Friction, Inclined Planes & Block-on-Block — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Friction, Inclined Planes & Block-on-Block MCQs with step-by-step solutions (35 questions). Part of Newton's Laws of Motion & Friction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Friction, Inclined Planes & Block-on-Block · medium · theory
Static friction is:
A. Greater than the limiting friction
B. Independent of the applied force
C. Always equal to μ_s N
D. Self-adjusting, taking any value up to a maximum of μ_s N ✓ Correct
Solution: Static friction adjusts itself to match the applied force up to a maximum (limiting) value f_max = μ_s N. Kinetic friction, once sliding starts, is essentially constant at μ_k N.
Q2 — Friction, Inclined Planes & Block-on-Block · medium · theory
For most surfaces, the relationship between the coefficients of static and kinetic friction is:
A. μ_s > μ_k ✓ Correct
B. μ_s = 0
C. μ_s < μ_k
D. μ_s = μ_k
Solution: The limiting static friction is generally larger than the kinetic friction, so μ_s > μ_k.
Q3 — Friction, Inclined Planes & Block-on-Block · medium · theory
The force of friction between two dry surfaces is approximately:
A. Independent of the normal force
B. Proportional to the area of contact
C. Independent of the (apparent) area of contact and proportional to the normal force ✓ Correct
D. Proportional to the speed
Solution: To a good approximation friction depends on the nature of the surfaces and the normal force (f = μN), but not on the apparent contact area.
Q4 — Friction, Inclined Planes & Block-on-Block · hard · theory
The angle of friction λ is related to the coefficient of static friction by:
A. sinλ = μ_s
B. cosλ = μ_s
C. tanλ = μ_s ✓ Correct
D. λ = μ_s
Solution: The angle of friction is the angle the resultant of N and limiting friction makes with N: tanλ = f_max/N = μ_s.
Q5 — Friction, Inclined Planes & Block-on-Block · hard · theory
The angle of repose (the incline angle at which a block just begins to slide) equals:
A. 90° − angle of friction
B. twice the angle of friction
C. The angle of friction, tan⁻¹(μ_s) ✓ Correct
D. tan⁻¹(μ_k)
Solution: A block on an incline just slides when mg sinθ = μ_s mg cosθ ⇒ tanθ = μ_s. So the angle of repose equals the angle of friction.
Q6 — Friction, Inclined Planes & Block-on-Block · medium · theory
Kinetic friction on a moving body always acts:
A. Perpendicular to the surface
B. Vertically upward
C. In the direction of motion
D. Opposite to the direction of relative sliding ✓ Correct
Solution: Kinetic friction opposes the relative sliding between the surfaces in contact.
Q7 — Friction, Inclined Planes & Block-on-Block · hard · theory
The minimum force required to move a block on a horizontal surface is least when the force is applied at an angle to the horizontal equal to:
A. 0° (horizontal)
B. The angle of friction, tan⁻¹(μ) ✓ Correct
C. 90° (vertical)
D. 45° always
Solution: F = μmg/√(1+μ²) is minimum when the pull is applied at the angle of friction θ = tan⁻¹(μ), because that best reduces the normal force while pulling forward.
Q8 — Friction, Inclined Planes & Block-on-Block · medium · theory
A block placed on a rough incline of angle θ will remain at rest (without any other force) provided:
A. tanθ ≥ μ_s
B. sinθ ≤ μ_s
C. θ ≥ 90°
D. tanθ ≤ μ_s ✓ Correct
Solution: It stays as long as the gravity component along the incline does not exceed the limiting friction: mg sinθ ≤ μ_s mg cosθ ⇒ tanθ ≤ μ_s.
Q9 — Friction, Inclined Planes & Block-on-Block · hard · theory
On an inclined plane, the direction of the static friction on a block:
A. Is always horizontal
B. Is always up the incline
C. Can be up or down the incline, depending on the block's tendency to slide ✓ Correct
D. Is always down the incline
Solution: Friction opposes the tendency of relative motion; if an external force tends to push the block up, friction acts down, and vice versa.
Q10 — Friction, Inclined Planes & Block-on-Block · medium · theory
The force of friction is:
A. Always doing positive work
B. A conservative force
C. A fundamental force of nature
D. A non-conservative (contact, electromagnetic in origin) force ✓ Correct
Solution: Friction is a non-conservative contact force (microscopically electromagnetic); the work it does depends on the path.
Q11 — Friction, Inclined Planes & Block-on-Block · hard · theory
A block rests on top of another block on a smooth floor, with friction between the two blocks. If a horizontal force is applied to the LOWER block, the two move together only if:
A. The upper block is heavier than the lower
B. The floor is rough
C. The friction needed to accelerate the upper block does not exceed μ_s m_upper g ✓ Correct
D. The applied force is zero
Solution: The upper block is accelerated only by friction; they move together while that friction stays below its limiting value μ_s m_upper g (max common acceleration = μ_s g).
Q12 — Friction, Inclined Planes & Block-on-Block · medium · theory
A block is pressed against a vertical wall by a horizontal force and does not slip. The friction on the block:
A. Is zero
B. Acts downward
C. Acts horizontally
D. Acts vertically upward and balances the weight ✓ Correct
Solution: The weight acts down, so static friction acts up along the wall to balance it (f = mg), while the normal force is horizontal.
Q13 — Friction, Inclined Planes & Block-on-Block · hard · theory
Two blocks resting on a rough floor are joined by a string and pulled. Whether they start moving depends on:
A. Whether the applied force exceeds the total limiting friction of both blocks ✓ Correct
B. Only the length of the string
C. The colour of the blocks
D. Only the mass of the heavier block
Solution: The system moves only when the pull exceeds the sum of the limiting frictional forces on both blocks.
Q14 — Friction, Inclined Planes & Block-on-Block · medium · theory
Rolling friction compared with sliding friction for the same bodies is:
A. Always zero
B. Exactly equal
C. Much larger
D. Much smaller ✓ Correct
Solution: Rolling friction is much smaller than sliding friction, which is why wheels and ball-bearings reduce resistance.
Q15 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A 5 kg block rests on a horizontal surface with μ_s = 0.4 (g = 10 m/s²). The maximum (limiting) static friction is:
A. 25 N
B. 2 N
C. 20 N ✓ Correct
D. 50 N
Solution: f_max = μ_s N = μ_s mg = 0.4 × 5 × 10 = 20 N.
Q16 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A 5 kg block (μ_s = 0.5, g = 10 m/s²) is pushed by a horizontal force of 15 N but does not move. The friction force acting on it is:
A. 0 N
B. 50 N
C. 15 N ✓ Correct
D. 25 N
Solution: Limiting friction = 0.5 × 50 = 25 N. Since 15 N < 25 N, the block stays still and static friction equals the applied force, 15 N. Pitfall: static friction is 15 N, not 25 N.
Q17 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A 10 kg block moving on a surface (μ_k = 0.3, g = 10 m/s²) is pulled by a horizontal force of 50 N. Its acceleration is:
A. 2 m/s² ✓ Correct
B. 5 m/s²
C. 8 m/s²
D. 3 m/s²
Solution: f_k = μ_k mg = 0.3 × 100 = 30 N; a = (50 − 30)/10 = 2 m/s².
Q18 — Friction, Inclined Planes & Block-on-Block · hard · numerical
The angle of repose for a surface with coefficient of friction 1/√3 is:
A. 45°
B. 15°
C. 30° ✓ Correct
D. 60°
Solution: θ = tan⁻¹(μ) = tan⁻¹(1/√3) = 30°.
Q19 — Friction, Inclined Planes & Block-on-Block · hard · numerical
The minimum force needed to move a 10 kg block on a horizontal floor (μ = 0.75, g = 10 m/s²), applied at the optimum angle, is:
A. 100 N
B. 80 N
C. 75 N
D. 60 N ✓ Correct
Solution: F_min = μmg/√(1+μ²) = (0.75 × 100)/√(1 + 0.5625) = 75/1.25 = 60 N. Pitfall: applying the force horizontally would require 75 N; the optimum angle reduces it to 60 N.
Q20 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A block just begins to slide down an incline when the incline angle is 45°. The coefficient of static friction is:
A. 1 ✓ Correct
B. 0.577
C. 0.5
D. 1.5
Solution: μ_s = tan(angle of repose) = tan45° = 1.
Q21 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A block slides down a rough incline of 37° with μ_k = 0.5 (sin37° = 0.6, cos37° = 0.8, g = 10 m/s²). Its acceleration is:
A. 1 m/s²
B. 4 m/s²
C. 6 m/s²
D. 2 m/s² ✓ Correct
Solution: a = g(sinθ − μ cosθ) = 10(0.6 − 0.5 × 0.8) = 10(0.6 − 0.4) = 2 m/s².
Q22 — Friction, Inclined Planes & Block-on-Block · hard · numerical
The force required to move a 2 kg block up a 30° rough incline (μ = 0.2, g = 10 m/s², cos30° ≈ 0.866) is:
A. ≈ 20 N
B. ≈ 13.5 N ✓ Correct
C. ≈ 6.5 N
D. ≈ 10 N
Solution: F = mg(sinθ + μcosθ) = 2 × 10(0.5 + 0.2 × 0.866) = 20(0.5 + 0.173) ≈ 13.5 N.
Q23 — Friction, Inclined Planes & Block-on-Block · hard · numerical
A 4 kg block rests on a 30° incline with μ_s = 0.6 (tan30° ≈ 0.577, g = 10 m/s²). The friction force on it is:
A. 24 N
B. 34.6 N
C. 0 N
D. 20 N ✓ Correct
Solution: Since tan30° = 0.577 < μ_s = 0.6, the block does not slide; static friction balances the down-slope component: f = mg sinθ = 40 × 0.5 = 20 N. Pitfall: friction is NOT μmgcosθ here (that is only the maximum).
Q24 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A body moving at 20 m/s on a rough horizontal floor (μ = 0.4, g = 10 m/s²) comes to rest after a distance of:
A. 50 m ✓ Correct
B. 100 m
C. 40 m
D. 25 m
Solution: Deceleration a = μg = 4 m/s²; s = v²/(2a) = 400/8 = 50 m.
Q25 — Friction, Inclined Planes & Block-on-Block · hard · numerical
A 2 kg block sits on a 4 kg block on a smooth floor; μ between them is 0.5 (g = 10 m/s²). The maximum horizontal force applied to the LOWER block so that they move together is:
A. 10 N
B. 60 N
C. 20 N
D. 30 N ✓ Correct
Solution: Upper block's max acceleration = μg = 5 m/s² (friction 10 N on 2 kg). Common a_max = 5 m/s² ⇒ F = (2 + 4)(5) = 30 N.
Q26 — Friction, Inclined Planes & Block-on-Block · hard · numerical
For the same stack (2 kg on 4 kg, smooth floor, μ = 0.5), the maximum force applied to the UPPER block so that they move together is:
A. 30 N
B. 10 N
C. 15 N ✓ Correct
D. 20 N
Solution: The lower block is driven only by friction (max 10 N), so a_max = 10/4 = 2.5 m/s². Then F = (2 + 4)(2.5) = 15 N. Pitfall: the limit differs from the lower-block case (30 N).
Q27 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A 2 kg block is held against a vertical wall by a horizontal push. If μ = 0.5 (g = 10 m/s²), the minimum push needed to prevent it from sliding down is:
A. 4 N
B. 20 N
C. 10 N
D. 40 N ✓ Correct
Solution: Friction must balance weight: μN ≥ mg ⇒ N ≥ mg/μ = 20/0.5 = 40 N.
Q28 — Friction, Inclined Planes & Block-on-Block · hard · numerical
Two blocks of 2 kg (μ = 0.5) and 3 kg (μ = 0.3) on a rough floor are joined by a string and pulled by 25 N (g = 10 m/s²). Their acceleration is:
A. 1.2 m/s² ✓ Correct
B. 2 m/s²
C. 5 m/s²
D. 0.5 m/s²
Solution: Total friction = 0.5(2×10) + 0.3(3×10) = 10 + 9 = 19 N; a = (25 − 19)/(2 + 3) = 6/5 = 1.2 m/s².
Q29 — Friction, Inclined Planes & Block-on-Block · medium · numerical
A 2 kg block moving at some speed on a floor with μ_k = 0.4 experiences a deceleration (g = 10 m/s²) of:
A. 8 m/s²
B. 2 m/s²
C. 4 m/s² ✓ Correct
D. 0.4 m/s²
Solution: a = μ_k g = 0.4 × 10 = 4 m/s² (independent of mass).
Q30 — Friction, Inclined Planes & Block-on-Block · hard · numerical
A block rests on the floor of a truck. The minimum coefficient of friction needed so it does not slide when the truck accelerates at 5 m/s² (g = 10 m/s²) is:
A. 0.5 ✓ Correct
B. 0.05
C. 1.0
D. 0.25
Solution: For no sliding μmg ≥ ma ⇒ μ ≥ a/g = 5/10 = 0.5.