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Static, kinetic and rolling friction: NEET MCQs with answers

58 questions on this topic of Laws of Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A block of mass 6 kg rests on a rough horizontal surface having coefficient of static friction μs = 0.5 and kinetic friction μk = 0.4. If a horizontal force of 20 N is applied on the block, what is the magnitude of the frictional force acting on the block? (Take g = 10 m/s²)

  1. 20 N
  2. 24 N
  3. 30 N
  4. 0 N
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Answer: (A) 20 N

The maximum possible static friction is fs(max) = μs N = 0.5 × 6 × 10 = 30 N. Since the applied horizontal force of 20 N is less than fs(max), the block does not move. Therefore, the static friction self-adjusts to match the applied force, so f = 20 N.

medium

2. A block of mass 2 kg is pulled across a rough horizontal surface by a force of 20 N acting at an angle of 30° above the horizontal. If the coefficient of kinetic friction is 0.25, the acceleration of the block is (take g = 10 m/s², √3 = 1.73):

  1. 6.32 m/s²
  2. 4.56 m/s²
  3. 7.41 m/s²
  4. 3.66 m/s²
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Answer: (C) 7.41 m/s²

Vertical equilibrium: N + F sin 30° = mg ⇒ N = 20 - 20(0.5) = 10 N. Kinetic friction fk = μk N = 0.25 × 10 = 2.5 N. Net horizontal force: F cos 30° - fk = 20 × (√3/2) - 2.5 = 17.32 - 2.5 = 14.82 N. Acceleration a = Fnet / m = 14.82 / 2 = 7.41 m/s².

hard

3. A block of mass m is at rest on a rough horizontal surface with angle of friction λ. What is the minimum magnitude of force required to move the block along the horizontal surface?

  1. mg tan λ
  2. mg sin λ
  3. mg cos λ
  4. mg / cos λ
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Answer: (B) mg sin λ

To move the block with the minimum force, the force F is applied at an angle θ to the horizontal. F = μ mg / (cos θ + μ sin θ). The denominator is maximized when tan θ = μ. Since μ = tan λ, the maximum value of (cos θ + μ sin θ) is √(1 + μ²) = 1 / cos λ. Thus, F_min = (μ mg) / √(1 + μ²) = mg sin λ.

medium

4. An object takes twice as much time to slide down a 45° rough inclined plane as it takes to slide down a frictionless inclined plane of the same length and inclination. The coefficient of kinetic friction μk between the object and the rough plane is:

  1. 0.75
  2. 0.50
  3. 0.25
  4. 0.80
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Answer: (A) 0.75

For a smooth incline, a1 = g sin θ. For a rough incline, a2 = g(sin θ - μk cos θ). Since s = (1/2) a t², t ∝ 1/√a. Given t2 = 2 t1, we have a1 / a2 = (t2 / t1)² = 4. With θ = 45°, sin 45° = cos 45°, so g sin 45° / [g sin 45°(1 - μk)] = 4. Thus 1 / (1 - μk) = 4 ⇒ 1 - μk = 0.25 ⇒ μk = 0.75.

easy

5. A block of mass 5 kg rests on a rough horizontal plane. The coefficient of static friction is μs = 0.4 and coefficient of kinetic friction is μk = 0.3. A horizontal force of 15 N is applied to the block. The frictional force acting on the block is (take g = 10 m/s²):

  1. 20 N
  2. 0 N
  3. 12 N
  4. 15 N
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Answer: (D) 15 N

Limiting static friction is fs(max) = μs·N = 0.4 × (5 × 10) = 20 N. Since the applied horizontal force (15 N) is less than the limiting friction (20 N), the block does not move. Static friction adjusts itself to equal the applied force, so f = 15 N.

hard

6. A body of mass 5 kg is sliding down a rough inclined plane of inclination 30° at a constant velocity. What extra force directed parallel to the incline must be applied to make it move up the plane with an acceleration of 2 m/s²? (Take g = 10 m/s²)

  1. 25 N
  2. 50 N
  3. 60 N
  4. 35 N
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Answer: (C) 60 N

When sliding down with constant velocity, mg sin 30° = fk, so fk = 5 × 10 × 0.5 = 25 N. When moving up, both the component of gravity down the plane (25 N) and kinetic friction opposing motion down the plane (25 N) act downwards. To accelerate up at a = 2 m/s², F - mg sin 30° - fk = m a. Thus, F - 25 - 25 = 5 × 2 = 10 N, giving F = 60 N.

medium

7. A 4 kg block on a horizontal floor (coefficient of kinetic friction 0.3) is pulled horizontally with a force of 47 N. Taking g = 10 m/s², its acceleration is:

  1. 8.75 m/s²
  2. 11.8 m/s²
  3. 3 m/s²
  4. 14.8 m/s²
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Answer: (A) 8.75 m/s²

Kinetic friction = μmg = 0.3 × 4 × 10 = 12 N. a = (F − f)/m = (47 − 12)/4 = 8.75 m/s².

medium

8. A 3 kg block on a horizontal floor (coefficient of kinetic friction 0.3) is pulled horizontally with a force of 14 N. Taking g = 10 m/s², its acceleration is:

  1. 7.67 m/s²
  2. 1.67 m/s²
  3. 3 m/s²
  4. 4.67 m/s²
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Answer: (B) 1.67 m/s²

Kinetic friction = μmg = 0.3 × 3 × 10 = 9 N. a = (F − f)/m = (14 − 9)/3 = 1.67 m/s².

medium

9. A 4 kg block on a horizontal floor (coefficient of kinetic friction 0.3) is pulled horizontally with a force of 42 N. Taking g = 10 m/s², its acceleration is:

  1. 7.5 m/s²
  2. 10.5 m/s²
  3. 3 m/s²
  4. 13.5 m/s²
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Answer: (A) 7.5 m/s²

Kinetic friction = μmg = 0.3 × 4 × 10 = 12 N. a = (F − f)/m = (42 − 12)/4 = 7.5 m/s².

medium

10. A 5 kg block on a horizontal floor (coefficient of kinetic friction 0.2) is pulled horizontally with a force of 20 N. Taking g = 10 m/s², its acceleration is:

  1. 1 m/s²
  2. 4 m/s²
  3. 2 m/s²
  4. 6 m/s²
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Answer: (C) 2 m/s²

Kinetic friction = μmg = 0.2 × 5 × 10 = 10 N. a = (F − f)/m = (20 − 10)/5 = 2 m/s².

hard

11. A 3 kg block is placed on top of a 7 kg slab which rests on a frictionless horizontal floor. The coefficient of static friction between the block and the slab is 0.4. What is the maximum horizontal force that can be applied to the slab so that both the block and the slab move together without slipping? (take g = 10 m/s²)

  1. 28 N
  2. 12 N
  3. 40 N
  4. 50 N
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Answer: (C) 40 N

The only horizontal force accelerating the 3 kg block is the static friction from the slab. The maximum acceleration of the 3 kg block without slipping is a_max = (μs m1 g) / m1 = μs g = 0.4 × 10 = 4 m/s². To accelerate the combined system (m1 + m2 = 3 + 7 = 10 kg) at this rate, the maximum force applied to the slab is F_max = (m1 + m2) a_max = 10 × 4 = 40 N.

hard

12. A block of mass 5 kg is placed against a vertical wall. A force F inclined at an angle of 45° to the horizontal pushes the block upwards against the wall. The coefficient of static friction between the block and the wall is μs = 0.25. What is the minimum magnitude of F required to prevent the block from sliding down? (Take g = 10 m/s²)

  1. 30√2 N
  2. 50√2 N
  3. 25√2 N
  4. 40√2 N
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Answer: (D) 40√2 N

Resolving F into components: normal reaction N = F cos 45° and vertical upward force F_up = F sin 45°. Since the block tends to slip downwards, the static friction acts upwards: fs = μs N = 0.25 F cos 45°. For equilibrium, F sin 45° + μs F cos 45° = mg. Since sin 45° = cos 45° = 1/√2, F(1 + 0.25)/√2 = 5 × 10 = 50. Thus, 1.25 F = 50√2, which yields F = 40√2 N.

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