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Subjects · Leaving Cert Applied Maths

Leaving Cert Applied Maths: Friction & inclined planes

How often Friction & inclined planes comes up on the Applied Maths papers, every year it was asked, and questions to try.

HL Asked on 3 of the last 3 Higher Level papers, most recently in 2025. banker

OL Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

Friction & inclined planes, Higher Level(7 marks)

Quick ones on Friction & inclined planes.

(a)Limiting friction between two surfaces is given by…?
  1. F = R/μ
  2. F = μR
  3. F = μ/R
(b)A block rests on a slope at angle θ, acted on only by weight, reaction and friction. Normal reaction?
  1. mg
  2. mg sin θ
  3. mg cos θ
(c)A 5 kg block is on rough horizontal ground, μ = 0.4 (g = 9.8). Maximum friction force?
  1. 19.6 N
  2. 49 N
  3. 2 N
Show the answers

(a) F = μR

(b) mg cos θ

(c) 19.6 N

Your turn: Higher Level questions on Friction & inclined planes.

Higher Level

Asked on 3 of the last 3 Higher Level papers, most recently in 2025. banker

Every paper, year by year

YearWhere it came up
2025Q3, Q5
2024Q2, Q7
2023Q5

Links open the State Examinations Commission’s paper for that year.

More Friction & inclined planes questions

Friction & inclined planes, 3 marks

A block moves up a rough slope with tan α = 3/4, μ = 0.5 (g = 9.8), with no driving force. Deceleration?

  1. 1.96 m/s²
  2. 5.88 m/s²
  3. 9.8 m/s²
Show the answer

9.8 m/s²

Moving up, friction acts down the slope along with the weight component: a = g(0.6 + 0.5 × 0.8) = 9.8 × 1 = 9.8 m/s². Friction always opposes the motion.

Friction & inclined planes, 3 marks

A horizontal force P pulls a 4 kg block on rough ground (μ = 0.25, g = 9.8) with acceleration 1.5 m/s². P?

  1. 9.8 N
  2. 15.8 N
  3. 6 N
Show the answer

15.8 N

Friction = 0.25 × 4 × 9.8 = 9.8 N. P − 9.8 = 4 × 1.5 = 6, so P = 15.8 N. 6 N is the resultant; 9.8 N only overcomes friction.

Friction & inclined planes, 2 marks

Component of a body's weight along a slope inclined at θ to the horizontal?

  1. mg sin θ
  2. mg cos θ
  3. mg tan θ
Show the answer

mg sin θ

Resolve W = mg along and perpendicular to the plane: mg sin θ down the slope, mg cos θ into it. Check with θ = 0: a flat surface has no pull along it, and sin 0 = 0.

Ordinary Level

Asked on 3 of the last 3 Ordinary Level papers, most recently in 2025. banker

Every paper, year by year

YearWhere it came up
2025Q1, Q2, Q5
2024Q8
2023Q9

Links open the State Examinations Commission’s paper for that year.

More Friction & inclined planes questions

Friction & inclined planes, 3 marks

A 4 kg block rests on a rough horizontal floor, μ = 0.5. Limiting friction? (g = 9.8 m s⁻²)

  1. 39.2 N
  2. 2 N
  3. 19.6 N
Show the answer

19.6 N

On level ground R = mg = 4 × 9.8 = 39.2 N. F = μR = 0.5 × 39.2 = 19.6 N. 39.2 N is the normal reaction.

Friction & inclined planes, 3 marks

A box is pulled by 20 N at constant speed across a rough floor. The friction force is…

  1. 40 N
  2. 20 N
  3. 0 N
Show the answer

20 N

Constant speed means zero acceleration, so the forces balance: friction equals the 20 N pull, acting the opposite way.

Friction & inclined planes, 3 marks

A 3 kg block on a level floor just starts to move under a 14.7 N pull. Find μ. (g = 9.8 m s⁻²)

  1. 0.5
  2. 2
  3. 0.2
Show the answer

0.5

R = 3 × 9.8 = 29.4 N. At the point of moving, 14.7 = μ × 29.4, so μ = 0.5. μ = 2 divides the wrong way round.

Other Applied Maths topics

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