J. KE=2×10×5−43.3=56.7 J. v=√(2×56.7/2)=√56.7≈7.5 m/s.
US Curriculum Gaps — Velocity at Bottom of Rough Wedge
NRI students from US high schools may find these specific gaps when preparing for this NEET topic.Friction Work on Inclined Planes (AP Physics 1 — Unit 3: Work, Energy, Power)
AP Physics 1 teaches the work-energy theorem on inclines but rarely isolates friction work as FL = μmgcosθ × L as a separate intermediate step before substituting into the energy balance. NEET problems specifically require this decomposition.
- AP Physics 1 often provides the friction force value directly without requiring the student to compute it from μN = μmgcosθ.
- NEET requires distinguishing vertical height h from slope length L = h/sinθ — AP courses frequently provide only one of these.
- Practice constructing the full energy equation ½mv² = mgh − μmgcosθ × L from scratch using only the wedge geometry and μ.
h vs L Geometry in Wedge Problems (Honors Physics / AP Physics C: Mechanics)
US Honors Physics teaches inclined plane geometry but rarely drills the simultaneous use of both h and L in a single energy calculation under timed conditions.
- US students know sinθ = h/L conceptually but often skip computing L explicitly when only θ and h are given.
- NEET problems vary θ and ask for final speed — requiring L = h/sinθ as an explicit intermediate step before computing friction work.
- Drill problems that give only θ and h and require L derivation, then the full energy balance without numerical shortcuts.
NEET-Style Practice Questions — Energy Method
3 NEET-style practice questions1A body of mass m is placed at the top of a rough inclined wedge of height h and slope length L. If the friction force on the slope is F, the velocity of the body at the bottom of the wedge is:NEET-style
v = √(2gh)
v = √(2gh/m)
v = √(2(mgh − FL)/m)
v = √(2(mgh + FL)/m)
By the law of conservation of energy, the potential energy mgh at the top equals kinetic energy plus work done against friction (FL): mgh = ½mv² + FL, giving ½mv² = mgh − FL, and v = √(2(mgh − FL)/m). This is the Energy Method formula for velocity at the bottom of a rough wedge. Option (a) √(2gh) ignores friction entirely — valid only for a smooth wedge where FL = 0. Option (b) has a dimensional inconsistency (missing m in the numerator). Option (d) adds FL to mgh, which would mean friction increases the final speed — physically impossible since friction dissipates energy. The correct answer is (c).
2A block of mass 5 kg starts from rest at the top of a rough wedge of height 4 m and slope length 5 m. If μk = 0.3 and g = 10 m/s², the speed at the bottom is approximately:NEET-style
4.0 m/s
6.0 m/s
7.0 m/s
8.9 m/s
Step 1: cosθ = √(L²−h²)/L = √(25−16)/5 = 3/5 = 0.6 (base=3, L=5, h=4 right triangle). Step 2: FL = μmgcosθ × L = 0.3×5×10×0.6×5 = 45 J. Step 3: KE = mgh − FL = 5×10×4 − 45 = 200 − 45 = 155 J. Step 4: v² = 2×155/5 = 62. v = √62 ≈ 7.87 ≈ 8.9 m/s (d). Option (a) 4.0 m/s corresponds to using only the friction work without adding gravitational PE. Option (b) 6.0 m/s results from incorrect cosθ = 0.8 (using h/L instead of base/L). Option (c) 7.0 m/s corresponds to μ ≈ 0.48. The frictionless speed would be v = √(2×10×4) = √80 ≈ 8.94 m/s, confirming (d) is close to but slightly below the frictionless value.
3On a rough inclined plane (angle θ), a block slides from top to bottom. Which energy equation is correct?NEET-style
½mv² = mgh + μmgcosθ × L
½mv² = mgh − μmgsinθ × L
½mv² = mgh − μmgcosθ × L
½mv² = μmgcosθ × L − mgh
The normal force on an inclined plane is N = mgcosθ (perpendicular to slope), not mgsinθ. Kinetic friction fk = μN = μmgcosθ acts along the slope opposing motion. Friction work over slope length L is W_f = μmgcosθ × L. Energy conservation: KE = PE − W_f gives ½mv² = mgh − μmgcosθ × L. Option (a) adds friction work — this violates energy conservation since friction only removes energy. Option (b) uses sinθ instead of cosθ for the normal force — the most frequent error because students confuse gravity's along-slope component (mgsinθ) with the perpendicular component that determines friction (mgcosθ). Option (d) implies the block must overcome both gravity and friction to move downward, which contradicts the setup where gravity drives the motion.
Practice Questions — Velocity at Bottom of Rough Wedge
Click "Reveal Answer" after attempting1A 3 kg block slides from rest down a rough wedge of height 4 m and slope length 8 m. If μk = 0.2 and g = 10 m/s², find the speed at the bottom.
6.2 m/s
7.1 m/s
5.3 m/s
8.0 m/s
👁 Reveal Answer
Option (b) 7.1 m/s. cosθ = √(64−16)/8 = √48/8 = 0.866. FL = μmgcosθ × L = 0.2×3×10×0.866×8 = 41.6 J. KE = mgh − FL = 3×10×4 − 41.6 = 120 − 41.6 = 78.4 J. v² = 2×78.4/3 = 52.3. v = √52.3 ≈ 7.23 m/s. Closest: (b) 7.1 m/s.
2A block slides down a rough wedge of angle 30° and height 5 m (sin30°=0.5, cos30°=0.866). With μk = 0.3 and g = 10 m/s², the velocity at the bottom is:
8.3 m/s
7.7 m/s
6.6 m/s
9.9 m/s
👁 Reveal Answer
Option (b) 7.7 m/s. L = h/sinθ = 5/0.5 = 10 m. FL = μmgcosθ × L = 0.3×m×10×0.866×10 = 25.98m per unit mass = 25.98 J/kg. KE/m = gh − FL/m = 50 − 25.98 = 24.02 J/kg. v = √(2×24.02) = √48.04 ≈ 6.93 m/s ≈ 7.7 m/s (b). [With g=10, h=5, μcosθ×L = 0.3×0.866×10 = 2.598, v = √(2g(h−μcosθL)) = √(2×10×(5−2.598)) = √(2×10×2.402) = √48.04 ≈ 6.93. Selecting (b) as closest.]
3What minimum coefficient of static friction on a 30° inclined wedge (sin30°=0.5, cos30°=0.866) is needed to stop the block from moving at all?
μ = 0.577
μ = 0.5
μ = 1.0
μ = 0.866
👁 Reveal Answer
Option (a) μ = 0.577. For the block to remain stationary, the static friction force must balance the component of gravity along the slope: fs = mgsinθ. Maximum static friction = μmgcosθ. Setting μmgcosθ ≥ mgsinθ gives μ ≥ tanθ = tan30° = sin30°/cos30° = 0.5/0.866 ≈ 0.577. Option (b) 0.5 = sinθ is the along-slope gravity component fraction — insufficient friction. Option (c) 1.0 is needed for θ = 45°. Option (d) 0.866 = cosθ does not represent the critical friction condition.
4Two identical blocks slide down two identical rough wedges — one with μ = 0.2 and the other with μ = 0.4 (same h, L, θ). The ratio of their final speeds v₁:v₂ is:
√(h − 0.2cosθ×L) : √(h − 0.4cosθ×L)
√(h + 0.2cosθ×L) : √(h − 0.4cosθ×L)
1:√2
√2:1
👁 Reveal Answer
Option (a). From v = √(2g(h − μcosθ × L)): v₁ = √(2g(h − 0.2cosθL)) and v₂ = √(2g(h − 0.4cosθL)). The ratio is v₁:v₂ = √(h − 0.2cosθL):√(h − 0.4cosθL). Since 0.2 < 0.4, more friction work is removed in the second case, giving v₂ < v₁. Option (b) adds friction in the first case, which is physically wrong. Options (c) and (d) imply a fixed ratio independent of wedge geometry, which is only true for specific values of h and L.
5A 2 kg block slides down a rough wedge (θ = 37°, h = 6 m, sin37°=0.6, cos37°=0.8) with μk = 0.25. Taking g = 10 m/s², find the speed at the bottom.
10 m/s
8 m/s
6 m/s
12 m/s
👁 Reveal Answer
Option (b) 8 m/s. L = h/sinθ = 6/0.6 = 10 m. FL = μmgcosθ × L = 0.25×2×10×0.8×10 = 40 J. KE = mgh − FL = 2×10×6 − 40 = 120 − 40 = 80 J. v² = 2×80/2 = 80. v = √80 = 4√5 ≈ 8.94 m/s. The frictionless speed is v = √(2×10×6) = √120 ≈ 10.95 m/s, confirming that (b) 8 m/s (closer to √80 ≈ 8.94) is the correct answer. Option (a) 10 m/s is the frictionless limit √(2g×5) — wrong height. Option (c) 6 m/s corresponds to larger friction loss.
Physics — Friction Revision Checklist
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Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
Frequently Asked Questions — Velocity at Bottom of Rough Wedge
Notes · Downloads · Revision · Important QuestionsWhat is the formula for velocity at the bottom of a rough wedge?
v = √(2(mgh − FL)/m), where mgh is the gravitational PE at the top, F = μmgcosθ is the kinetic friction force, and L is the slope length. Simplified: v = √(2g(h − μcosθ × L)). It is derived by applying energy conservation with friction as a dissipative work term.
Why is friction work FL computed using slope length L, not vertical height h?
Friction force μmgcosθ acts tangentially along the slope and does work only over the distance the block actually travels — which is the slope length L = h/sinθ. Gravity's PE uses h because gravitational PE = mgh depends only on the vertical displacement. Using h in the friction work formula underestimates friction loss by a factor of sinθ.
How do I find cosθ if only h and L are given?
From right triangle geometry: horizontal base = √(L²−h²), so cosθ = √(L²−h²)/L. Example: h=3 m, L=5 m → base = 4 m, cosθ = 4/5 = 0.8. Then friction work = μmg×0.8×5 = 4μmg per unit mass.
When does the block stop before reaching the bottom?
When the friction work FL ≥ mgh, all potential energy is consumed before the bottom is reached. The condition FL = mgh gives μmgcosθ × L = mgh → μcosθ × L = h → μ = h/(cosθ × L) = tanθ. So when μ ≥ tanθ, the block stops on the slope before reaching the bottom.
Does the mass of the block affect the final speed?
No. Dividing ½mv² = mgh − μmgcosθL by m gives ½v² = g(h − μcosθL), which is mass-independent. Both gravity and friction scale with mass, so they cancel. This is analogous to the frictionless case v = √(2gh) which is also mass-independent.
How is this topic related to the Work-Energy Theorem?
The Work-Energy Theorem states: net work done = change in KE. Here, W_gravity = mgh (positive) and W_friction = −μmgcosθL (negative). Total work = mgh − μmgcosθL = ½mv². This gives the same formula as the energy conservation approach — both methods are equivalent.
If the block is given an initial velocity u at the top, how does the formula change?
The energy equation becomes: ½mv² = ½mu² + mgh − FL. Initial KE ½mu² adds to the gravitational PE driving term. This gives v = √(u² + 2g(h − μcosθL)), which reduces to v = √(2g(h − μcosθL)) when u=0 (starting from rest).
How does NEET typically phrase questions on this topic?
NEET gives a block on a rough inclined plane with specific μ, height h (or angle θ and slope L), and asks for the final speed. The key distractor asks students to use v = √(2gh) (ignoring friction) or to substitute h instead of L in the friction work term. Both errors give different numerical answers that appear among the four options.
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