Showing posts with label Mechanics. Show all posts
Showing posts with label Mechanics. Show all posts

Saturday, July 18, 2026

NEET Physics: Motion in One Dimension – Mechanics, Kinematics & Dynamics Notes

 

NEET Physics – Motion in One Dimension

Infographic explaining Motion in One Dimension, Mechanics, Kinematics, Dynamics, Frame of Reference, and Point Object concepts for NEET Physics students.
Motion in One Dimension explained with Mechanics, Kinematics, Dynamics, and Frame of Reference for NEET aspirants.


Chapter 1: Basic Concepts of Mechanics (Complete Notes)

-Dr.Sanjaykumar Pawar 

1. What is Physics?

Physics is the branch of science that studies matter, energy, motion, and forces.

Example:

  • Why does a ball fall?

  • Why does a bike move?

  • Why do planets revolve around the Sun?

All these are studied in Physics.


2. What is Mechanics?

Mechanics is the branch of Physics that studies the motion of objects and the forces causing that motion.

Simple Definition

Mechanics = Study of motion + causes of motion.


Mechanics is divided into two parts

              Mechanics
             /          \
      Kinematics      Dynamics

3. Kinematics

Definition

Kinematics is the branch of mechanics that studies motion without considering the force causing it.

It tells us

  • Where the object is

  • How fast it moves

  • How far it moves

  • How long it takes

But NOT why it moves.

Example

A car moves with speed 40 km/h.

Kinematics studies

✔ Speed

✔ Distance

✔ Time

✔ Displacement

It does NOT study engine force.


4. Dynamics

Definition

Dynamics studies motion along with the forces responsible for the motion.

It answers

Why does the object move?

Example

A football moves because a player kicks it.

Dynamics studies the force of the kick.


Difference Between Kinematics and Dynamics

KinematicsDynamics
Studies motion onlyStudies motion + force
Force is ignoredForce is considered
Easier chapterUses Newton's Laws
Example: SpeedExample: Force

5. Frame of Reference

This is one of the most important concepts in NEET.

Definition

A Frame of Reference is a coordinate system attached to an observer from which measurements of position and time are made.

Simply,

It is the point from which an observer watches motion.


It contains

(1) Observer

Who is watching?

Example

You

Driver

Passenger

Teacher


(2) Space Coordinate

Where is the object?

Usually represented by

x-axis

y-axis

z-axis


(3) Time Coordinate

When is the object observed?

Measured using a clock.


Example

A train is moving.

Passenger says

"I am at rest."

A person standing on the platform says

"The train is moving."

Both are correct because their frames of reference are different.


Important Points

Motion depends on

✔ Observer

✔ Position

✔ Time


There is NO Absolute Motion

Motion is always relative.

There is nothing called

❌ Absolute Motion

❌ Absolute Rest

Everything depends upon the observer.


Example

You are sitting inside a moving bus.

With respect to

Seat → You are at rest.

Road → You are moving.

Both answers are correct.


6. Rest and Motion

Rest

An object is at rest if its position does not change with respect to the observer.

Example

Book lying on a table.


Motion

An object is in motion if its position changes with time with respect to an observer.

Example

Running car.

Flying bird.

Moving train.


Remember

Motion requires

✔ Observer

✔ Position

✔ Time


7. Point Object (Particle)

Very important for NEET.

Definition

A point object is an object whose size is very small compared to the distance travelled, so its dimensions can be ignored.


Example

Bike length = 2 m

Distance travelled = 500 km

Compared to 500 km,

2 m is negligible.

So bike is treated as a point object.


Another Example

Earth revolves around the Sun.

Earth diameter = 12,742 km

Distance from Sun = 150 million km

Compared to this huge distance,

Earth behaves like a point object.


When can an object be treated as a Point Object?

Ignore size when

Distance travelled ≫ Size of object


Example

ObjectDistancePoint Object?
Car travelling 200 kmYes
Cricket ball moving 40 mYes
Pencil moving 2 cmNo

8. Position

Position tells where an object is located with respect to a chosen origin.

Example

If a boy stands 5 m to the right of the origin,

Position = +5 m

If he stands 3 m to the left,

Position = −3 m


9. Origin

Origin is the reference point from which distances are measured.

Usually represented by

O

Example

<------|------|------|------>

     -2    -1    O    +1   +2

10. Coordinate System

Used to locate objects.

One-dimensional (1D)

Only x-axis

Example

Train on straight track.


Two-dimensional (2D)

x-axis and y-axis

Example

Football ground.


Three-dimensional (3D)

x-axis

y-axis

z-axis

Example

Flying aeroplane.


Important NEET Concepts

Observer

Person who measures motion.


Coordinate

Shows position.


Time

Shows when motion occurs.


Motion depends on observer.

Always remember

Motion is Relative.


Frequently Asked NEET Facts

SI Unit of Distance

metre (m)


SI Unit of Time

second (s)


SI Unit of Speed

m/s


SI Unit of Force

Newton (N)


Previous Year NEET Concepts

Example 1

A passenger sitting inside a moving train is

A) Moving with respect to train

B) At rest with respect to train

C) Moving with respect to seat

D) None

Answer

B


Example 2

A car moves from Delhi to Jaipur.

The car is treated as

A) Rigid body

B) Point object

C) Fluid

D) None

Answer

B


Example 3

Motion is

A) Absolute

B) Relative

C) Constant

D) None

Answer

B


Formula Sheet (This Topic)

There are no numerical formulas in this introductory topic. Focus on understanding the concepts.


One-Page Revision

Mechanics

  • Study of motion and the causes of motion.

Kinematics

  • Motion only.

  • No force.

Dynamics

  • Motion + force.

Frame of Reference

  • Observer + coordinate system + time.

Motion

  • Position changes with time relative to an observer.

Rest

  • Position does not change relative to an observer.

Point Object

  • Size is negligible compared to the distance travelled.

Key Rule

  • There is no absolute rest or absolute motion; all motion is relative to the observer.

These notes provide a strong conceptual foundation for Motion in One Dimension, a high-weightage topic in the NEET Physics syllabus. 


Internal Links

  1. Units and Measurements Notes for NEET
  2. Vector and Scalar Quantities
  3. Motion in Two Dimensions
  4. Laws of Motion
  5. Work, Energy and Power
  6. System of Particles and Rotational Motion
  7. Gravitation Notes
  8. Oscillations and SHM
  9. Mechanical Properties of Solids
  10. Mechanical Properties of Fluids
  11. Thermal Physics Complete Notes
  12. Waves and Sound Notes
  13. Complete NEET Physics Formula Sheet
  14. NEET Physics PYQs Chapter-wise
  15. Motion in One Dimension MCQs with Answers
  16. Motion in One Dimension Numericals for NEET 
  17. Motion in One Dimension Revision Notes
  18. Class 11 Physics Complete Notes
  19. NEET Physics Mock Tests
  20. NEET Physics Study Plan
NEET Physics - Motion in One Dimension | Part 1

🔥 NEET Physics – Motion in One Dimension

Part 1 : Ultra-Hard NEET Numericals

Instructions

  • Total Questions : 25
  • Difficulty : NEET Advanced
  • Use SI Units.
  • Take g = 10 m/s² whenever required.
  • Do not use calculator unless instructed.

Q1.

A car starts from rest and moves with a constant acceleration of 4 m/s² for 15 s. Calculate
  1. Final velocity
  2. Total displacement
  3. Average velocity

Q2.

A train moving at 20 m/s accelerates uniformly to 50 m/s in 15 s. Find
  1. Acceleration
  2. Distance travelled

Q3.

A body travels
  • 40 m East
  • 30 m West
  • 20 m East
Calculate
  1. Total distance
  2. Displacement

Q4.

A particle moves with velocity v = 5 + 2t where t is in seconds. Find
  1. Velocity after 8 s
  2. Displacement in first 8 s

Q5.

A ball is thrown vertically upward with speed 40 m/s. Calculate
  1. Maximum height
  2. Time to reach highest point
  3. Total time of flight

Q6.

A runner completes
  • First half distance at 10 m/s
  • Second half distance at 15 m/s
Find average speed.

Q7.

A particle covers
  • 100 m in North direction
  • 100 m in East direction
Calculate
  1. Distance
  2. Displacement

Q8.

A car moving at 25 m/s stops uniformly in 5 seconds. Calculate
  1. Retardation
  2. Stopping distance

Q9.

Two cars move towards each other. Car A = 30 m/s Car B = 20 m/s Initially separated by 500 m. Find time to meet.

Q10.

A particle travels
  • 2 hours at 40 km/h
  • 3 hours at 60 km/h
Find
  1. Total distance
  2. Average speed

Q11.

A body starts from rest with acceleration 3 m/s² Find displacement in 12 seconds.

Q12.

A train moving at 72 km/h crosses a 300 m platform in 30 s. Find length of train.

Q13.

A cyclist increases speed uniformly from 5 m/s to 15 m/s over a distance of 100 m. Find acceleration.

Q14.

A body moves 10 m East 10 m North 10 m West Find displacement.

Q15.

A particle has velocity v = 20 − 5t Find
  1. Time when particle stops
  2. Distance travelled till then

Q16.

A stone falls freely from a tower. It reaches ground in 6 s. Find
  1. Height of tower
  2. Velocity on reaching ground

Q17.

A bus travels 80 km at 40 km/h and 120 km at 60 km/h. Find average speed.

Q18.

A particle moves according to s = 4t² + 3t Find velocity after 5 seconds.

Q19.

Two trains Length = 200 m each Speed = 15 m/s and 25 m/s move in opposite directions. Find time taken to cross each other.

Q20.

A particle covers first one-third distance with speed 10 m/s remaining distance with speed 20 m/s Find average speed.

Q21.

A body starts with velocity 15 m/s Acceleration 2 m/s² Find distance covered in 20 s.

Q22.

A particle moves 50 m East 120 m West 70 m East Calculate displacement.

Q23.

A train moving at 54 km/h crosses a man standing on platform in 12 seconds. Length of train?

Q24.

A particle starts from rest. Acceleration = 5 m/s² Find distance travelled during 4th second.

Q25. ⭐ NEET Challenge

A particle moves according to x = 5 + 4t + 2t² Find
  1. Position after 6 s
  2. Velocity after 6 s
  3. Acceleration
  4. Distance covered between 2 s and 6 s
Practice Rule
  • Attempt all questions without looking at formulas.
  • Write complete steps.
  • Time yourself (45 minutes).
  • Target Score : 25/25
NEET Physics - Motion in One Dimension | Supplementary Part 2

🔥 Supplementary Part 2

Ultra-Hard Motion in One Dimension Numericals (Q26–Q50)

Instructions

  • Difficulty: NEET Advanced
  • Attempt without calculator.
  • Take g = 10 m/s² unless otherwise stated.

Q26

A car accelerates uniformly from 10 m/s to 40 m/s in 12 seconds. Find:
  1. Acceleration
  2. Distance travelled

Q27

A particle moves with velocity v = 8 + 3t. Find displacement during first 10 seconds.

Q28

A train moving at 90 km/h crosses a pole in 18 s. Find length of train.

Q29

A ball is projected vertically upward with speed 50 m/s. Find:
  1. Maximum height
  2. Total time of flight

Q30

A body covers:
  • 120 m at 6 m/s
  • 180 m at 9 m/s
Find average speed.

Q31

Two cars move in the same direction. Car A = 18 m/s Car B = 24 m/s Initial gap = 240 m. Find time taken by B to overtake A.

Q32

A particle starts from rest with acceleration 4 m/s². Find distance travelled during the 8th second.

Q33

A cyclist slows uniformly from 12 m/s to rest in 6 s. Find stopping distance.

Q34

A particle moves:
  • 40 m North
  • 30 m East
  • 40 m South
Find displacement.

Q35

Velocity equation: v = 24 − 4t Find:
  1. Time to stop
  2. Distance travelled

Q36

A freely falling stone reaches the ground with velocity 60 m/s. Find height of the tower.

Q37

A train of length 180 m crosses a bridge of length 420 m in 24 s. Find speed of train.

Q38

A body moves according to x = 3t² + 2t + 5. Find velocity at t = 8 s.

Q39

A car travels:
  • 100 km at 50 km/h
  • 150 km at 75 km/h
Find average speed.

Q40

Two trains of lengths 250 m and 150 m move in opposite directions at 18 m/s and 22 m/s. Find crossing time.

Q41

A particle starts from rest. Acceleration = 6 m/s². Find displacement after 15 seconds.

Q42

A ball is thrown downward from a building with velocity 15 m/s. It reaches ground in 5 s. Find height of building.

Q43

A particle covers one-fourth distance at 12 m/s and remaining distance at 24 m/s. Find average speed.

Q44

Position equation: s = 2t³ − 5t² + 8. Find velocity at t = 3 s.

Q45

A train moving at 54 km/h overtakes another moving at 36 km/h. Both trains are 150 m long. Find overtaking time.

Q46

A body starts with velocity 20 m/s and acceleration 3 m/s². Find velocity after travelling 150 m.

Q47

A particle moves:
  • 100 m East
  • 60 m West
  • 20 m East
  • 40 m West
Find:
  1. Total distance
  2. Displacement

Q48

A stone is thrown upward from a cliff of height 80 m with speed 30 m/s. Find time taken to hit the ground.

Q49

A particle has acceleration a = 4 m/s². Initial velocity = 6 m/s. Find displacement in first 12 s.

Q50 ⭐ NEET Challenge

Position equation: x = 10 + 6t + 3t² Find:
  1. Position at t = 8 s
  2. Velocity at t = 8 s
  3. Acceleration
  4. Distance travelled between 4 s and 8 s

Self Evaluation

Score Performance
23–25 Excellent (NEET Top Rank Level)
18–22 Very Good
12–17 Good, Needs Practice
Below 12 Revise Theory & Formulae
NEET Physics Mock Test - Motion in One Dimension (Part 3)

🔥 NEET Physics Mock Test

Motion in One Dimension (Part 3)

Total Questions : 45
Time : 45 Minutes
Marks : 180
Correct : +4
Wrong : −1

Section A (MCQs)

Q1. SI unit of displacement is

A) m/s
B) m
C) km
D) s

Q2. Distance is a

A) Vector
B) Scalar
C) Tensor
D) None

Q3. Velocity can become zero during upward motion of a projectile at

A) Highest point
B) Lowest point
C) Throughout
D) Never

Q4. Average velocity equals average speed when

A) Circular path
B) Straight line without changing direction
C) Closed path
D) Random motion

Q5. Which quantity may be zero even when distance is not zero?

A) Speed
B) Time
C) Displacement
D) Velocity

Q6. A body starts from rest. Initial velocity is

A) 1 m/s
B) 10 m/s
C) 0 m/s
D) −1 m/s

Q7. Uniform velocity means

A) Constant speed only
B) Constant velocity
C) Variable speed
D) Variable direction

Q8. Unit of acceleration

A) m
B) m/s
C) m/s²
D) km/h

Q9. Graph of uniform motion is

A) Straight line
B) Circle
C) Parabola
D) Hyperbola

Q10. Which is a vector?

A) Speed
B) Distance
C) Velocity
D) Time

Q11. A car moves 20 m East then 20 m West. Displacement is

A)40 m B)20 m C)0 D)10 m

Q12. Average speed is

A)Total Distance/Total Time B)Displacement/Time C)v²-u² D)None

Q13. Motion along straight line is called

A)Rectilinear B)Circular C)Rotational D)Projectile

Q14. Retardation is

A)Positive acceleration B)Negative acceleration C)Zero acceleration D)Infinite acceleration

Q15. Speedometer measures

A)Velocity B)Distance C)Instantaneous Speed D)Average Speed

Q16. Odometer measures

A)Speed B)Distance C)Acceleration D)Velocity

Q17. Motion is always

A)Absolute B)Relative C)Constant D)Uniform

Q18. A freely falling body has acceleration

A)g B)0 C)2g D)Depends on mass

Q19. Unit of velocity

A)m B)m/s C)m/s² D)km

Q20. Equation v=u+at is valid for

A)Variable acceleration B)Constant acceleration C)Circular motion D)Random motion

Q21. Which quantity is never negative?

A)Velocity B)Displacement C)Speed D)Acceleration

Q22. Area under velocity-time graph gives

A)Acceleration B)Distance/Displacement C)Speed D)Force

Q23. Slope of displacement-time graph gives

A)Acceleration B)Velocity C)Force D)Momentum

Q24. Slope of velocity-time graph gives

A)Displacement B)Acceleration C)Time D)Distance

Q25. A body moving with constant velocity has

A)Zero acceleration B)Maximum acceleration C)Infinite acceleration D)Variable acceleration

Q26. Graph of rest on displacement-time graph is

A)Horizontal line B)Vertical line C)Curve D)Circle

Q27. SI unit of time

A)Hour B)Second C)Minute D)Day

Q28. Which is scalar?

A)Velocity B)Acceleration C)Distance D)Displacement

Q29. Initial velocity symbol

A)s B)t C)u D)v

Q30. Final velocity symbol

A)u B)a C)v D)s


Section B (Assertion-Reason)

Q31. Motion is relative.
Reason: There is no absolute rest.
Q32. Average speed is always less than average velocity.
Reason: Speed is scalar.
Q33. Displacement may be zero while distance is not zero.
Reason: Distance is scalar.
Q34. Constant speed always means constant velocity.
Reason: Direction may change.
Q35. Uniform acceleration means equal change in velocity in equal intervals of time.

Section C (Numericals)

Q36. A body starts from rest with acceleration 5 m/s². Find velocity after 8 s.
Q37. A train moving at 20 m/s travels for 30 s. Find distance.
Q38. A body moves 60 m East then 80 m West. Find displacement.
Q39. A stone is dropped from height 45 m. Find time to reach ground.
Q40. Find average speed if a car travels 120 km in 2 hours.
Q41. Find displacement using s=ut+½at² where u=10 m/s, a=2 m/s², t=5 s.
Q42. Find acceleration if velocity changes from 10 m/s to 30 m/s in 4 s.
Q43. A train of length 180 m crosses a pole in 12 s. Find speed.
Q44. A ball is thrown upward with 20 m/s. Find maximum height.
Q45. A body moves with velocity equation v=6+2t. Find velocity after 8 s.

OMR Answer Sheet

QAns QAns QAns
1___16___31___
2___17___32___
3___18___33___
4___19___34___
5___20___35___
6___21___36___
7___22___37___
8___23___38___
9___24___39___
10___25___40___
11___26___41___
12___27___42___
13___28___43___
14___29___44___
15___30___45___
NEET Physics - Motion in One Dimension | Part 4 Answer Key

🔥 NEET Physics

Part 4 – Answer Key & Step-by-Step Solutions


Section A (MCQs)

Q Answer Q Answer Q Answer
1B11C21C
2B12A22B
3A13A23B
4B14B24B
5C15C25A
6C16B26A
7B17B27B
8C18A28C
9A19B29C
10C20B30C

Section B (Assertion–Reason)

Question Correct Option
31 A (Both true, Reason correctly explains)
32 D (Assertion false, Reason true)
33 B (Both true, Reason not correct explanation)
34 D (Assertion false, Reason true)
35 A (Both true, Reason correctly explains)

Section C (Numericals)

Q36

Given u = 0 m/s a = 5 m/s² t = 8 s

v = u + at
v = 0 + 5 × 8

Answer = 40 m/s
Q37

Distance = Speed × Time
= 20 × 30

Answer = 600 m
Q38

East = +60 West = -80

Displacement = 60 - 80 = -20 m

Answer = 20 m West
Q39

s = 45 m u = 0 g =10

45 = 5t²
t² =9
t =3 s

Answer = 3 s
Q40

Average Speed =120/2

Answer =60 km/h
Q41

s = ut + ½at²
=10×5 + ½×2×25
=50+25

Answer =75 m
Q42

a=(30−10)/4
=20/4

Answer =5 m/s²
Q43

Speed =180/12

Answer =15 m/s
Q44

Maximum Height H=u²/2g
=20²/(20)
=20 m

Answer =20 m
Q45

v=6+2t
=6+16
Answer =22 m/s

Score Analysis

Marks Performance
170–180 Outstanding ⭐⭐⭐⭐⭐
150–169 Excellent
120–149 Very Good
90–119 Good
Below 90 Needs Revision
Common Mistakes in NEET Motion in One Dimension
  • Confusing distance with displacement.
  • Using average velocity instead of average speed.
  • Incorrect sign convention (+ and -).
  • Using wrong kinematic equation.
  • Mixing km/h and m/s units.
  • Ignoring direction in vector quantities.
  • Calculation mistakes in free-fall problems.
NEET Physics - Motion in One Dimension | Part 5 (PYQs)

🔥 Part 5

Previous Year Question Bank (NEET / AIPMT / AIIMS Style)

This section contains NEET-style questions inspired by previous exam patterns. They are designed for practice and concept building.


Q1. A body moves 10 m east and then 10 m west. What is its displacement?

A) 20 m   B) 10 m   C) 0 m   D) 5 m
Answer: C (0 m)
Q2. Which of the following is a scalar quantity?

A) Velocity B) Acceleration C) Distance D) Displacement
Answer: C
Q3. SI unit of acceleration is

A) m B) m/s C) m/s² D) km/h
Answer: C
Q4. A car starts from rest with acceleration 4 m/s². Find velocity after 5 s.
Answer: 20 m/s
Q5. A train moving at 25 m/s crosses a pole in 8 s. Find length of train.
Answer: 200 m
Q6. A body falls freely for 4 s. Find its velocity on reaching the ground.
Answer: 40 m/s
Q7. Average speed is defined as A) Distance / Time B) Displacement / Time C) Velocity / Time D) None
Answer: A
Q8. Which graph represents uniform motion?
Answer: Straight line on displacement–time graph
Q9. A body moves with constant velocity. Its acceleration is
Answer: Zero
Q10. A stone is thrown upward with speed 30 m/s. Find maximum height.
Answer: 45 m

🔥 PYQ Practice Set (Without Answers)

  1. A body covers 240 m in 30 s. Find average speed.
  2. A car accelerates from 10 m/s to 30 m/s in 5 s. Find acceleration.
  3. Define displacement with one example.
  4. Differentiate between speed and velocity.
  5. A particle moves according to x = 4 + 5t + 2t². Find velocity at t = 6 s.
  6. A train of length 250 m crosses a bridge of length 350 m in 24 s. Find speed.
  7. A freely falling body reaches the ground in 8 s. Find height.
  8. State any two differences between distance and displacement.
  9. Find stopping distance of a car moving at 30 m/s if retardation is 5 m/s².
  10. Find average speed if equal distances are covered at 20 km/h and 30 km/h.

🔥 Formula Revision Sheet

  • v = u + at
  • s = ut + ½at²
  • v² = u² + 2as
  • Average Speed = Total Distance / Total Time
  • Average Velocity = Total Displacement / Total Time
  • Distance = Speed × Time
  • Acceleration = (v − u) / t

🎯 NEET Rapid Revision

  • Motion is relative.
  • Distance is scalar.
  • Displacement is vector.
  • Speed is scalar.
  • Velocity is vector.
  • Acceleration can be positive or negative.
  • Retardation = Negative acceleration.
  • Slope of displacement–time graph = Velocity.
  • Slope of velocity–time graph = Acceleration.
  • Area under velocity–time graph = Displacement.

Sunday, June 21, 2026

Motion of an Object Under Free Fall | NEET Physics Notes with Graphs & Formulas

CBSE Class 11 Physics (Motion Under Free Fall) – Question Bank with Answers  

Educational diagram explaining free fall motion with acceleration-time, velocity-time and displacement-time graphs for NEET Physics.
Motion of an object under free fall showing acceleration, velocity, and displacement-time graphs.


- Dr.Sanjaykumar Pawar 

A. Multiple Choice Questions (MCQs)

1. A body is said to be in free fall when:

a) It falls with constant velocity
b) It falls under gravity only
c) It falls in vacuum only
d) It moves downward

Answer: b) It falls under gravity only


2. The acceleration due to gravity near Earth's surface is:

a) 8.9 m/s²
b) 9.8 m/s²
c) 10.8 m/s²
d) 12 m/s²

Answer: b) 9.8 m/s²


3. The velocity-time graph for a freely falling body is:

a) Horizontal line
b) Parabola
c) Straight line
d) Circle

Answer: c) Straight line


4. The slope of a velocity-time graph represents:

a) Velocity
b) Distance
c) Acceleration
d) Momentum

Answer: c) Acceleration


5. Distance travelled in successive equal intervals of time during free fall follows:

a) 1 : 2 : 3 : 4
b) 2 : 4 : 6 : 8
c) 1 : 3 : 5 : 7
d) 1 : 4 : 9 : 16

Answer: c) 1 : 3 : 5 : 7


6. Stopping distance of a vehicle is proportional to:

a) Speed
b) Square of speed
c) Cube of speed
d) Inverse of speed

Answer: b) Square of speed


7. For a freely falling body released from rest:

a) u = g
b) u = 1
c) u = 0
d) u = 10

Answer: c) u = 0


8. Which equation represents free fall motion?

a) v = u + gt
b) v = u − gt
c) v = u/t
d) v = gt²

Answer: b) v = u − gt


9. The SI unit of acceleration due to gravity is:

a) m
b) m/s
c) m/s²
d) kg

Answer: c) m/s²


10. If speed doubles, stopping distance becomes:

a) Double
b) Triple
c) Four times
d) Eight times

Answer: c) Four times


B. Very Short Answer Questions (1 Mark)

1. Define free fall.

Answer: Motion of a body under the influence of gravity alone is called free fall.

2. What is the value of g near Earth's surface?

Answer: 9.8 m/s².

3. What is the acceleration of a freely falling body?

Answer: g downward.

4. Who proposed the law of odd numbers?

Answer: Galileo Galilei.

5. What is stopping distance?

Answer: Distance travelled by a vehicle after brakes are applied until it comes to rest.

6. What is the slope of a v–t graph?

Answer: Acceleration.

7. What is the shape of displacement-time graph in free fall?

Answer: Parabola.

8. What is the initial velocity of a body released from rest?

Answer: Zero.


C. Short Answer Questions (2–3 Marks)

1. Why is acceleration constant during free fall?

Answer: The only force acting on the body is gravity. Near Earth's surface, gravity remains nearly constant. Therefore acceleration remains constant and equal to g.


2. Write the equations of motion for free fall.

Answer:


3. State Galileo's Law of Odd Numbers.

Answer: The distances covered by a freely falling body during successive equal intervals of time are proportional to odd numbers:

1 : 3 : 5 : 7 : 9 ...


4. Why is the displacement-time graph parabolic?

Answer: Displacement in free fall is proportional to the square of time.


s = \frac{1}{2}gt^2

Since displacement depends on , the graph is a parabola.


5. Explain why stopping distance increases with speed.

Answer: Stopping distance is given by:


d_s = \frac{v_0^2}{2a}

Hence stopping distance is proportional to the square of velocity. Therefore higher speed results in much larger stopping distance.


D. Long Answer Questions (5 Marks)

1. Explain the variation of acceleration, velocity and displacement with time during free fall.

Answer:

(i) Acceleration-Time Graph

  • Acceleration remains constant.
  • Value = –g.
  • Graph is a horizontal straight line.

(ii) Velocity-Time Graph

  • Velocity changes uniformly with time.
  • Equation:

v = u - gt
  • Graph is a straight line with negative slope.

(iii) Displacement-Time Graph

  • Displacement increases as square of time.

s = ut - 1/2gt²
  • Graph is parabolic.

Thus acceleration is constant, velocity changes uniformly, and displacement changes non-uniformly.


2. Prove Galileo's Law of Odd Numbers.

Answer:

For free fall:


s=1/2gt²

After time t:


s_1=1/2gt²

After 2t:


s_2=4s_1

After 3t:


s_3=9s_1

Distance in successive intervals:

First interval:


s_1

Second interval:


s_2-s_1=3s_1

Third interval:


s_3-s_2=5s_1

Fourth interval:


s_4-s_3=7s_1

Hence ratio:


1:3:5:7

Thus proved.


E. Assertion and Reason Questions

1.

Assertion (A): A freely falling body has constant acceleration.

Reason (R): Gravity acts uniformly near Earth's surface.

Answer: Both A and R are true, and R is the correct explanation of A.


2.

Assertion (A): Stopping distance depends on velocity.

Reason (R): Stopping distance is proportional to velocity squared.

Answer: Both A and R are true, and R is the correct explanation of A.


3.

Assertion (A): Velocity-time graph in free fall is a straight line.

Reason (R): Acceleration remains constant.

Answer: Both A and R are true, and R correctly explains A.


4.

Assertion (A): Displacement-time graph in free fall is linear.

Reason (R): Displacement is proportional to time squared.

Answer: Assertion is false but Reason is true.


F. Fill in the Blanks

  1. Motion under gravity alone is called free fall.

  2. The value of acceleration due to gravity is approximately 9.8 m/s².

  3. The slope of a velocity-time graph gives acceleration.

  4. Galileo's law follows the ratio 1 : 3 : 5 : 7.

  5. Stopping distance is proportional to the square of velocity.

  6. The SI unit of acceleration is m/s².

  7. The displacement-time graph of free fall is a parabola.

  8. A body released from rest has initial velocity zero.


G. Case Study Questions

Case Study

A ball is dropped from the top of a tower. It falls freely under gravity. The acceleration remains constant throughout the motion. The velocity increases uniformly while displacement increases rapidly with time.

Questions

1. What is the acceleration acting on the ball?

Answer: g = 9.8 m/s² downward.


2. Which force acts on the ball during free fall?

Answer: Gravitational force.


3. What is the shape of the velocity-time graph?

Answer: Straight line.


4. What is the shape of the displacement-time graph?

Answer: Parabola.


5. Which law explains distances covered in successive seconds?

Answer: Galileo's Law of Odd Numbers.


H. Statement-Based Questions

Statement 1

Acceleration due to gravity remains constant during free fall.

Statement 2

Velocity changes uniformly with time.

a) Both statements are true.
b) Both statements are false.
c) Statement 1 true, Statement 2 false.
d) Statement 1 false, Statement 2 true.

Answer: a) Both statements are true.


Statement 1

Stopping distance is proportional to speed.

Statement 2

Stopping distance is proportional to square of speed.

Answer: Statement 1 is false and Statement 2 is true.


I. Match the Columns

Column A Column B
1. Free Fall a. Gravity only
2. g b. 9.8 m/s²
3. v–t graph slope c. Acceleration
4. Galileo d. Odd number law
5. Stopping Distance e. Depends on v²

Answers

1 → a

2 → b

3 → c

4 → d

5 → e


CBSE Exam Important Questions

1. Define free fall and explain its characteristics.

2. Draw and explain acceleration-time, velocity-time and displacement-time graphs for free fall.

3. State and prove Galileo's Law of Odd Numbers.

4. Derive the formula for stopping distance.

5. Explain why stopping distance increases with speed.

6. Write equations of motion for a freely falling body.

7. Differentiate between velocity and acceleration during free fall.

These questions cover MCQs, competency-based questions, assertion-reason, case study, fill in the blanks, statement-based questions, match the columns, short answers, and long answers as per the latest CBSE Class 11 examination pattern


 Internal Links

Motion in a Straight Line Notes

Acceleration and Velocity Concepts

Kinematics Formula Sheet

Free Fall Concepts

Acceleration Due to Gravity Notes

Newton's Law of Universal Gravitation

Projectile Motion Basics

Graph Section

Velocity-Time Graph Explained

Acceleration-Time Graph Problems

Position-Time Graph Interpretation

Galileo Law Section

Motion Under Constant Acceleration

Important NEET Kinematics Questions

NCERT Kinematics Solutions

Stopping Distance Section

Newton's Laws of Motion

Friction Notes Class 11

Braking Force and Retardation Problems

Revision Section

NEET Physics Formula Handbook

Most Important Kinematics Numericals

NEET Physics Previous Year Questions

Motion of an Object Under Free Fall - NEET Notes

Motion of an Object Under Free Fall

NEET Physics Easy Notes

1. Free Fall – Basic Idea

  • When an object falls under the effect of gravity only, the motion is called free fall.
  • Air resistance is neglected in free fall problems.
  • Acceleration due to gravity is represented by g.

Near Earth’s surface:

g ≈ 9.8 m/s²
  • Direction of gravity is always downward.
  • If upward direction is taken positive, then acceleration becomes negative.

Fig. 2.7 : Motion of Object Under Free Fall

(a) Variation of Acceleration with Time

  • Acceleration remains constant throughout the motion.
  • The graph is a horizontal straight line.
  • Value of acceleration is always:
a = -g

Important NEET Point

Constant acceleration means velocity changes uniformly with time.

(b) Variation of Velocity with Time

  • Initial velocity for a freely falling body released from rest:
u = 0
  • Velocity increases linearly with time.
  • Equation of velocity:
v = u - gt

Since (u = 0),

v = -gt

Graph Understanding

  • Straight line with negative slope.
  • Slope of v–t graph = acceleration = (-g).

Important NEET Concepts

  • Velocity becomes more negative with time.
  • Body gains speed while falling downward.

(c) Variation of Distance (Position) with Time

  • Distance covered in free fall is proportional to square of time.
  • Equation of motion:
y = ut - 1/2 gt²

For (u = 0),

y = -1/2 gt²

Graph Understanding

  • Graph is a parabola.
  • Distance increases rapidly with time.
  • Motion is non-uniform because velocity changes continuously.

Galileo’s Law of Odd Numbers

Statement

  • “The distances travelled during successive equal intervals of time by a freely falling body are in the ratio of odd numbers.”

Ratio is:

1 : 3 : 5 : 7 : 9 : ...

Proof in Simple Steps

Step 1: Position after Different Times

For free fall:

y = -1/2 gt²

After time (t):

y₁ = 1/2 gt²

After time (2t):

y₂ = 1/2 g(2t)² = 4y₁

After time (3t):

y₃ = 1/2 g(3t)² = 9y₁

After time (4t):

y₄ = 16y₁

Step 2: Distance in Successive Intervals

First Interval

y₁ = 1y₁

Second Interval

y₂ - y₁ = 4y₁ - y₁ = 3y₁

Third Interval

y₃ - y₂ = 9y₁ - 4y₁ = 5y₁

Fourth Interval

y₄ - y₃ = 16y₁ - 9y₁ = 7y₁

Hence ratios become:

1 : 3 : 5 : 7

Important Result for NEET

For a body starting from rest under gravity:

sâ‚™ ∝ (2n - 1)

where:

  • (sâ‚™) = distance travelled in nth second.

Example 2.6 – Stopping Distance of Vehicles

Definition

  • Distance travelled by a vehicle before coming to rest after brakes are applied is called stopping distance.

Derivation

Using equation of motion:

v² = u² + 2as

For stopping:

  • Final velocity (v = 0)
  • Initial velocity (u = v₀)

So,

0 = v₀² + 2adâ‚›

Therefore,

dâ‚› = -v₀² / 2a

Conclusions

  • Stopping distance is proportional to square of initial velocity.
dâ‚› ∝ v₀²

NEET Important Points

  • If speed doubles → stopping distance becomes 4 times.
  • Stronger brakes mean larger retardation and smaller stopping distance.

Quick Revision Formula Sheet

Concept Formula
Velocity-Time Relation v = u - gt
Position-Time Relation s = ut - 1/2 gt²
Velocity-Position Relation v² = u² - 2gs

One-Line NEET Tricks

  • Acceleration due to gravity is constant.
  • v–t graph slope gives acceleration.
  • Distance in nth second follows odd number rule.
  • Stopping distance depends on square of speed.
  • Free fall graphs are very important for NEET numericals.
NEET Physics Easy Notes © 2026

Wednesday, June 17, 2026

Work and Kinetic Energy Explained: Work-Energy Theorem for NEET

- Dr.Sanjaykumar Pawar 

Educational diagram explaining Work, Kinetic Energy, and the Work-Energy Theorem with force, displacement, and energy formulas for NEET Physics students.
Work-Energy Theorem showing how work done by a force changes the kinetic energy of an object.


Work-Energy Theorem Notes for NEET |


Kinetic Energy Made Easy

NOTIONS OF WORK AND KINETIC ENERGY

├── 1. Kinematics Relation

│   │

│   ├── Equation

│   │   └── v² − u² = 2as

│   │

│   ├── u = Initial Velocity

│   ├── v = Final Velocity

│   ├── a = Acceleration

│   └── s = Displacement

├── 2. Multiply by m/2

│   │

│   ├── (m/2)(v² − u²) = (m/2)(2as)

│   │

│   ├── ½mv² − ½mu² = mas

│   │

│   ├── Newton's Second Law

│   │   └── F = ma

│   │

│   └── Therefore

│       └── ½mv² − ½mu² = Fs

├── 3. Three-Dimensional Form

│   │

│   ├── v² − u² = 2(a·d)

│   │

│   ├── a = Acceleration Vector

│   ├── d = Displacement Vector

│   └── a·d = Dot Product

├── 4. Kinetic Energy (K)

│   │

│   ├── Formula

│   │   └── K = ½mv²

│   │

│   ├── Definition

│   │   └── Energy due to motion

│   │

│   ├── Unit

│   │   └── Joule (J)

│   │

│   └── Properties

│       ├── Always Positive

│       ├── Depends on Mass

│       └── Depends on Velocity²

├── 5. Work Done (W)

│   │

│   ├── Formula

│   │   └── W = F·d

│   │

│   ├── General Formula

│   │   └── W = Fd cosθ

│   │

│   ├── Definition

│   │   └── Force × Displacement

│   │

│   └── Unit

│       └── Joule (J)

├── 6. Work-Energy Equation

│   │

│   ├── Initial Kinetic Energy

│   │   └── Ki = ½mu²

│   │

│   ├── Final Kinetic Energy

│   │   └── Kf = ½mv²

│   │

│   └── Relation

│       └── Kf − Ki = W

├── 7. Work-Energy Theorem

│   │

│   ├── Statement

│   │   └── Change in Kinetic Energy

│   │       = Net Work Done

│   │

│   ├── Formula

│   │   └── Wnet = ΔK

│   │

│   └── Alternative Form

│       └── Wnet = Kf − Ki

├── 8. Types of Work

│   │

│   ├── Positive Work

│   │   ├── W > 0

│   │   ├── Force along Motion

│   │   └── Kinetic Energy Increases

│   │

│   ├── Negative Work

│   │   ├── W < 0

│   │   ├── Force opposite Motion

│   │   └── Kinetic Energy Decreases

│   │

│   └── Zero Work

│       ├── W = 0

│       ├── Force ⟂ Displacement

│       └── Kinetic Energy Constant

└── 9. NEET Formula Box

    │

    ├── K = ½mv²

    ├── W = Fd cosθ

    ├── W = F·d

    ├── Wnet = ΔK

    └── Wnet = Kf − Ki


FINAL CONCEPT

└── Net Work Done on a Body

    └── Produces Equal Change in Kinetic Energy

        └── Wnet = ΔK 

CBSE Class 11 Physics

Work, Energy and Power

Topic: Notions of Work and Kinetic Energy – Work-Energy Theorem


A. Multiple Choice Questions (MCQs)

1. The kinetic energy of a body is given by:

(a) mv² (b) ½mv² (c) mv (d) m²v

Answer: (b) ½mv²


2. SI unit of work is:

(a) Newton (b) Watt (c) Joule (d) Pascal

Answer: (c) Joule


3. Work done is maximum when angle between force and displacement is:

(a) 0° (b) 45° (c) 90° (d) 180°

Answer: (a) 0°


4. If force is perpendicular to displacement, work done is:

(a) Positive (b) Negative (c) Zero (d) Infinite

Answer: (c) Zero


5. Work-Energy theorem states:

(a) Work done equals momentum (b) Work done equals force (c) Work done equals change in kinetic energy (d) Work done equals acceleration

Answer: (c)


6. Kinetic energy depends on:

(a) Mass only (b) Velocity only (c) Mass and velocity (d) Density

Answer: (c)


7. If velocity becomes twice, kinetic energy becomes:

(a) Two times (b) Four times (c) Six times (d) Eight times

Answer: (b)


8. The dimension of work and energy is:

(a) MLT⁻¹ (b) ML²T⁻² (c) ML²T⁻¹ (d) MLT⁻²

Answer: (b)


9. Negative work is done when:

(a) Force and displacement are same (b) Force is perpendicular (c) Force opposes displacement (d) Force is zero

Answer: (c)


10. The unit of kinetic energy is:

(a) Newton (b) Joule (c) Watt (d) kg

Answer: (b)


B. Very Short Answer Questions (1 Mark)

Q1. Define kinetic energy.

Answer: Kinetic energy is the energy possessed by a body due to its motion.


Q2. Write the formula of kinetic energy.

Answer: K = ½mv²


Q3. State SI unit of work.

Answer: Joule (J)


Q4. What is the work done when force is perpendicular to displacement?

Answer: Zero.


Q5. State Work-Energy theorem.

Answer: Change in kinetic energy equals the net work done on a body.


C. Short Answer Questions (2–3 Marks)

Q1. Define work done. Write its mathematical expression.

Answer:

Work done is the product of force and displacement in the direction of force.

W = Fd cosθ

where θ is the angle between force and displacement.


Q2. Why is kinetic energy always positive?

Answer:

K = ½mv²

Since mass is positive and square of velocity is always positive, kinetic energy is always positive.


Q3. Distinguish between positive and negative work.

Positive Work Negative Work
Force acts along displacement Force acts opposite displacement
Energy increases Energy decreases
Example: Pulling a cart Example: Braking a bicycle

Q4. Write any two applications of Work-Energy theorem.

Answer:

  1. Finding velocity without using time.
  2. Calculating work done by forces.

D. Long Answer Questions (5 Marks)

Q1. Derive Work-Energy theorem.

Answer:

From equation of motion:

v² − u² = 2as

Multiplying both sides by m/2,

½m(v² − u²) = mas

Since,

F = ma

Therefore,

½mv² − ½mu² = Fs

Now,

Kf = ½mv²

Ki = ½mu²

Thus,

Kf − Ki = W

Hence,

W = ΔK

This proves that net work done on a particle equals change in its kinetic energy.


Q2. Explain positive, negative and zero work with examples.

Answer:

  1. Positive Work:

    • Force and displacement in same direction.
    • Example: Pulling a trolley.
  2. Negative Work:

    • Force opposite displacement.
    • Example: Brakes on bicycle.
  3. Zero Work:

    • Force perpendicular to displacement.
    • Example: Centripetal force in circular motion.

E. Assertion and Reason Questions

Q1.

Assertion (A): Kinetic energy is always positive.

Reason (R): Kinetic energy depends on square of velocity.

Answer: Both A and R are true and R is the correct explanation.


Q2.

Assertion (A): Work done by centripetal force is zero.

Reason (R): Centripetal force is perpendicular to displacement.

Answer: Both A and R are true and R is the correct explanation.


Q3.

Assertion (A): Negative work increases kinetic energy.

Reason (R): Negative work opposes motion.

Answer: Assertion is false but Reason is true.


Q4.

Assertion (A): SI unit of work and energy is same.

Reason (R): Both are measured in Joules.

Answer: Both A and R are true and R is the correct explanation.


F. Fill in the Blanks

  1. Kinetic energy of a body is ______ due to its motion.

Answer: energy


  1. Formula of kinetic energy is ______.

Answer: ½mv²


  1. SI unit of work is ______.

Answer: Joule


  1. Work done is zero when force is ______ to displacement.

Answer: perpendicular


  1. According to Work-Energy theorem,

W = ______

Answer: ΔK


G. Statement Based Questions

State whether True or False.

  1. Kinetic energy can be negative.

Answer: False


  1. Work and energy have same units.

Answer: True


  1. Work done is maximum when θ = 90°.

Answer: False


  1. Work-Energy theorem relates work and kinetic energy.

Answer: True


  1. Negative work decreases kinetic energy.

Answer: True


H. Match the Columns

Column A

A. Kinetic Energy

B. Work Done

C. Positive Work

D. Negative Work

E. SI Unit

Column B

  1. Joule

  2. ½mv²

  3. Force opposite displacement

  4. Fd cosθ

  5. Force along displacement

Answers

A → 2

B → 4

C → 5

D → 3

E → 1


I. Case Study Questions

Case Study

A student pushes a 5 kg box along a horizontal floor with a force of 20 N. The box moves 4 m in the direction of force.

Q1. What is the work done?

W = Fd

= 20 × 4

= 80 J

Answer: 80 J


Q2. If all work converts into kinetic energy, what is change in kinetic energy?

Answer: 80 J


Q3. Which theorem relates work and kinetic energy?

Answer: Work-Energy theorem.


Q4. Is the work positive or negative?

Answer: Positive.


Q5. Why?

Answer: Force and displacement are in the same direction.


J. Important Board Exam Questions

  1. Define kinetic energy.
  2. State and prove Work-Energy theorem.
  3. Write SI unit of work and energy.
  4. Differentiate positive and negative work.
  5. Explain zero work with example.
  6. Define work done and derive its expression.
  7. Explain Work-Energy theorem with suitable example.
  8. Why is kinetic energy always positive?
  9. Give practical applications of Work-Energy theorem.
  10. Derive K = ½mv² from Work-Energy theorem.

One-Line Revision

• K = ½mv²

• W = Fd cosθ

• Wnet = ΔK

• Positive Work → KE increases

• Negative Work → KE decreases

• Zero Work → KE remains constant

• SI Unit of Work & Energy = Joule (J)

• Work-Energy Theorem: Net work done = Change in kinetic energy 

Internal Links

Laws of Motion Explained for NEET

Work, Energy and Power Complete Notes

Conservation of Mechanical Energy

Motion in a Straight Line Notes

Motion in a Plane and Vectors

Newton's Laws of Motion Questions

Circular Motion for NEET

Units and Dimensions Physics Notes

Kinematics Formula Sheet

NEET Physics Chapter-wise Revision Notes


Work and Kinetic Energy - NEET Notes

NOTIONS OF WORK AND KINETIC ENERGY
THE WORK–ENERGY THEOREM

1. Equation from Kinematics

For rectilinear motion under constant acceleration:

v² − u² = 2as

Where:

  • u = Initial velocity
  • v = Final velocity
  • a = Acceleration
  • s = Displacement

This equation relates velocity, acceleration and displacement.


2. Multiplying by m/2

Multiply both sides by m/2:

(m/2)(v² − u²) = (m/2)(2as)
½mv² − ½mu² = mas

From Newton's Second Law:

F = ma

Therefore:

½mv² − ½mu² = Fs
Important: The left side involves mass and velocity, while the right side involves force and displacement.

3. Generalisation to Three Dimensions

For motion in three dimensions, vectors are used.

v² − u² = 2(a · d)

Where:

  • a = Acceleration vector
  • d = Displacement vector
  • · = Dot product

Multiplying by m/2:

½mv² − ½mu² = m(a · d)

Since:

F = ma

We get:

½mv² − ½mu² = F · d

4. Kinetic Energy (K)

The quantity

K = ½mv²

is called Kinetic Energy.

Definition

Kinetic Energy is the energy possessed by a body due to its motion.

SI Unit

Joule (J)

Important Facts

  • Kinetic energy is always positive.
  • It depends on mass and velocity.
  • If velocity doubles, kinetic energy becomes four times.
  • If velocity becomes three times, kinetic energy becomes nine times.

5. Work Done (W)

The quantity

W = F · d

is called Work Done.

Definition

Work is said to be done when a force produces displacement in an object.

Formula

W = Fd cosθ

Where:

  • F = Force
  • d = Displacement
  • θ = Angle between force and displacement

SI Unit

Joule (J)

6. Work-Energy Equation

Initial kinetic energy:

Ki = ½mu²

Final kinetic energy:

Kf = ½mv²

Substituting in the equation:

Kf − Ki = W
Change in Kinetic Energy = Work Done

7. Work-Energy Theorem

Statement

The change in kinetic energy of a particle is equal to the work done on it by the net force acting on it.

Mathematically:

Wnet = ΔK

or

Wnet = Kf − Ki

8. Physical Meaning of Work-Energy Theorem

Case 1: Positive Work

W > 0

Force acts in the direction of motion.

Kf > Ki

Result: Speed increases.

Example: Pushing a moving cart.


Case 2: Negative Work

W < 0

Force acts opposite to the direction of motion.

Kf < Ki

Result: Speed decreases.

Example: Applying brakes on a bicycle.


Case 3: Zero Work

W = 0

Force acts perpendicular to displacement.

Kf = Ki

Result: No change in speed.

Example: Centripetal force in circular motion.


9. Quick NEET Revision Box

Kinetic Energy

K = ½mv²

Work Done

W = Fd cosθ

Work-Energy Theorem

Wnet = Kf − Ki = ΔK

SI Unit

  • Work → Joule (J)
  • Kinetic Energy → Joule (J)

Key Idea

Net work done on a body changes its kinetic energy.


NEET One-Line Summary

Whenever a net force does work on an object, its kinetic energy changes by exactly the same amount.
Wnet = ΔK
Prepared for NEET Physics Revision

Thursday, June 4, 2026

Coefficient of Static Friction from Angle of Repose: Solved Example

 How to Calculate Static Friction Using Inclined Plane Angle

-Dr.Sanjaykumar Pawar 



Internal Links

Laws of Friction Explained

Static vs Kinetic Friction

Free Body Diagrams in Physics

Inclined Plane Problems and Solutions

Newton's Laws of Motion

Force Equilibrium Concepts

Applications of Friction in Daily Life

JEE Mechanics Important Questions

NEET Physics Friction Chapter Notes

Work, Energy and Power Fundamentals

Physics diagram showing a block at rest on an inclined plane with weight, normal force, static friction, force components, and calculation of coefficient of static friction using θmax = 15°.
Free-body diagram of a block on an inclined plane showing the relationship between angle of repose and coefficient of static friction.


Example 4.8 - Coefficient of Static Friction

Example 4.8

A mass of 4 kg rests on a rough horizontal plane. The plane is gradually inclined. When the angle becomes 15°, the block just begins to slide. Find the coefficient of static friction (μₛ).

Given Data

  • Mass of block, m = 4 kg
  • Angle of inclination, θ = 15°
  • Find μₛ

Step 1: Understand the Situation

As the plane is tilted, the component of weight acting down the plane increases.

Static friction opposes this motion and keeps the block at rest.

At θ = 15°, the block is just about to move. At this point, static friction reaches its maximum value.

Maximum Static Friction = μₛN

Step 2: Forces Acting on the Block

The following forces act on the block:

  • Weight (mg) acting vertically downward
  • Normal reaction (N) acting perpendicular to the plane
  • Static friction (fâ‚›) acting upward along the plane

Step 3: Resolve Weight into Components

Weight mg is resolved into two components:

Along the plane = mg sin θ

This component tends to pull the block downward.

Perpendicular to the plane = mg cos θ

This component presses the block against the surface.

Step 4: Apply Equilibrium Conditions

Since the block is still at rest:

fₛ = mg sin θ
N = mg cos θ

Step 5: Use Maximum Static Friction Formula

fₛ = μₛN

Substitute the values:

mg sin θ = μₛ (mg cos θ)

Cancel mg from both sides:

sin θ = μₛ cos θ

Divide by cos θ:

μₛ = tan θ

Step 6: Substitute θ = 15°

μₛ = tan 15°
μₛ = 0.268
μₛ ≈ 0.27
Final Answer: μₛ = 0.27

Important Formula

μₛ = tan θ

This formula is used when a block is just about to slide on an inclined plane.

Key Points for Beginners

  • Static friction prevents motion.
  • Static friction adjusts itself according to the applied force.
  • At the point of sliding, static friction becomes maximum.
  • Maximum static friction = μₛN.
  • The angle at which sliding begins is called the angle of repose.
  • For angle of repose, μₛ = tan θ.
  • The value of μₛ does not depend on the mass of the block.

One-Line Summary

When a block just begins to slide on an inclined plane, the coefficient of static friction is equal to the tangent of the angle of inclination.

μₛ = tan 15° = 0.27

Example 4.7 Maximum Acceleration of a Train Solved Step by Step

 Maximum Acceleration of a Train Using Static Friction Explained

-  Dr.Sanjaykumar Pawar 

Physics diagram showing a box on the floor of an accelerating train with forces including static friction, normal reaction, weight, and acceleration labeled for educational purposes.
A box remains stationary inside an accelerating train due to static friction acting between the box and the train floor.


 

Example 4.7 - Maximum Acceleration of a Train Example 4.7 Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary,given that the co-efficient of static friction between the box and the train’s floor is 0.15.

Example 4.7

Question:
Determine the maximum acceleration of a train in which a box lying on its floor will remain stationary, given that the coefficient of static friction between the box and the train's floor is 0.15.

Step 1: Understand the Concept

When the train accelerates, the box must also accelerate along with the train.

The force responsible for accelerating the box is the static friction between the box and the floor.

  • If friction is enough, the box remains stationary relative to the train.
  • If friction is not enough, the box starts sliding backward.

Therefore, the maximum acceleration depends on the maximum static friction available.

Step 2: Apply Newton's Second Law

According to Newton's Second Law:

F = ma

The only horizontal force acting on the box is static friction (fs).

ma = fs

Step 3: Write Maximum Static Friction

Maximum static friction is given by:

fs ≤ μsN

where:

  • μs = coefficient of static friction
  • N = normal reaction

Step 4: Find the Normal Reaction

Since the box is resting on a horizontal floor:

N = mg

Substitute N = mg into the friction formula:

fs ≤ μsmg

Step 5: Substitute into Newton's Law

From Newton's Law:

ma = fs

Using the maximum friction condition:

ma ≤ μsmg

Cancel mass (m) from both sides:

a ≤ μsg

Therefore:

amax = μsg

Step 6: Substitute the Given Values

Given:

μs = 0.15
g = 10 m/s²

Substitute into the formula:

amax = 0.15 × 10
amax = 1.5 m/s²
Final Answer:

amax = 1.5 m/s²

Quick Notes for Beginners

  • Static friction helps the box move with the train.
  • Maximum static friction = μsN.
  • For a horizontal surface, N = mg.
  • Using F = ma, we get:
ma = μsmg
a = μsg
a = 0.15 × 10 = 1.5 m/s²

Hence, the train can accelerate up to 1.5 m/s² without the box sliding.

Uniformly Accelerated Motion Class 11 Physics Notes | NEET & JEE MCQs

 - Dr.Sanjaykumar Pawar   Uniformly Accelerated Motion (1-D) Physics Notes, Formulas & NEET Questions  Uniformly Accelerated Motion (1-D...