NCERT Physics Class 11 Example 5.8 & 5.9 Solutions | Spring Compression Explained

 - Dr.Sanjaykumar Pawar 

A moving car compresses a spring, demonstrating conservation of mechanical energy and work-energy theorem.

Internal Links

• NCERT Class 11 Physics Chapter 5 Notes
• Work, Energy and Power Formulas
• Conservation of Mechanical Energy
• Work-Energy Theorem Explained
• Hooke's Law and Spring Force
• Potential Energy and Kinetic Energy
• NCERT Physics Example 5.1–5.7 Solutions
• NCERT Physics Chapter 5 Exercise Solutions
• JEE Physics Work Energy Questions
• NEET Physics Chapter-wise MCQs
• Energy Conservation Numerical Problems
• Class 11 Physics Formula Sheet

NCERT Physics Class 11

Chapter 5 - Work, Energy and Power

Example 5.8 & Example 5.9

Question 1

A car of mass 1000 kg is moving with a speed of 18 km h^-1 on a smooth horizontal road and collides with a spring of spring constant 5.25 × 10³ N m^-1. Find the maximum compression of the spring.

Answer

Given

• Mass (m) = 1000 kg
• Speed (v) = 18 km h^-1 = 5 m s^-1
• Spring constant (k) = 5.25 × 10³ N m^-1

Formula

At maximum compression,

Kinetic Energy = Spring Potential Energy

½mv² = ½kx²

Step 1 : Calculate Kinetic Energy

K = ½ × 1000 × 5²

K = 12500 J

Step 2 : Calculate Compression

12500 = ½ × 5.25 × 10³ × x²

12500 = 2625x²

x² = 12500 / 2625

x² = 4.76

x = √4.76

x ≈ 2.18 m

Maximum Compression = 2.0 m (Approx.)

---

Question 2

Using Example 5.8, if the coefficient of friction is 0.5, calculate the maximum compression of the spring.

Answer

Given

• Mass = 1000 kg
• Speed = 5 m s^-1
• Spring constant = 5.25 × 10³ N m^-1
• Coefficient of friction = 0.5
• g = 10 m s^-2

Formula

Work-Energy Theorem

ΔK = W

½mv² = ½kx² + μmgx

Step 1

12500 = 2625x² + 5000x

Step 2

2625x² + 5000x − 12500 = 0

Step 3

Using quadratic formula,

x = 1.35 m

Maximum Compression = 1.35 m

---

Practice Questions

Practice Question 1

A spring of spring constant 400 N m^-1 is compressed by 0.5 m. Calculate the elastic potential energy stored.

Formula

U = ½kx²

U = ½ × 400 × (0.5)²

U = 50 J

Answer = 50 J

Practice Question 2

A body of mass 2 kg moves with speed 10 m s^-1. Find its kinetic energy.

Formula

K = ½mv²

K = ½ × 2 × 10²

K = 100 J

Answer = 100 J

Practice Question 3

A spring stores 100 J of energy. Its spring constant is 500 N m^-1. Find the compression.

100 = ½ × 500 × x²

100 = 250x²

x² = 0.4

x = 0.63 m

Answer = 0.63 m

Practice Question 4

A 500 kg car moving at 10 m s^-1 hits a spring of spring constant 10000 N m^-1. Find the maximum compression.

K = ½mv²

K = ½ × 500 × 10²

K = 25000 J

25000 = ½ × 10000 × x²

25000 = 5000x²

x² = 5

x = 2.24 m

Answer = 2.24 m

---

Important Formulae

Formula	Expression
Kinetic Energy	K = ½mv²
Spring Potential Energy	U = ½kx²
Work-Energy Theorem	ΔK = W
Mechanical Energy	KE + PE = Constant (Without Friction)

---

Important Viva Questions

1. Why does the car stop at maximum compression?
   Because all kinetic energy is converted into spring potential energy.


2. Why is conservation of mechanical energy not applicable when friction is present?
   Because friction is a non-conservative force and converts mechanical energy into heat.


3. What is the formula of spring potential energy?
   U = ½kx²


4. What is the formula of kinetic energy?
   K = ½mv²

NEET Physics Notes

Chapter: Work, Energy and Power

Topics Covered

• Spring Force (Hooke's Law)
• Spring Potential Energy
• Kinetic Energy
• Conservation of Mechanical Energy
• Work-Energy Theorem
• Friction
• NCERT Example 5.8
• NCERT Example 5.9
• Important Formulae
• NEET MCQs

1. Spring Force (Hooke's Law)

When a spring is stretched or compressed, it tries to return to its original length. This restoring force is called Spring Force.

Formula

F = -kx

Symbol	Meaning
F	Spring Force (N)
k	Spring Constant (N/m)
x	Compression or Extension (m)

Remember

• Negative sign shows restoring force.
• Force acts opposite to displacement.
• Larger k means stronger spring.
• Smaller k means softer spring.

2. Spring Potential Energy

Energy stored inside a compressed or stretched spring.

Formula

U = ½ kx²

Symbol	Meaning
U	Potential Energy
k	Spring Constant
x	Compression

3. Kinetic Energy

Energy possessed by a moving object.

Formula

K = ½ mv²

Symbol	Meaning
m	Mass
v	Velocity

4. Conservation of Mechanical Energy

When there is no friction, the total mechanical energy remains constant.

½mv² = ½kx²

Used to calculate maximum compression of a spring.

Example 5.8

To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs and other constraints. Consider a typical simulation with a car of mass 1000 kg moving at a speed of 18.0 km/h on a smooth road and colliding with a horizontal mounted spring of spring constant 5.25 × 10³ N m⁻¹. What is the maximum compression of the spring?

Example 5.9

Consider Example 5.8 taking the coefficient of friction, μ = 0.5, and calculate the maximum compression of the spring.

NCERT Example 5.8

Question

Mass = 1000 kg
Speed = 18 km/h
Spring Constant = 5.25 × 10³ N/m
Find the maximum compression.

Solution

Step 1 18 km/h = 5 m/s

Kinetic Energy

K = ½ ×1000×5²

K = 12500 J

Using Conservation of Energy

½kx² = 12500

x = 2.0 m

Answer: Maximum Compression = 2.0 m

5. Work-Energy Theorem

The work done by all forces acting on a body equals the change in kinetic energy.

Formula

W = ΔK

6. Friction

Friction always opposes motion.

Formula

F = μN

On horizontal surface,

N = mg

Therefore,

F = μmg

NCERT Example 5.9

Question

Mass = 1000 kg
Speed = 5 m/s
k = 5.25 ×10³ N/m
μ = 0.5

Solution

Apply Work-Energy Theorem

−½mv² = −½kx² − μmgx

Rearranging,

kx² + 2μmgx − mv² = 0

x = 1.35 m

Answer: Maximum Compression = 1.35 m

Important Formula Sheet

Formula	Equation
Kinetic Energy	½mv²
Spring Force	F = -kx
Spring Potential Energy	½kx²
Friction	μmg
Work-Energy Theorem	W = ΔK
Conservation of Energy	½mv² = ½kx²

NEET Important Points

• Hooke's law is valid only within elastic limit.
• Spring force is a restoring force.
• Potential energy is always positive.
• Friction is a non-conservative force.
• Mechanical energy is conserved only without friction.
• Work done by friction is negative.
• Spring energy depends on x².
• Kinetic energy depends on velocity².

NEET Practice MCQs

Q1. The force exerted by a spring is

• A. Constant
• B. Proportional to displacement
• C. Inversely proportional
• D. Zero

Answer: B

Q2. Potential energy stored in a spring is

• A. kx
• B. kx²
• C. ½kx²
• D. ½kx

Answer: C

Q3. Which force is non-conservative?

• A. Gravity
• B. Spring
• C. Electrostatic
• D. Friction

Answer: D

Q4. Mechanical energy remains constant when

• A. Friction acts
• B. Air resistance acts
• C. Only conservative forces act
• D. External work is done

Answer: C

Quick Revision

• Hooke's Law → F = -kx
• Spring Energy → ½kx²
• Kinetic Energy → ½mv²
• Work-Energy Theorem → W = ΔK
• Without friction → Energy conserved.
• With friction → Use Work-Energy Theorem.
• Maximum compression occurs when kinetic energy becomes zero.
• Friction reduces spring compression.

---

Prepared for NEET Aspirants

Easy Notes • NCERT Based • Beginner Friendly

Example 4.7 Maximum Acceleration of a Train Solved Step by Step

 Maximum Acceleration of a Train Using Static Friction Explained

-  Dr.Sanjaykumar Pawar 

A box remains stationary inside an accelerating train due to static friction acting between the box and the train floor.

 

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 (f_s).

ma = f_s

Step 3: Write Maximum Static Friction

Maximum static friction is given by:

f_s ≤ μ_sN

where:

• μ_s = 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:

f_s ≤ μ_smg

Step 5: Substitute into Newton's Law

From Newton's Law:

ma = f_s

Using the maximum friction condition:

ma ≤ μ_smg

Cancel mass (m) from both sides:

a ≤ μ_sg

Therefore:

a_max = μ_sg

Step 6: Substitute the Given Values

Given:

μ_s = 0.15

g = 10 m/s²

Substitute into the formula:

a_max = 0.15 × 10

a_max = 1.5 m/s²

Final Answer:

a_max = 1.5 m/s²

Quick Notes for Beginners

• Static friction helps the box move with the train.
• Maximum static friction = μ_sN.
• For a horizontal surface, N = mg.
• Using F = ma, we get:

ma = μ_smg

a = μ_sg

a = 0.15 × 10 = 1.5 m/s²

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