Here is the trap that costs marks on friction questions: treating static friction as always equal to μₛN.
Static friction is self-adjusting. When you place a 10 kg block on a rough surface (μₛ = 0.4) and push it with 5 N, the friction force is 5 N — not μₛmg = 39.2 N. Static friction matches the applied force exactly, up to a ceiling of f_{s,max} = μₛN. Only when the applied force reaches or exceeds that ceiling does the block begin to slide. This is the single most tested conceptual point in NEET friction questions.
Once sliding begins, kinetic friction takes over: f_k = μ_k N, where μ_k is typically less than μₛ for the same surface pair (NCERT Class 11 Physics Chapter 4, page 60). Kinetic friction is approximately constant — it does not self-adjust. Its direction is opposite to the instantaneous relative velocity of the sliding surfaces.
The three empirical laws of friction (Coulomb-Amontons laws) are:
Watch out on inclines. On a rough incline at angle θ, the normal force is N = mg cos θ, not mg. A common error is writing friction as μmg (flat-surface normal) instead of μmg cos θ. This drops the cos θ factor entirely. For a block sliding down a rough incline: a = g(sin θ − μ_k cos θ). For a smooth incline (μ = 0), a = g sin θ. The difference between these two expressions is exactly the friction-limited deceleration μ_k g cos θ.
Another high-frequency confusion: when a question asks for the maximum acceleration a vehicle can have before an object on its floor slides, the answer is a_max = μₛg — no mass term. The friction force is μₛmg, but dividing by mass m gives acceleration μₛg. Mixing up force and acceleration costs a mark plus a negative-marking penalty.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
A 5 kg block rests on a horizontal surface with μₛ = 0.4 and μ_k = 0.3. A horizontal force of 10 N is applied. What is the friction force acting on the block? (Take g = 10 m/s²)
Answer: D. D is correct. Maximum static friction = μₛmg = 0.4 × 5 × 10 = 20 N. Since the applied force (10 N) is below this ceiling, static friction self-adjusts to exactly match it: f_s = 10 N. The block does not move (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because 19.6 N would be μₛ × m × 9.8, using g = 9.8 instead of the given g = 10 m/s², and more fundamentally, it uses f_s = μₛN regardless of the applied force — static friction self-adjusts below its maximum (trap: assuming static friction always equals μₛN).
Why B is wrong: B is wrong because 15 N = μ_k × m × 10, which applies kinetic friction. The block has not started sliding, so kinetic friction is irrelevant.
Why C is wrong: C is wrong because 14.7 N = μ_k × m × 9.8, which applies kinetic friction with g = 9.8. The block is not sliding (applied force < f_{s,max}), so kinetic friction does not apply.
Which of the following statements about the laws of friction is correct?
Answer: B. B is correct. The Coulomb-Amontons laws state that maximum static friction is proportional to the normal force and independent of the apparent area of contact (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because the Coulomb-Amontons law explicitly states that maximum static friction is independent of the apparent contact area.
Why C is wrong: C is wrong because kinetic friction is typically LESS than maximum static friction for the same pair of surfaces (μ_k < μₛ).
Why D is wrong: D is wrong because μ_k is typically less than μₛ, not greater. This reversal is a common confusion.
The coefficient of static friction between the tyres of a car and a road surface is 0.5. What is the maximum acceleration the car can achieve on a level road without wheel spin? (Take g = 10 m/s²)
Answer: A. A is correct. The maximum friction force is f_{s,max} = μₛmg. By Newton's second law, ma = μₛmg, so a_max = μₛg = 0.5 × 10 = 5 m/s². Mass cancels (NCERT Class 11 Physics Chapter 4, page 60).
Why B is wrong: B is wrong because 0.5 m/s² mistakes the coefficient itself for the acceleration. The coefficient is dimensionless; the acceleration is μₛ × g.
Why C is wrong: C is wrong because 50 m/s² = μₛ × g × 10, which introduces a spurious factor — likely confusing with μₛmg for a 10 kg body and not dividing by mass (trap: confusing friction force with friction-limited acceleration).
Why D is wrong: D is wrong because 5000 N is a force (μₛmg for some assumed mass), not an acceleration. The question asks for acceleration in m/s², not force in N (trap: reporting force when acceleration is asked).
A block of mass 2 kg is placed on a rough horizontal surface (μₛ = 0.3, μ_k = 0.2). A horizontal force is gradually increased from zero. At what applied force does the block just begin to move? (Take g = 10 m/s²)
Answer: C. C is correct. The block begins to move when the applied force equals the maximum static friction: f_{s,max} = μₛ × m × g = 0.3 × 2 × 10 = 6 N (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because 4 N = μ_k × m × g = 0.2 × 2 × 10. This uses kinetic friction, which applies only after the block is already sliding — not at the onset of motion (trap: using μ_k instead of μₛ for impending motion).
Why B is wrong: B is wrong because 2 N = μ_k × g = 0.2 × 10, which both uses the kinetic coefficient and leaves out the mass; the onset of motion needs μₛ × m × g = 6 N.
Why D is wrong: D is wrong because 3 N = μₛ × g = 0.3 × 10, omitting the mass. This gives the friction-limited acceleration, not the friction force (trap: confusing acceleration and force).
A block slides down a rough incline of angle 30° with μ_k = 0.2. What is the acceleration of the block? (Take g = 10 m/s², cos 30° = √3/2 ≈ 0.866, sin 30° = 0.5)
Answer: A. A is correct. On a rough incline, a = g(sin θ − μ_k cos θ) = 10(0.5 − 0.2 × 0.866) = 10(0.5 − 0.1732) = 10 × 0.3268 = 3.27 m/s² (NCERT Class 11 Physics Chapter 4, page 60).
Why B is wrong: B is wrong because 5.00 m/s² = g sin 30° = 10 × 0.5. This is the smooth-incline answer — the friction term μ_k g cos θ has been entirely omitted (trap: missing μ cos θ term on rough incline).
Why C is wrong: C is wrong because 3.00 m/s² = g(sin 30° − μ_k) = 10(0.5 − 0.2). This subtracts μ_k directly from sin θ instead of μ_k cos θ, dropping the cos θ factor from the friction component.
Why D is wrong: D is wrong because 6.73 m/s² = g(sin 30° + μ_k cos 30°), which ADDS the friction term instead of subtracting. Friction opposes the sliding motion (down the incline), so it must be subtracted.
A crate rests on the floor of a truck. The coefficient of static friction between the crate and the truck floor is 0.6. What is the maximum acceleration the truck can have so that the crate does not slide? (Take g = 10 m/s²)
Answer: C. C is correct. Static friction provides the horizontal force on the crate: f_s = ma. At the limit, μₛmg = ma_max, so a_max = μₛg = 0.6 × 10 = 6 m/s². Mass cancels (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because 0.6 m/s² treats the dimensionless coefficient μₛ as the acceleration itself, without multiplying by g.
Why B is wrong: B is wrong because 60 N is μₛmg for some assumed mass (e.g. 10 kg), which gives force, not acceleration. The question asks for acceleration in m/s² (trap: reporting friction force instead of friction-limited acceleration).
Why D is wrong: D is wrong because 16.7 m/s² = g/μₛ = 10/0.6, which inverts the relationship. The correct formula is a_max = μₛg, not g/μₛ.
A 4 kg block rests on a rough horizontal surface with μₛ = 0.5. A horizontal force of 25 N is applied. What is the maximum friction force the surface can provide, and does the block move? (Take g = 10 m/s²)
Answer: B. B is correct. Maximum static friction = μₛmg = 0.5 × 4 × 10 = 20 N. The applied force (25 N) exceeds this ceiling, so the block moves. Once moving, kinetic friction acts, but at the instant of breaking free, the maximum static friction the surface could provide was 20 N — insufficient to prevent motion (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because if static friction self-adjusted to 25 N, that would exceed f_{s,max} = 20 N. Static friction cannot exceed its maximum. The block must move (trap: assuming static friction always equals the applied force without checking the ceiling).
Why C is wrong: C is wrong because while 20 N is the correct maximum static friction, the block DOES move since the applied 25 N exceeds this limit.
Why D is wrong: D is wrong because 12 N = μ_k × m × g with an assumed μ_k = 0.3 not given in the problem. The question asks about the friction force at the onset — which is the maximum static friction of 20 N.
Two identical inclines have the same angle θ and length L. One is smooth and the other is rough with coefficient of kinetic friction μ_k. If a block slides from rest down each incline, the ratio of the time taken on the rough incline (t_r) to the time on the smooth incline (t_s) is:
Answer: C. C is correct. Smooth incline: a_s = g sin θ. Rough incline: a_r = g(sin θ − μ_k cos θ). Using L = ½at², we get t = √(2L/a). The ratio t_r/t_s = √(a_s/a_r) = √(sin θ / (sin θ − μ_k cos θ)) (NCERT Class 11 Physics Chapter 4, page 60).
Why A is wrong: A is wrong because sin θ + μ_k cos θ in the denominator adds friction instead of subtracting it from g sin θ. On a downward slide, friction opposes motion and reduces acceleration, so the denominator must be sin θ − μ_k cos θ.
Why B is wrong: B is wrong because it inverts the ratio. Since the rough incline has smaller acceleration, t_r > t_s, so the ratio t_r/t_s > 1. Option B gives a value < 1 (trap: inverting a_r/a_s instead of a_s/a_r under the square root).
Why D is wrong: D is wrong because it omits the square root. From L = ½at², time goes as 1/√a, so the ratio involves √(a_s/a_r), not a_r/a_s.
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Pattern: Body on an accelerating vehicle — maximum acceleration before sliding (NEET pattern: static friction limit body on vehicle, observed 2023).
Given
A box of mass m = 20 kg rests on the floor of a truck. The coefficient of static friction between the box and the truck floor is μₛ = 0.40. Take g = 10 m/s² (exact, problem-defined).
Required
Find the maximum acceleration of the truck such that the box does not slide on the floor.
Concept
Static friction between the box and the truck floor is the only horizontal force on the box. This friction accelerates the box along with the truck. If the truck's acceleration exceeds the friction-limited value, the box slides backward relative to the truck.
Formula
At the sliding threshold: f_{s,max} = ma_max
f_{s,max} = μₛN = μₛmg
Therefore: μₛmg = ma_max
Substitution
μₛ × m × g = m × a_max
0.40 × 20 × 10 = 20 × a_max
Calculation
80 = 20 × a_max
a_max = 80 / 20 = 4.0 m/s²
Note: mass m cancels algebraically (μₛg = a_max), so the result is independent of the box's mass. The value g = 10 m/s² is an exact problem-defined constant and does not limit significant figures.
Final answer
a_max = 4.0 m/s²
The truck can accelerate at up to 4.0 m/s² without the box sliding. Beyond this, kinetic friction takes over and the box slides backward relative to the truck.
Common trap
Computing μₛmg = 0.40 × 20 × 10 = 80 N and reporting 80 N as the answer. The question asks for acceleration (m/s²), not force (N). Always divide by mass — or better, recognise algebraically that a_max = μₛg and skip the mass entirely (trap: friction force vs accel).
Similar NEET-style question
A glass is placed on the dashboard of a car. The coefficient of static friction between the glass and dashboard is 0.30. What is the maximum deceleration the car can have during braking so that the glass does not slide forward? (Take g = 10 m/s²)
Answer: a_max = μₛg = 0.30 × 10 = 3.0 m/s².
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Static friction f_s opposes impending motion between surfaces in contact and is self-adjusting up to a maximum f_s ≤ μ_s N (where N is normal force, μ_s coefficient of static friction). Kinetic friction f_k = μ_k N opposes actual sliding motion (μ_k < μ_s typically). Rolling friction is much smaller than sliding friction.
-- NCERT Class 11 Physics, Ch. 4, p. 60When two surfaces slide relative to each other, kinetic friction opposes the motion with magnitude proportional to the normal force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_k | Kinetic friction force | N |
| mu_k | Coefficient of kinetic friction | (dimensionless) |
| N | Normal force | N |
The maximum value of static friction between two surfaces in contact equals the coefficient of static friction times the normal force. Below f_s_max, static friction self-adjusts to whatever value is needed to prevent relative motion.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_s_max | Maximum static friction | N |
| mu_s | Coefficient of static friction | (dimensionless) |
| N | Normal force | N |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Sign Convention
Student writes a = g sin θ for a rough incline (which is the smooth-incline answer); forgets to subtract μ g cos θ.
Question contrasts rough vs smooth inclines, or asks for acceleration on a rough incline.
On a rough incline (block sliding down): a = g(sin θ - μ_k cos θ). On a rough incline (block sliding up): a = -g(sin θ + μ_k cos θ). Smooth case (μ = 0): just g sin θ.
Root cause: formula misuse
Static friction is SELF-ADJUSTING: f_s exactly cancels the applied tangential force up to a ceiling f_s_max = mu_s * N. Below the ceiling, f_s = applied force. At the ceiling, motion is impending. Substituting mu_s * N too early over-estimates the friction.
Distractor uses mu_s * N for the static friction force in a no-slip scenario where the applied force is well below threshold.
More in Laws of Motion: 14 exam traps and mistakes · 7 formulas · 9 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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