Friction problems: Class 11 practice
Kinetic friction f = μN opposes motion. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Newton's second law
- Trigonometric ratios
The idea behind friction problems
Kinetic friction f = μN opposes motion. On a horizontal surface N = mg; on an incline N = mg cos θ. A block slides down a rough incline with a = g(sin θ − μ cos θ) when tan θ > μ.
Step-by-step method
- Draw a free-body diagram.
- Find N.
- Friction = μN; net force = applied − friction.
- a = net force ÷ m.
Worked example 1 (Basic)
A 18 kg block is pulled with 84 N on a floor with μ = 0.3. Find its acceleration (g = 10).
- f = μmg = 54 N
- a = (84 − 54)/18 = 1.67 m/s²
- Answer: 1.67 m/s²
Worked example 2 (Standard)
A block slides down a 30° incline with μ = 0.25. Find its acceleration (g = 10).
- a = g(sin θ − μ cos θ)
- = 10(0.5 − 0.25 × 0.866) = 2.83 m/s²
- Answer: 2.83 m/s²
Worked example 3 (Challenge)
Find the angle of repose for a surface with μ = 0.4.
- tan θ = μ
- θ = 21.8°
- Answer: 21.8°
Common mistakes and misconceptions
- Using N = mg on an incline
- Letting friction exceed the applied force when at rest
Practice questions
- (Basic) A 17 kg block is pulled with 81 N on a floor with μ = 0.25. Find its acceleration (g = 10).
- (Basic) A 5 kg block is pulled with 32 N on a floor with μ = 0.1. Find its acceleration (g = 10).
- (Standard) A block slides down a 53° incline with μ = 0.2. Find its acceleration (g = 10).
- (Standard) A block slides down a 60° incline with μ = 0.1. Find its acceleration (g = 10).
- (Challenge) Find the angle of repose for a surface with μ = 0.2.
- (Challenge) Find the angle of repose for a surface with μ = 0.3.
Answer key with full working
- Q1: f = μmg = 42.5 N → a = (81 − 42.5)/17 = 2.26 m/s² → Answer: 2.26 m/s²
- Q2: f = μmg = 5 N → a = (32 − 5)/5 = 5.4 m/s² → Answer: 5.4 m/s²
- Q3: a = g(sin θ − μ cos θ) → = 10(0.799 − 0.2 × 0.602) = 6.78 m/s² → Answer: 6.78 m/s²
- Q4: a = g(sin θ − μ cos θ) → = 10(0.866 − 0.1 × 0.5) = 8.16 m/s² → Answer: 8.16 m/s²
- Q5: tan θ = μ → θ = 11.31° → Answer: 11.31°
- Q6: tan θ = μ → θ = 16.7° → Answer: 16.7°
Syllabus note
This skill is listed under "Laws of Motion" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is friction problems in?
"Laws of Motion", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.