Kinetic Friction — NEET Physics MCQs with Solutions
Free NEET Physics Kinetic Friction MCQs with step-by-step solutions (16 questions). Part of Friction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Kinetic Friction · medium · numerical
If the normal force is doubled, the co-efficient of friction is:
A. not changed ✓ Correct
B. doubled
C. tripled
D. halved
Solution: The coefficient of friction depends only on the nature of the two surfaces in contact, not on the normal force. Doubling the normal force doubles the friction force but leaves μ unchanged.
Q2 — Kinetic Friction · medium · numerical
If the coefficient of friction of a plane inclined at 45° is 0.5, then the acceleration of a body sliding freely down it is (g = 9.8 m/s²):
A. 9.8/√2 m/s²
B. 4.9 m/s²
C. 9.8/(2√2) m/s² ✓ Correct
D. 9.8 m/s²
Solution: a = g(sinθ − μcosθ) = 9.8 × (1/√2) × (1 − 0.5) = 9.8/(2√2) ≈ 3.46 m/s².
Q3 — Kinetic Friction · medium · numerical
A body of mass 100 g is sliding down an inclined plane of inclination 30°. What is the frictional force experienced if μ = 0.7? (g = 10 m/s²)
A. 0.7 × √2 × (1/√3) N
B. 0.7 × √3 × (1/2) N ✓ Correct
C. 0.7 × √2 × (1/3) N
D. 0.7 × √3 N
Solution: Kinetic friction f = μmg cosθ = 0.7 × (0.1 × 10) × cos30° = 0.7 × √3 × (1/2) N.
Q4 — Kinetic Friction · medium · numerical
A car is moving along a straight horizontal road with a speed v₀. If the coefficient of friction between the tyres and the road is μ, then the shortest distance in which the car can be stopped is:
A. v₀/μ
B. v₀/(μg)
C. v₀²/(2μg) ✓ Correct
D. v₀²/(μg)
Solution: Using v² = u² − 2as with final velocity 0 and retardation μg: 0 = v₀² − 2μg·s, so s = v₀²/(2μg).
Q5 — Kinetic Friction · medium · numerical
A block of mass 10 kg is moving up an inclined plane of inclination 30° with an initial speed of 5 m/s. It stops after 0.5 s. What is the value of the coefficient of kinetic friction? (g = 10 m/s²)
A. 1/√3 ✓ Correct
B. √3
C. 0.6
D. 0.5
Solution: Retardation up the incline = v/t = 5/0.5 = 10 m/s² = g(sinθ + μcosθ) = 10(0.5 + μ·√3/2). Solving gives μ = 1/√3.
Q6 — Kinetic Friction · medium · numerical
A particle is projected up a rough plane inclined at 45° to the horizontal. If the coefficient of friction is 1/2, then the retardation is:
A. (g/√2)(1 + 1/2) ✓ Correct
B. g/√2
C. (g/√2)(1 − 1/2)
D. g/2
Solution: Retardation while moving up = g(sinθ + μcosθ) = (g/√2)(1 + 1/2) at θ = 45°, μ = 1/2.
Q7 — Kinetic Friction · medium · numerical
A body of mass 10 kg lies on a rough horizontal surface. When a horizontal force of F newtons acts on it, it gets an acceleration of 5 m/s². When the horizontal force is doubled, it gets an acceleration of 18 m/s². The coefficient of friction between the body and the surface is (take g = 10 m/s²):
A. 0.8 ✓ Correct
B. 0.2
C. 0.4
D. 0.6
Solution: F − μmg = 10 × 5 and 2F − μmg = 10 × 18. Subtracting gives F = 130 N, then 130 − 100μ = 50, so μ = 0.8.
Q8 — Kinetic Friction · medium · numerical
On the horizontal surface of a truck, a block of mass 1 kg is placed (μ = 0.6) and the truck is moving with acceleration 5 m/s². The frictional force on the block will be:
A. 8 N
B. 5 N ✓ Correct
C. 6 N
D. 5.88 N
Solution: The friction needed to accelerate the block with the truck is ma = 1 × 5 = 5 N, which is less than the maximum static friction μmg = 5.88 N, so the friction force is 5 N.
Q9 — Kinetic Friction · medium · numerical
A 10 kg box is placed on a surface. The coefficient of friction between the surface and the box is μ = 0.5. If a horizontal force of 100 N is applied, the acceleration of the box will be (g = 10 m/s²):
A. 7.5 m/s²
B. 5 m/s² ✓ Correct
C. none
D. 2.5 m/s²
Solution: Kinetic friction = μmg = 0.5 × 10 × 10 = 50 N; net force = 100 − 50 = 50 N; a = 50/10 = 5 m/s².
Q10 — Kinetic Friction · medium · numerical
A block B is pushed momentarily along a horizontal surface with an initial velocity v. If μ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time:
A. μg/v
B. v/(μg) ✓ Correct
C. v/g
D. g/v
Solution: Retardation a = μg, so the time to stop is t = v/a = v/(μg).
Q11 — Kinetic Friction · medium · numerical
A marble block of mass 2 kg lying on ice, when given a velocity of 6 m/s, is stopped by friction in 10 s. Then the coefficient of friction is:
A. 0.03
B. 0.01
C. 0.06 ✓ Correct
D. 0.02
Solution: Retardation a = v/t = 6/10 = 0.6 m/s²; μ = a/g = 0.6/10 = 0.06.
Q12 — Kinetic Friction · medium · numerical
Consider a car moving on a straight road with a speed of 100 m/s. The distance at which the car can be stopped is (μk = 0.5, g = 10 m/s²):
A. 100 m
B. 400 m
C. 1000 m ✓ Correct
D. 800 m
Solution: s = v²/(2μg) = 100²/(2 × 0.5 × 10) = 1000 m.
Q13 — Kinetic Friction · medium · numerical
Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:
A. 0.33
B. 0.25
C. 0.80
D. 0.75 ✓ Correct
Solution: For the same distance, doubling the time means a₂ = a₁/4. With a₁ = g sinθ and a₂ = g(sinθ − μcosθ) at 45°: 1 − μ = 1/4, so μ = 0.75.
Q14 — Kinetic Friction · medium · numerical
A wooden block of mass m resting on a rough horizontal table (coefficient of friction = μ) is pulled by a force F applied at an angle θ above the horizontal. The acceleration of the block moving horizontally is:
A. (F sinθ)/m
B. none
C. (F/m)(cosθ + μsinθ) − μg ✓ Correct
D. (F cosθ)/m
Solution: Normal reaction N = mg − F sinθ; net horizontal force = F cosθ − μN; hence a = (F/m)(cosθ + μsinθ) − μg.
Q15 — Kinetic Friction · medium · numerical
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40 N is applied at an angle of 30° above the horizontal, the acceleration of the block will be (g = 10 m/s²):
A. 10.0 m/s²
B. 5.73 m/s² ✓ Correct
C. 3.17 m/s²
D. 8.0 m/s²
Solution: N = Mg − F sin30° = 50 − 20 = 30 N; a = (F cos30° − μN)/M = (40 × √3/2 − 0.2 × 30)/5 ≈ 5.73 m/s².
Q16 — Kinetic Friction · medium · numerical
A body is projected up a rough inclined plane from the bottom with some velocity. It travels up the incline and then returns back. If the time of ascent is ta and the time of descent is td, then:
A. ta > td
B. ta = td
C. ta < td ✓ Correct
D. data insufficient
Solution: The retardation during ascent, g(sinθ + μcosθ), is greater than the acceleration during descent, g(sinθ − μcosθ), so the return journey is slower and ta < td.