OCR A Level Mathematics A: Friction and rough surfaces (H240) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover friction, references 3.03r to 3.03v of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026). Question 2 uses Stage 1 content (3.03r). The rest is Stage 2 content, which is A Level only. In every question take g = 9.8 m s⁻². Learn the methods first from the study guide or the revision notes. See also the course hub and the checklist.

Questions

1. The normal contact force on a block resting on a rough surface is 40 N. The coefficient of friction is 0.35. Find the greatest possible frictional force. [2]

2. A particle of mass 1.5 kg moves on a horizontal surface, with i and j perpendicular unit vectors in the surface. The applied force is (10i + 9j) N. The friction force has magnitude 5 N and acts in the direction of −(4i + 3j). Find the acceleration as a vector, and its magnitude. [3]

3. The contact force between a crate and rough ground has a normal component of 68 N and a frictional component of 21 N.

(a) Find the magnitude of the contact force. [2] (b) The crate is on the point of slipping. Find the coefficient of friction. [2]

4. A sledge of mass 18 kg is on rough horizontal snow with coefficient of friction 0.2. A rope pulls it with tension 35 N at 35° above the horizontal. Show that the sledge does not move, and state the frictional force. [5]

5. A block of mass 9 kg is on rough horizontal ground with coefficient of friction 0.4. A force P acts on it at 20° below the horizontal, pushing it. The block is on the point of sliding.

(a) Find P. [5] (b) Find the normal contact force. [1]

6. A block of mass 5 kg is on a rough plane inclined at 28° to the horizontal. The coefficient of friction is 0.6.

(a) Show that the block stays at rest, and find the friction force. [4] (b) The plane is slowly tilted. Find the greatest inclination at which the block stays at rest. [2]

7. A block of mass 8 kg is on a rough plane inclined at 25° to the horizontal. The coefficient of friction is 0.3. A force P acts up a line of greatest slope.

(a) Show that the block would slide down if P were zero. [2] (b) Find the least P for which the block stays at rest. [3] (c) Find the greatest P for which the block stays at rest. [3]

8. A body of mass 12 kg is at rest on rough horizontal ground, with coefficient of friction 0.3. A horizontal force of 50 N acts for 4 s and is then removed. Find the total distance moved until the body stops. [7]

9. A body of mass 2 kg is sent down the line of greatest slope on a rough plane at 15° to the horizontal, starting at 5 m s⁻¹. The coefficient of friction is 0.35.

(a) Show that the body decelerates, and find the deceleration. [3] (b) Find the distance travelled before it stops. [2] (c) Explain whether it then remains at rest. [2]

10. A block of mass 6 kg is on a rough plane inclined at 20° to the horizontal, with coefficient of friction 0.2. A force of 45 N acts up a line of greatest slope, and the block starts from rest.

(a) Find the acceleration. [4] (b) After 2 s the force is removed. Find the further distance the block moves up the plane. [3] (c) Show that the block then slides back down. [2]

11. A block of mass 7 kg is pulled across rough horizontal ground by a force of 40 N at 30° above the horizontal. The block accelerates at 1.1 m s⁻².

(a) Find the coefficient of friction. [4] (b) Find the greatest force at this angle for which the block stays in contact with the ground. [2]

Answers

1. Greatest friction = μR = 0.35 × 40 [1] = 14 N [1] [2] Examiner insight: write μR explicitly; a bare 14 gives nothing to credit if it is wrong.

2. Friction = −(4i + 3j) N, since |4i + 3j| = 5. Resultant = (10 − 4)i + (9 − 3)j = (6i + 6j) N [1]. a = (6i + 6j) ÷ 1.5 = (4i + 4j) m s⁻² [1]. |a| = 4√2 = 5.66 m s⁻² [1] [3] Examiner insight: turning “magnitude 5 N in a given direction” into a vector by dividing by the length of the direction vector is the step to show clearly.

3. (a) S = √(68² + 21²) [1] = √5065 = 71.2 N [1] (b) Limiting: F = μR, so μ = 21 ÷ 68 [1] = 0.309 [1] [4] Examiner insight: the contact force is the resultant of two perpendicular components, so adding 68 and 21 loses the method.

4. Vertical: R = 18 × 9.8 − 35 sin 35° = 176.4 − 20.08 = 156.3 N [1] [1]. Maximum friction = 0.2 × 156.3 = 31.3 N [1]. Horizontal pull = 35 cos 35° = 28.7 N, which is less than 31.3 N, so no motion [1]. Friction balances the pull: F = 28.7 N [1] [5] Examiner insight: in a “show that” you must compare the two forces explicitly and conclude. Quoting 31.3 N as the friction is a common error.

5. (a) R = 88.2 + P sin 20° [1]. Limiting: P cos 20° = 0.4R [1]. P cos 20° = 35.28 + 0.4P sin 20° [1]. P(0.9397 − 0.1368) = 35.28 [1]. P = 43.9 N [1] (b) R = 88.2 + 43.94 sin 20° = 103 N [1] [6] Examiner insight: the push adds to R. If your R is below the weight, the sign of the P sin 20° term is wrong.

6. (a) Along the plane the friction needed is 5 × 9.8 × sin 28° = 23.0 N [1]. R = 5 × 9.8 × cos 28° = 43.3 N [1]. Maximum friction = 0.6 × 43.3 = 26.0 N [1]. Since 23.0 < 26.0, the block stays at rest, with F = 23.0 N [1] (b) Limiting: mg sin θ = μ mg cos θ [1], so tan θ = 0.6 and θ = 31.0° [1] [6] Examiner insight: the friction is 23.0 N, not 26.0 N. State which number is the actual friction.

7. (a) R = 8 × 9.8 × cos 25° = 71.05 N, so μR = 21.3 N [1]. Weight component down the slope = 8 × 9.8 × sin 25° = 33.1 N, which exceeds 21.3 N, so it would slide [1] (b) About to slide down, friction up the slope: P + μR = mg sin 25° [1]. P = 33.13 − 21.32 [1] = 11.8 N [1] (c) About to slide up, friction down the slope: P = mg sin 25° + μR [1] = 33.13 + 21.32 [1] = 54.4 N [1] [8] Examiner insight: draw a separate diagram for each limiting case. The friction arrow reverses between (b) and (c).

8. R = 117.6 N, so F = 0.3 × 117.6 = 35.28 N [1]. 12a = 50 − 35.28 [1], a = 1.227 m s⁻². After 4 s, v = 4.907 m s⁻¹ [1]. s₁ = ½ × 1.227 × 16 = 9.81 m [1]. After removal, deceleration = 0.3 × 9.8 = 2.94 m s⁻² [1]. s₂ = 4.907² ÷ (2 × 2.94) = 4.09 m [1]. Total = 13.9 m [1] [7] Examiner insight: carry unrounded values from stage 1 into stage 2. Early rounding of v can shift the third significant figure.

9. (a) R = 2 × 9.8 × cos 15° = 18.93 N, F = 0.35R = 6.63 N [1]. Weight component down the slope = 2 × 9.8 × sin 15° = 5.07 N, so 2a = 5.07 − 6.63 [1]. a = −0.777, so it decelerates at 0.777 m s⁻² [1] (b) 0 = 5² − 2(0.777)s [1], s = 16.1 m [1] (c) tan 15° = 0.268 < 0.35 (or 5.07 N < 6.63 N) [1], so friction can balance the weight component and it stays at rest [1] [7] Examiner insight: “explain” needs the comparison and a conclusion. Saying “friction is big enough” without figures is weaker.

10. (a) R = 6 × 9.8 × cos 20° = 55.25 N [1]. F = 0.2R = 11.05 N [1]. Weight component = 6 × 9.8 × sin 20° = 20.11 N. 6a = 45 − 20.11 − 11.05 [1]. a = 2.31 m s⁻² [1] (b) v = 2 × 2.306 = 4.61 m s⁻¹ [1]. Deceleration = 9.8(sin 20° + 0.2 cos 20°) = 5.19 m s⁻² [1]. s = 4.613² ÷ (2 × 5.194) = 2.05 m [1] (c) tan 20° = 0.364 > 0.2 [1], so the weight component exceeds the maximum friction and it slides back [1] [9] Examiner insight: moving up the slope, friction and the weight component both act down it. Using g sin 20° − 0.2g cos 20° in (b) is the classic slip.

11. (a) R = 7 × 9.8 − 40 sin 30° = 68.6 − 20 = 48.6 N [1]. Horizontal: 40 cos 30° − F = 7 × 1.1 [1]. F = 34.64 − 7.7 = 26.94 N [1]. μ = 26.94 ÷ 48.6 = 0.554 [1] (b) Contact is lost when R = 0: P sin 30° = 68.6 [1], P = 137 N [1] [6] Examiner insight: find R first, from the vertical forces including the rope’s vertical component, before using F = μR.

Where marks are usually lost

• Using R = mg when a rope or push has a vertical component, or on a slope.
• Taking F = μR for a stationary body that is not close to slipping.
• Reversing the friction direction wrongly between the least and greatest force cases.
• Adding the normal and frictional components of the contact force.
• Not concluding in words after comparing forces in a “show that” or “explain”.
• Rounding intermediate values and losing accuracy at the end.
• Using the up-slope deceleration formula for motion down the slope, or the reverse.
• Treating the stopped body as sliding back without comparing tan α with μ.

Next steps

• Revisit the revision notes and the study guide.
• Go back to the course hub and tick off the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. These questions follow references 3.03r to 3.03v.