Laws of Motion — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Laws of Motion" — 7 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Laws of Motion" — 7 important questions with detailed answers for CBSE board exam prep…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

State Newton's three laws of motion and explain the second law with a mathema…
A block of mass 5 kg is placed on a horizontal surface. If a horizontal force…
Two blocks of masses 4 kg and 6 kg are connected by a light string and are pu…
A lift with a person inside accelerates upward at 2 m/s². If the person has a…
Define static friction and kinetic friction. A block of mass 10 kg is on a ho…
A car of mass 1000 kg is moving on a horizontal road with coefficient of fric…
+ 1 more questions in the full chapter

Solutions Summary:

Question	Status
State Newton's three laws of motion and explain the secon…	✓ Solved
A block of mass 5 kg is placed on a horizontal surface. I…	✓ Solved
Two blocks of masses 4 kg and 6 kg are connected by a lig…	✓ Solved
A lift with a person inside accelerates upward at 2 m/s².…	✓ Solved
Define static friction and kinetic friction. A block of m…	✓ Solved
A car of mass 1000 kg is moving on a horizontal road with…	✓ Solved

Showing 6 of 7 questions

Q1: State Newton's three laws of motion and explain the second law with a mathematical equation.

Newton's Three Laws of Motion: First Law (Law of Inertia): Every object continues to be in a state of rest or uniform motion in a straight line unless compelled by an external force to change that state. Key points: - An object at rest tends to remain at rest - An object in motion tends to remain in motion - This tendency to resist change in motion is called inertia - Inertia is directly proportional to mass Second Law (Law of Acceleration): The rate of change of momentum of a body is directl...

Q2: A block of mass 5 kg is placed on a horizontal surface. If a horizontal force of 20 N is applied and the block accelerates at 3 m/s², find the friction force acting on the block.

Given: Mass m = 5 kg Applied force F = 20 N (horizontal) Acceleration a = 3 m/s² Find: Friction force f Step 1: Apply Newton's second law The net force on the block is: F_net = ma = 5 × 3 = 15 N Step 2: Identify forces in horizontal direction In the horizontal direction: - Applied force F = 20 N (forward) - Friction force f (backward, opposing motion) Step 3: Apply Newton's second law in horizontal direction F_net = F - f 15 = 20 - f f = 20 - 15 = 5 N Step 4: Verify using free body diagr...

Q3: Two blocks of masses 4 kg and 6 kg are connected by a light string and are pulled horizontally by a force of 50 N. Find (a) acceleration of the system (b) tension in the string connecting the two blocks (take g = 10 m/s²).

Given: Mass m₁ = 4 kg (leading block) Mass m₂ = 6 kg (trailing block) Applied force F = 50 N Assuming frictionless surface and g = 10 m/s² Part (a): Acceleration of the system Step 1: Consider the entire system as one body Total mass M = m₁ + m₂ = 4 + 6 = 10 kg Step 2: Apply Newton's second law to the system F_net = Ma 50 = 10 × a a = 50/10 = 5 m/s² Part (b): Tension in the string Step 1: Consider only the second block (6 kg) Forces on the 6 kg block: - Tension T (pulling forward) - No o...

Q4: A lift with a person inside accelerates upward at 2 m/s². If the person has a mass of 60 kg, calculate the normal force exerted by the floor of the lift on the person. (Take g = 10 m/s²)

Given: Mass of person m = 60 kg Acceleration of lift a = 2 m/s² (upward) g = 10 m/s² Find: Normal force N exerted by floor on person Step 1: Identify forces acting on the person - Weight W = mg (downward) - Normal force N (upward, from floor) Step 2: Set up coordinate system Let upward direction be positive (+) Downward direction be negative (-) Step 3: Calculate weight W = mg = 60 × 10 = 600 N Step 4: Apply Newton's second law in vertical direction F_net = ma (upward) N - W = ma N - 60...

Q5: Define static friction and kinetic friction. A block of mass 10 kg is on a horizontal surface with coefficient of static friction μ_s = 0.5 and coefficient of kinetic friction μ_k = 0.3. (g = 10 m/s²). What is the minimum force needed to start moving the block, and what force is needed to keep it moving at constant velocity?

Friction Definitions: Static Friction (f_s): Static friction is the friction force that acts between two surfaces when there is no relative motion between them. It prevents an object from moving when an external force is applied. Properties: - Acts when there is a tendency to move but no actual motion - Can have any value from 0 to μ_s × N - Maximum static friction f_s(max) = μ_s × N - Direction: opposite to the applied force - μ_s is usually greater than μ_k Kinetic Friction (f_k): Kinetic f...

Q6: A car of mass 1000 kg is moving on a horizontal road with coefficient of friction 0.8. The car suddenly applies brakes. Calculate (a) maximum deceleration possible (b) minimum distance to stop from 20 m/s (c) time taken to stop.

Given: Mass m = 1000 kg Coefficient of friction μ = 0.8 (kinetic, during braking) Initial velocity u = 20 m/s Final velocity v = 0 m/s (car stops) g = 10 m/s² Part (a): Maximum deceleration possible Step 1: Calculate normal force N = mg = 1000 × 10 = 10,000 N Step 2: Calculate maximum friction force When braking, kinetic friction acts: f_max = μ × N = 0.8 × 10,000 = 8000 N Step 3: Apply Newton's second law F = ma 8000 = 1000 × a_max a_max = 8000/1000 = 8 m/s² Maximum deceleration = 8 m/s² (...

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