Friction Solved Examples (Class 8 Science)
Friction numericals develop understanding of friction force, limiting friction, and coefficient of friction. These calculations are essential for solving r
TL;DR: Friction numericals develop understanding of friction force, limiting friction, and coefficient of friction. These calculations are essential for solv…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Friction numericals develop understanding of friction force, limiting friction, and coefficient of friction. These calculations are essential for solving r
Friction — Solved Numerical Examples (Step by Step)
Example 1: A wooden block of mass 5 kg is placed on a horizontal surface. The coefficient of static friction is 0.4. Calculate the maximum static friction. (g = 10 m/s²)
Solution: Mass = 5 kg, g = 10 m/s², so Normal force N = 5 × 10 = 50 N. Coefficient of static friction (μs) = 0.4. Maximum static friction = μs × N = 0.4 × 50 = 20 N.
Example 2: A 10 kg box is dragged along a floor with coefficient of kinetic friction 0.2. Find the friction force. (g = 10 m/s²)
Solution: Mass = 10 kg, g = 10 m/s², Normal force = 10 × 10 = 100 N. Coefficient of kinetic friction (μk) = 0.2. Friction force = μk × N = 0.2 × 100 = 20 N.
Example 3: A book weighing 2 kg rests on an inclined plane at 30 degrees. The coefficient of static friction is 0.5. Will the book slide? (g = 10 m/s²)
Solution: Component along plane = mg sin(30°) = 2 × 10 × 0.5 = 10 N. Normal force = mg cos(30°) = 2 × 10 × 0.866 = 17.32 N. Maximum static friction = 0.5 × 17.32 = 8.66 N. Since 10 N > 8.66 N, the book will slide.
Example 4: A car of mass 1000 kg brakes on a road with coefficient of kinetic friction 0.8. What is the maximum braking force? (g = 10 m/s²)
Solution: Mass = 1000 kg, g = 10 m/s², Normal force = 1000 × 10 = 10,000 N. Coefficient of kinetic friction = 0.8. Maximum braking force = 0.8 × 10,000 = 8000 N.
Example 5: A 3 kg block is pulled on a surface with coefficient of kinetic friction 0.25 by a horizontal force of 10 N. Calculate the net force and acceleration. (g = 10 m/s²)
Solution: Normal force = 3 × 10 = 30 N. Friction force = 0.25 × 30 = 7.5 N. Applied force = 10 N. Net force = 10 - 7.5 = 2.5 N. Acceleration = Net force / mass = 2.5 / 3 = 0.833 m/s².
Tips
- Static friction can vary from 0 to μs × N, while kinetic friction is constant at μk × N.
- Kinetic friction coefficient is always less than static friction coefficient.
- Normal force is perpendicular to the surface; on horizontal surfaces, N = mg.
Frequently Asked Questions
Why is it easier to keep a box moving than to start moving it?
Because static friction (which opposes starting motion) is greater than kinetic friction (which opposes ongoing motion). Once moving, less force is needed to maintain motion.
Does friction depend on surface area?
No. Friction depends only on the normal force and coefficient of friction, not on the contact area between surfaces.
More Science Solved Examples
- Light - Previous Year Questions
- Electricity - Previous Year Questions
- Life Processes - Previous Year Questions
- Chemical Reactions and Equations - Previous Year Questions
- Acids Bases and Salts - Previous Year Questions
- Matter in Our Surroundings - Previous Year Questions
🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →