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 5, page 12). 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)
Which of the following statements about the laws of friction is correct?
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)
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)
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)
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)
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 friction force on the block and does the block move? (Take g = 10 m/s²)
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:
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: 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. ---
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. 12An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.
| Symbol | Quantity | SI Unit |
|---|---|---|
| a_c | Centripetal acceleration | m/s^2 |
| v | Tangential speed | m/s |
| r | Radius of circle | m |
| omega | Angular speed | rad/s |
On a road banked at angle theta from horizontal with tyre-road friction coefficient mu_s, this is the maximum speed for safe negotiation of a curve of radius r.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed | m/s |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the curve | m |
| mu_s | Coefficient of static friction | (dimensionless) |
| theta | Banking angle | rad/deg |
The net inward force required to keep a body of mass m moving in a circle of radius r at speed v is m*v^2/r. This 'centripetal' force is NOT a new fundamental force — it is whichever real force (tension, friction, gravity, etc.) provides the inward acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_c | Centripetal force | N |
| m | Mass of the body | kg |
| v | Tangential speed | m/s |
| r | Radius of the circle | m |
| omega | Angular speed | rad/s |
If the net external force on a system of particles is zero, the total linear momentum of the system is conserved (vector equality of total p before and after any internal interaction). Basis of all collision and recoil analysis.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F_ext | Net external force on system | N |
| p_total | Sum of m_i*v_i over all particles | kg*m/s |
Impulse equals the product of force and the time interval over which it acts. By Newton's Second Law, impulse equals the change in linear momentum during that interval.
| Symbol | Quantity | SI Unit |
|---|---|---|
| J | Impulse (vector) | N*s = kg*m/s |
| F | Force (treated as average) | N |
| Delta_t | Time interval | s |
| Delta_p | Change in momentum | kg*m/s |
When 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 |
Static friction is the only force available to provide centripetal acceleration on a level road. Setting mu_s*N = m*v^2/r and N = m*g gives this maximum-safe speed bound.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v_max | Maximum safe speed (no skid) | m/s |
| mu_s | Coefficient of static friction (tyre vs road) | (dimensionless) |
| g | Gravitational acceleration | m/s^2 |
| r | Radius of the circular path | m |
The net external force on a body equals the rate of change of its linear momentum. For a body of constant mass, this reduces to F = m*a — net force equals mass times acceleration. Both F and a are vectors; the acceleration is in the direction of the net force.
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Net external (vector) force | N |
| m | Mass of the body | kg |
| a | Acceleration (vector) | m/s^2 |
| p | Linear momentum (= m*v) | kg*m/s |
| t | Time | s |
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: Similar Terms
Student claims velocity is constant in uniform circular motion (it's not — direction changes).
Question asks 'in uniform circular motion at constant speed, which is also constant?'
In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.
Category: Sign Convention
Student picks the direction of MOTION as the direction of the net force, instead of the direction of the change in momentum (Δp). When velocity changes direction at constant speed, the force is perpendicular to BOTH the initial and final velocity vectors (in the limit) — it's the direction of Δv.
Question describes a body changing direction (e.g. turning) and asks for the force direction or the direction of Δp.
F ∝ Δp = m Δv = m (v_f − v_i). Always draw v_i and v_f as vectors and subtract; the result (v_f − v_i) is the direction of net force.
Category: Unit Conversion
Student computes μmg (friction FORCE in Newtons) when asked for the friction-limited ACCELERATION. The two differ by a factor of m: a = μg, F = μmg.
Question gives μ, g, and a body's mass and asks for the maximum acceleration of the supporting surface OR the friction force on the body.
Read carefully: 'maximum acceleration of the vehicle so the body stays still' = μg (no mass). 'Friction force on the body' = μmg (mass present). Units expose the error: N for force, m/s² for acceleration.
Category: Sign Convention
Student forgets that velocity DIRECTION reverses on bounce; computes |v₁ - v₂| instead of |v₁ + v₂|.
Question describes ball dropped from height h₁, rebounding to height h₂; asks for impulse on ball.
Impulse J = Δp = m(v_after - v_before). Take down as positive: v_before = +v₁, v_after = -v₂. So J = m(-v₂ - v₁) = -m(v₁ + v₂); magnitude = m(v₁ + v₂).
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.
Category: Overthinking
Student takes torque about the centre of mass (introducing all 4 forces with non-zero moment arms) instead of about the floor contact (where 2 forces have zero moment arm and the equation simplifies).
Question gives a uniform rod or ladder leaning against a wall; asks for friction coefficient or limiting condition.
Pick the pivot to ELIMINATE unknown forces from the torque equation. Floor-contact pivot: normal force and friction at floor contribute zero torque; only weight (mid-length) and wall normal (top) appear. Result: μ_min = 1 / (2 tan θ).
Category: Sign Convention
Student adds magnitudes of momenta of fragments instead of vectors; ignores cancellation when fragments fly perpendicular or in opposite directions.
Question describes a body at rest exploding into multiple fragments with given mass ratios and partial velocity info.
Total momentum is a VECTOR. Initial p = 0; therefore Σ m_i v_i = 0 as a vector equation. Decompose along chosen axes (often natural symmetry axes); sum = 0 in each.
Category: Overthinking
Student tries to apply F=ma to the system as a whole (using net force = (m1-m2)g and total mass m1+m2) but loses track of the tension. The correct approach writes Newton's Second Law SEPARATELY for each mass and treats T as an unknown in two simultaneous equations.
Question contains a frictionless pulley with two unequal masses tied to a string. Asks for tension T or acceleration a (or both).
Draw a free-body diagram for EACH mass. Write F=ma per body, treating T as the same magnitude on both sides of the string. Solve simultaneously: a = (m1 - m2) g / (m1 + m2); T = 2 m1 m2 g / (m1 + m2).
Category: Negative Marking
Multi-mass pulley problem requires computing acceleration first, then tension. T = 2 m1 m2 g/(m1+m2). Sign errors in m1−m2 propagate.
Atwood machine or pulley system with multiple masses.
Compute a = (m1-m2)g/(m1+m2) first with m1 the heavier mass and downward as positive. Then T from F=ma on either mass. Always re-check by plugging back into both Newton's 2nd Law equations.
Category: Overthinking
Student applies the full external force F to a single block instead of recognising the system needs to be analysed for the contact (internal) force.
Question gives horizontal force F on block A which pushes block B; asks for contact force between A and B or acceleration.
System acceleration: a = F / (m_A + m_B). Contact force on B from A = m_B × a. The full F acts on the system, not on each block independently.
Category: Overthinking
Student computes P = Mgv (just lifting against gravity) and ignores the friction-opposing-motion term.
Question describes a lift moving at constant speed with explicit friction force on cable or guides.
At constant speed, net force = 0, so cable tension T = Mg + f_friction. Power = T × v = (Mg + f) × v. Always add friction when stated.
Root cause: concept gap
Action-reaction pairs ALWAYS act on DIFFERENT bodies. The pair to the book's weight is the gravitational pull the book exerts on the Earth. The pair to the normal force from table on book is the force the book exerts on the table. Equal-and-opposite forces on the SAME body are an equilibrium statement, not third-law statement.
Distractor labels two forces on the same body as a Newton's-third-law pair.
Root cause: concept gap
Centripetal force is NOT a new fundamental force. It is the NET inward radial component of the real forces (tension, friction, normal, gravity, etc.). On a free-body diagram, draw only the real forces; their net inward component must equal m*v^2/r.
Distractor sums tension + 'centripetal force' as separate inward forces.
Root cause: unit error
Impulse J has dimensions of momentum (kg*m/s) and equals F*Delta_t. Force has dimensions kg*m/s^2. Confusing them inflates or deflates an answer by a factor of seconds. Always check units before declaring an answer.
Distractor reports an answer in newtons where the correct answer is in N*s (or vice versa).
Root cause: concept gap
Linear momentum is conserved only when the net EXTERNAL force is zero. Gravity over a finite time changes momentum. For collision problems we use conservation because the collision happens over a brief time interval where external impulses are negligible compared to internal collision forces.
Distractor sets initial momentum = final momentum for a free-fall problem where gravity has acted for several seconds.
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.
treats velocity as scalar
Conflates speed (scalar) with velocity (vector)
scalar sum of momenta
Treating momentum as scalar; ignores vector cancellation
force along final velocity
Default to thinking force points in direction of motion
force along initial velocity
Newton-1 misread: object 'wants' to keep moving in original direction
subtracts v2 from v1 instead of adding
Forgetting velocity DIRECTION reversal at bounce
ignores mu cos theta term
Forgetting the friction-along-incline component
torque about wrong pivot
Default pivot at center of mass adds complexity
uses g where a belongs
Forgetting that the system accelerates, so weight is balanced by net force minus T
confuses tension with weight
Treating T = m·g for one of the masses (which is true only when a=0)
uses mu times g times mass
Confusing force (μ·N = μmg) with acceleration
treats each block with full F
Forgetting Newton's 3rd law internal force decomposition
forgets friction term
Ignoring opposing force
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