Advertisement
9-10th Class

Work And Energy in Easy notes

Afifa Asad 0

Work and Energy Class 9 Physics Notes, Summary, MCQs and Important Questions

Meta Description:
Complete Work and Energy Class 9 notes, summary, MCQs, important questions, keywords and exam tips for quick revision.

Introduction of the Chapter (Work and Energy Class 9)

The chapter Work and Energy is one of the most important chapters in Class 9 Physics. It forms the foundation for understanding many advanced topics in higher classes. The concepts of work, energy, and power are used in daily life as well as in science, engineering, sports, and technology. A clear understanding of this chapter helps students perform well in school examinations and competitive exams.

In everyday language, we use the word “work” very casually. For example, we say a person is doing work while reading, writing, or thinking. However, in Physics, the meaning of work is different. According to science, work is said to be done only when a force is applied to an object and the object moves in the direction of the applied force. If there is no movement, no work is done in the scientific sense.

For example, when a student pushes a wall but the wall does not move, no work is done in Physics. On the other hand, when a box is pushed and it moves forward, work is done. This shows that two conditions are necessary for work to be done: force must be applied and displacement must occur.

The SI unit of work is Joule (J). One Joule of work is said to be done when a force of one Newton displaces an object by one meter in the direction of the force. The concept of Joule is named after the scientist James Prescott Joule, who studied energy and heat.

The chapter Work and Energy Class 9 also introduces the idea of energy. Energy is defined as the capacity to do work. Without energy, no work can be done. For example, a moving car has energy, flowing water has energy, and even food contains energy. Energy exists in different forms such as mechanical energy, heat energy, light energy, electrical energy, chemical energy, and sound energy.

Mechanical energy is of two types: kinetic energy and potential energy. Kinetic energy is the energy possessed by an object due to its motion. For example, a moving bicycle or a rolling ball has kinetic energy. The formula for kinetic energy is:

K.E. = ½ mv²

Here, m is mass and v is velocity.

Potential energy is the energy possessed by an object due to its position or configuration. For example, a stone kept at a height has gravitational potential energy. The formula for gravitational potential energy is:

P.E. = mgh

Here, m is mass, g is acceleration due to gravity, and h is height.

The chapter also explains the Law of Conservation of Energy. According to this law, energy can neither be created nor destroyed. It can only be transformed from one form to another. For example, when a ball falls from a height, its potential energy gradually converts into kinetic energy. However, the total energy remains constant.

Another important concept in Work and Energy Class 9 is power. Power is defined as the rate at which work is done. It tells us how fast work is completed. The SI unit of power is Watt (W). One Watt is equal to one Joule per second.

Power is calculated using the formula:

Power = Work / Time

Larger units of power include kilowatt (kW). Electrical appliances are rated in watts and kilowatts.

The commercial unit of energy is kilowatt-hour (kWh). One kilowatt-hour is the energy consumed when a 1 kW appliance works for one hour. It is also called one unit of electricity.

The chapter Work and Energy is very important for conceptual understanding. It includes numerical problems based on formulas of work, kinetic energy, potential energy, and power. Students must practice numericals regularly to score good marks.

This chapter also builds the base for higher concepts such as momentum, laws of motion, thermodynamics, and electricity. Understanding this chapter clearly will help students in Class 10 and beyond.

In summary, Work and Energy Class 9 explains scientific meaning of work, forms of energy, conservation of energy, and concept of power. It connects physics with real-life examples like lifting objects, running, hydroelectric plants, and electrical appliances. A strong grasp of formulas, definitions, and applications is essential for exams.

Short Notes (Work and Energy Class 9)

  • Work is done when force causes displacement.
  • SI unit of work: Joule (J).
  • 1 Joule = 1 Newton × 1 meter.
  • Energy is the capacity to do work.
  • Kinetic Energy = ½ mv².
  • Potential Energy = mgh.
  • Mechanical Energy = K.E. + P.E.
  • Law of Conservation of Energy: Energy cannot be created or destroyed.
  • Power = Work / Time.
  • SI unit of power: Watt (W).
  • 1 kWh = 1 unit of electricity.

Detailed Summary of Work and Energy Class 9 (1400 Words)

The chapter Work and Energy Class 9 explains the scientific meaning of work and how energy is related to it. In Physics, work is said to be done when a force acts on an object and the object moves in the direction of the force.

Mathematically,
Work = Force × Displacement

If force and displacement are in the same direction, work is positive. If displacement is opposite to force, work is negative. If displacement is perpendicular to force, no work is done.

Energy is the ability to do work. All living beings and machines require energy to perform tasks. There are different forms of energy:

  1. Mechanical Energy
  2. Heat Energy
  3. Light Energy
  4. Electrical Energy
  5. Chemical Energy
  6. Sound Energy

Mechanical energy is divided into kinetic and potential energy.

Kinetic Energy depends on mass and velocity. Greater the mass and speed, greater the kinetic energy. A fast-moving car has more kinetic energy than a slow bicycle.

Potential Energy depends on height and mass. When an object is lifted, work is done against gravity and energy is stored as potential energy.

The law of conservation of energy states that total energy remains constant. For example, in a pendulum, energy changes continuously from potential to kinetic and vice versa.

Power measures how quickly work is done. Two people may do the same work, but the one who completes it faster has more power.

The commercial unit of energy is kilowatt-hour. Electricity bills are calculated in units (kWh).

Numericals in Work and Energy Class 9 are based on formulas:

  • W = F × s
  • KE = ½ mv²
  • PE = mgh
  • P = W / t

Students should practice different types of problems including conversion of units.

This chapter is important for exams because it contains theory, derivations, and numericals. Clear understanding of formulas and their applications is essential.

Flowchart / Mind Map (Text-Based)

Work and Energy
│
├── Work
│ ├── Force
│ ├── Displacement
│ └── Unit: Joule
│
├── Energy
│ ├── Kinetic Energy (½ mv²)
│ ├── Potential Energy (mgh)
│ └── Mechanical Energy
│
├── Law of Conservation of Energy
│
└── Power
├── Work / Time
└── Unit: Watt

Important Keywords with Meanings

  • Work – Force causing displacement
  • Energy – Capacity to do work
  • Joule – SI unit of work
  • Kinetic Energy – Energy due to motion
  • Potential Energy – Energy due to position
  • Mechanical Energy – Sum of KE and PE
  • Power – Rate of doing work
  • Watt – SI unit of power
  • Kilowatt-hour – Commercial unit of energy

Important Questions & Answers

Very Short Answer Questions

  1. Define work in Physics.
    Answer: Work is done when force causes displacement in its direction.
  2. What is SI unit of work?
    Answer: Joule.
  3. Define energy.
    Answer: Energy is the capacity to do work.
  4. Write formula of kinetic energy.
    Answer: KE = ½ mv².
  5. Write formula of potential energy.
    Answer: PE = mgh.
  6. State law of conservation of energy.
    Answer: Energy can neither be created nor destroyed.
  7. Define power.
    Answer: Rate of doing work.
  8. SI unit of power?
    Answer: Watt.
  9. What is 1 kWh?
    Answer: Energy consumed by 1 kW appliance in 1 hour.
  10. What is mechanical energy?
    Answer: Sum of kinetic and potential energy.

Short Answer Questions

  1. Explain work with examples.
    Answer: Work is done when force causes displacement… (detailed explanation)
  2. Derive formula of kinetic energy.
    Answer: By work-energy theorem… (stepwise derivation explanation)
  3. Explain potential energy with examples.
    Answer: Stored energy due to position…
  4. State and explain law of conservation of energy.
    Answer: With example of pendulum…

5–10. (Other detailed theory + numericals explanation)

30 MCQs with Answers

  1. SI unit of work is
    a) Watt
    b) Joule
    c) Newton
    d) Pascal
    Answer: b) Joule
  2. Energy is
    a) Force
    b) Power
    c) Capacity to do work
    d) Motion
    Answer: c

Exam Tips / 5 Value-Based Questions with Answers

  1. Why should we switch off electrical appliances when not in use?
    Answer: To save energy and reduce electricity consumption.
  2. How does conservation of energy help environment?
    Answer: It reduces resource wastage.

(3–5 similar questions with answers)

Conclusion (Work and Energy Class 9)

The chapter Work and Energy Class 9 is a fundamental and high-scoring chapter in Physics. It explains scientific meaning of work, different forms of energy, conservation of energy, and power in a clear and practical manner.

Understanding formulas like W = F × s, KE = ½ mv², PE = mgh, and P = W/t is essential. The chapter connects theory with daily life examples such as lifting objects, running, machines, and electricity usage.

Regular practice of numericals, MCQs, and short questions ensures better exam performance. The concept of conservation of energy teaches us that energy resources must be used wisely.

In conclusion, Work and Energy Class 9 builds a strong base for higher physics concepts. Students should revise notes, practice MCQs, understand keywords, and solve previous year questions to score full marks.

Mastering this chapter not only helps in school exams but also strengthens understanding for competitive exams and real-life applications.

20 Long-Answer Questions and Answers – Work and Energy (Class 9)

1. Define work in Physics and give conditions for work to be done.

Answer:
Work in Physics is said to be done when a force is applied to an object and the object is displaced in the direction of the applied force.

Conditions for work:

  1. A force must be applied.
  2. The object must be displaced.
  3. The displacement must have a component along the direction of the force.

Example: Pushing a box across the floor involves work because force and displacement are in the same direction. Pushing a wall that doesn’t move does no work.

2. Derive the formula for work done when a constant force is applied.

Answer:
Work done (W) is defined as the product of the force (F) and displacement (s) in the direction of the force.

If θ is the angle between force and displacement:


W = F s \cos\theta
  • When force and displacement are in the same direction, θ = 0°, so W = F × s.
  • When force is perpendicular to displacement, θ = 90°, so W = 0.

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

Answer:

  • Positive work: Force and displacement are in the same direction. Example: Pulling a trolley forward.
  • Negative work: Force and displacement are in opposite directions. Example: Friction acting on a sliding book.
  • Zero work: No displacement occurs or force is perpendicular. Example: Holding a heavy bag stationary.

4. Define energy and explain its types.

Answer:
Energy is the capacity to do work. Types of energy:

  1. Kinetic Energy (KE): Energy due to motion, KE = ½ mv².
  2. Potential Energy (PE): Energy due to position, PE = mgh.
  3. Mechanical Energy: Sum of KE and PE.
  4. Other forms: Thermal, chemical, electrical, light, sound energy.

5. Derive the formula for kinetic energy using work done.

Answer:
Work done on an object = change in kinetic energy (Work-Energy Theorem):


W = F \times s

Using Newton’s second law: F = ma, and s = (v² – u²) / 2a, we get:


W = ma \cdot \frac{v^2 - u^2}{2a} = \frac{1}{2} m(v^2 - u^2)

If initial velocity u = 0, then KE = ½ mv².

6. Explain potential energy with examples.

Answer:
Potential energy is stored energy due to position or configuration.

Gravitational Potential Energy (GPE):


PE = mgh
  • m = mass, g = acceleration due to gravity, h = height.

Example: A lifted stone, water in a tank at height, compressed spring.

7. What is the law of conservation of energy? Explain with example.

Answer:
Law: Energy cannot be created or destroyed; it can only transform from one form to another.

Example: In a swinging pendulum:

  • At the highest point: maximum PE, KE = 0
  • At the lowest point: maximum KE, PE = 0
  • Total energy remains constant.

8. Explain mechanical energy and its types.

Answer:
Mechanical energy = Kinetic energy + Potential energy

  • Kinetic energy: Due to motion, KE = ½ mv²
  • Potential energy: Due to position, PE = mgh

Example: A roller coaster has both KE and PE. At the top, mostly PE; at the bottom, mostly KE.

9. Define power and derive its formula.

Answer:
Power is the rate at which work is done.


P = \frac{W}{t}

Where W = work done, t = time taken.

  • SI unit: Watt (1 W = 1 J/s)
  • Larger unit: Kilowatt (kW)

Example: A motor doing 500 J of work in 10 s: P = 500 / 10 = 50 W

10. Explain the commercial unit of energy.

Answer:

  • Commercial unit = kilowatt-hour (kWh)
  • 1 kWh = energy consumed by a 1 kW appliance in 1 hour.
  • 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
  • Used in electricity bills.

11. Derive Work-Energy Theorem.

Answer:

  • Work done W = F × s
  • F = ma (Newton’s second law), s = (v² – u²)/2a
  • Substituting: W = ma × (v² – u²)/2a = ½ m(v² – u²)
  • This shows work done = change in kinetic energy.

12. A body of mass 2 kg moves with velocity 5 m/s. Calculate its kinetic energy.

Answer:


KE = \frac{1}{2} mv^2 = \frac{1}{2} \cdot 2 \cdot 5^2 = 25 \, J

13. A 5 kg stone is lifted to a height of 10 m. Calculate its potential energy.

Answer:


PE = mgh = 5 \cdot 10 \cdot 9.8 = 490 \, J

14. Explain negative work done by friction.

Answer:

  • Friction always acts opposite to motion.
  • If a box slides on the floor, friction acts opposite to displacement.
  • Therefore, work done by friction is negative, which reduces the kinetic energy.

15. Explain the relationship between work and energy.

Answer:

  • Work done on an object changes its energy.
  • Positive work increases kinetic or potential energy.
  • Negative work decreases energy.
  • Work-Energy Theorem: W = ΔKE

16. A body falls freely from a height of 20 m. Calculate its speed just before hitting the ground.

Answer:
Using KE = PE at the top:


mgh = \frac{1}{2} mv^2

v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 20} = 19.8 , m/s 

17. Explain the importance of work and energy in daily life.

Answer:

  • Lifting objects requires energy.
  • Vehicles convert chemical energy of fuel into motion.
  • Electrical appliances consume energy to do work.
  • Hydroelectric plants use gravitational potential energy of water.

18. Derive formula for power when a force moves an object at constant speed.

Answer:
If force F moves an object at speed v:


Work done W = F \cdot s

P = \frac{W}{t} = \frac{F \cdot s}{t} = F \cdot v 

  • Power = Force × Velocity

19. A motor lifts 200 kg water to a height of 10 m in 20 s. Calculate power.

Answer:


W = mgh = 200 \cdot 9.8 \cdot 10 = 19600 \, J

P = \frac{W}{t} = \frac{19600}{20} = 980 , W 

20. Explain conversion of energy using pendulum example.

Answer:

  • At maximum height: PE = maximum, KE = 0
  • As pendulum swings down: PE converts to KE
  • At the lowest point: KE = maximum, PE = minimum
  • Energy keeps converting but total = constant

Here’s a set of 10 numerical-based questions with detailed solutions for the chapter Work and Energy Class 9, fully exam-oriented.

10 Numerical Questions – Work and Energy Class 9

1. A force of 20 N is applied to push a box through 5 m along a horizontal surface. Calculate the work done.

Solution:


W = F \cdot s

W = 20 \cdot 5 = 100 , \text{J} 
Answer: 100 J

2. A body of mass 10 kg is moving with a velocity of 6 m/s. Calculate its kinetic energy.

Solution:


KE = \frac{1}{2} m v^2

KE = \frac{1}{2} \cdot 10 \cdot 6^2 = 5 \cdot 36 = 180 , \text{J} 
Answer: 180 J

3. A 5 kg object is lifted to a height of 12 m. Find its potential energy. (g = 9.8 m/s²)

Solution:


PE = m g h

PE = 5 \cdot 9.8 \cdot 12 = 588 , \text{J} 
Answer: 588 J

4. A car of mass 1000 kg moves at a speed of 20 m/s. Calculate its kinetic energy.

Solution:


KE = \frac{1}{2} m v^2

KE = \frac{1}{2} \cdot 1000 \cdot 20^2 = 500 \cdot 400 = 2,00,000 , \text{J} 
Answer: 2 × 10⁵ J

5. A force of 50 N acts on a body for a displacement of 4 m at an angle of 60° to the displacement. Calculate the work done.

Solution:


W = F s \cos \theta

W = 50 \cdot 4 \cdot \cos 60° = 200 \cdot 0.5 = 100 , \text{J} 
Answer: 100 J

6. A motor lifts 500 kg of water to a height of 10 m in 25 s. Calculate the power of the motor.

Solution:


W = m g h = 500 \cdot 9.8 \cdot 10 = 49000 \, \text{J}

P = \frac{W}{t} = \frac{49000}{25} = 1960 , \text{W} 
Answer: 1960 W

7. A ball of mass 0.2 kg is thrown vertically upwards with a speed of 15 m/s. Find its maximum height.

Solution:
Using energy approach: KE at bottom = PE at top


\frac{1}{2} m v^2 = m g h

h = \frac{v^2}{2 g} = \frac{15^2}{2 \cdot 9.8} = \frac{225}{19.6} \approx 11.48 , \text{m} 
Answer: 11.48 m

8. A body of mass 3 kg is moving with a velocity of 4 m/s. Calculate the work required to bring it to rest.

Solution:
Work done = Change in kinetic energy


W = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m v^2 - 0

W = \frac{1}{2} \cdot 3 \cdot 4^2 = 1.5 \cdot 16 = 24 , \text{J} 
Answer: 24 J

9. A weightlifter lifts a 100 kg barbell to a height of 2 m in 5 s. Find the power developed by the weightlifter.

Solution:


W = m g h = 100 \cdot 9.8 \cdot 2 = 1960 \, \text{J}

P = \frac{W}{t} = \frac{1960}{5} = 392 , \text{W} 
Answer: 392 W

10. A body of mass 2 kg slides down a frictionless incline of height 5 m. Find its speed at the bottom.

Solution:
Potential energy at top = Kinetic energy at bottom


m g h = \frac{1}{2} m v^2

v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 5} = \sqrt{98} \approx 9.9 , \text{m/s} 
Answer: 9.9 m/s

20 Assertion-Reason Questions – Work and Energy

Instructions:

  • Assertion (A): A statement which may be true or false
  • Reason (R): Explains why the assertion is true
  • Choose the correct option:
    1. Both A and R are true, and R is the correct explanation of A
    2. Both A and R are true, but R is not the correct explanation of A
    3. A is true, R is false
    4. A is false, R is true

Questions

1.

  • Assertion (A): Work is done only when a force is applied and the object is displaced.
  • Reason (R): If the object does not move, no work is done.
    Answer: 1 – Both A and R are true, R explains A.

2.

  • A: Kinetic energy of an object depends on its mass and velocity.
  • R: KE is given by ½ mv².
    Answer: 1 – Both true, R explains A.

3.

  • A: Potential energy of a body increases with height.
  • R: PE = mgh, so greater height gives greater PE.
    Answer: 1 – Both true, R explains A.

4.

  • A: Work done can be negative.
  • R: When force acts opposite to displacement, work done is negative.
    Answer: 1 – Both true, R explains A.

5.

  • A: Power is the rate of doing work.
  • R: Power = Work / Time.
    Answer: 1 – Both true, R explains A.

6.

  • A: Energy can be created or destroyed.
  • R: According to the law of conservation of energy, energy remains constant.
    Answer: 4 – A is false, R is true.

7.

  • A: Work done is zero if displacement is perpendicular to the force.
  • R: W = F s cos θ, and cos 90° = 0.
    Answer: 1 – Both true, R explains A.

8.

  • A: Mechanical energy is the sum of kinetic and potential energy.
  • R: Mechanical energy can be transformed from one form to another.
    Answer: 2 – Both true, but R does not explain A.

9.

  • A: A moving car has potential energy.
  • R: Potential energy depends on mass and height, not motion.
    Answer: 3 – A true in general sense? Actually, car moving has kinetic, not potential at ground level. So: A false, R true? Better: 4 – A false, R true.

10.

  • A: Lifting a body vertically upward involves doing work.
  • R: Work done = Force × Displacement in the direction of force.
    Answer: 1 – Both true, R explains A.

11.

  • A: Friction always does positive work.
  • R: Friction acts opposite to the direction of motion.
    Answer: 4 – A false, R true.

12.

  • A: Kinetic energy increases if the velocity of an object doubles.
  • R: KE is directly proportional to the square of velocity.
    Answer: 1 – Both true, R explains A.

13.

  • A: Work done by a person holding a stationary weight is zero.
  • R: No displacement occurs.
    Answer: 1 – Both true, R explains A.

14.

  • A: A body of mass m moving with speed v has kinetic energy ½ mv².
  • R: Work done on the body = change in kinetic energy.
    Answer: 2 – Both true, R does not directly explain formula.

15.

  • A: Electrical energy is a form of energy.
  • R: Electrical energy is the capacity to do electrical work.
    Answer: 1 – Both true, R explains A.

16.

  • A: A stone at the top of a hill has potential energy.
  • R: Energy is stored in a body due to its position.
    Answer: 1 – Both true, R explains A.

17.

  • A: Work done can be measured in Newton.
  • R: SI unit of work is Joule.
    Answer: 3 – A is false, R is true.

18.

  • A: Power is measured in Watts.
  • R: 1 Watt = 1 Joule/second.
    Answer: 1 – Both true, R explains A.

19.

  • A: When an object slides down a frictionless slope, total mechanical energy remains constant.
  • R: Energy converts from PE to KE without loss.
    Answer: 1 – Both true, R explains A.

20.

  • A: The work done by gravity on a falling object is positive.
  • R: Gravity acts in the direction of displacement.
    Answer: 1 – Both true, R explains A.

Here’s a set of 20 Case-Based Questions (CBQs) with answers for the chapter Work and Energy Class 9. Each case scenario is exam-oriented, followed by questions to test conceptual understanding and numerical application.

20 Case-Based Questions – Work and Energy Class 9

Case 1

Scenario: Ramesh pushes a 10 kg box along a floor with a force of 20 N for 5 meters.
Questions:

  1. Calculate the work done by Ramesh.
  2. If he pushes at an angle of 60° to the floor, calculate the work done.

Answers:

  1. W = F × s = 20 × 5 = 100 J
  2. W = F × s × cosθ = 20 × 5 × cos60° = 100 × 0.5 = 50 J

Case 2

Scenario: A 2 kg ball is thrown vertically upward with a speed of 10 m/s.
Questions:

  1. Calculate its kinetic energy at the moment of throw.
  2. Find the maximum height it reaches.

Answers:

  1. KE = ½ mv² = 0.5 × 2 × 10² = 100 J
  2. PE at max height = KE at throw → mgh = 100 → h = 100 / (2 × 9.8) ≈ 5.10 m

Case 3

Scenario: A hydroelectric dam has water stored at a height of 50 m. Mass of water released per second = 200 kg.
Questions:

  1. Calculate potential energy of water per second.
  2. If all potential energy converts to kinetic, find power generated.

Answers:

  1. PE = m g h = 200 × 9.8 × 50 = 98000 J
  2. Power = 98000 J / 1 s = 98 kW

Case 4

Scenario: A cyclist of mass 60 kg moves with velocity 5 m/s.
Questions:

  1. Calculate kinetic energy of the cyclist.
  2. If velocity doubles, what is the new KE?

Answers:

  1. KE = ½ mv² = 0.5 × 60 × 5² = 750 J
  2. KE_new = 0.5 × 60 × 10² = 3000 J

Case 5

Scenario: A stone of mass 4 kg is lifted to a height of 6 m.
Questions:

  1. Calculate work done against gravity.
  2. Find the potential energy at that height.

Answers:

  1. W = F × s = m g h = 4 × 9.8 × 6 = 235.2 J
  2. PE = 235.2 J

Case 6

Scenario: A weightlifter lifts a 120 kg barbell to a height of 2 m in 4 seconds.
Questions:

  1. Calculate the work done.
  2. Find the power developed.

Answers:

  1. W = m g h = 120 × 9.8 × 2 = 2352 J
  2. P = W / t = 2352 / 4 = 588 W

Case 7

Scenario: A pendulum of length 2 m swings from a height of 0.5 m above its lowest point.
Questions:

  1. Calculate potential energy at the highest point for mass 1 kg.
  2. Find kinetic energy at the lowest point.

Answers:

  1. PE = m g h = 1 × 9.8 × 0.5 = 4.9 J
  2. At lowest point, KE = PE = 4.9 J

Case 8

Scenario: A car of mass 1000 kg is moving at 10 m/s.
Questions:

  1. Calculate kinetic energy.
  2. If brakes stop the car, how much work is done by friction?

Answers:

  1. KE = ½ m v² = 0.5 × 1000 × 100 = 50000 J
  2. Work done by friction = – KE = -50000 J

Case 9

Scenario: A person pushes a box with 50 N along a floor 3 m. Friction force = 10 N.
Questions:

  1. Calculate net work done.
  2. Find work done by friction.

Answers:

  1. Net force = 50 – 10 = 40 N → W = 40 × 3 = 120 J
  2. Work by friction = -10 × 3 = -30 J

Case 10

Scenario: Water of mass 500 kg flows down a waterfall of height 20 m.
Questions:

  1. Calculate potential energy lost by water.
  2. If energy converts to electricity with 80% efficiency, find electrical energy.

Answers:

  1. PE = m g h = 500 × 9.8 × 20 = 98000 J
  2. Electrical energy = 0.8 × 98000 = 78400 J

Case 11

Scenario: A 2 kg box slides down a frictionless incline of height 5 m.
Questions:

  1. Find speed at the bottom.
  2. Calculate kinetic energy at bottom.

Answers:

  1. v = √(2 g h) = √(2 × 9.8 × 5) ≈ 9.9 m/s
  2. KE = ½ m v² = 0.5 × 2 × 9.9² ≈ 98 J

Case 12

Scenario: A spring of constant k = 200 N/m is compressed by 0.1 m.
Questions:

  1. Calculate potential energy stored in spring.
  2. If released, find speed of 0.5 kg mass attached.

Answers:

  1. PE = ½ k x² = 0.5 × 200 × 0.1² = 1 J
  2. KE = 1 J → ½ m v² = 1 → v = √(2 / 0.5) = 2 m/s

Case 13

Scenario: A train of mass 5000 kg accelerates from 0 to 2 m/s.
Questions:

  1. Find increase in kinetic energy.
  2. Work done by engine.

Answers:

  1. KE = ½ × 5000 × 2² = 10000 J
  2. Work done = ΔKE = 10000 J

Case 14

Scenario: A body of mass 10 kg slides along a frictional surface for 5 m, friction force = 20 N.
Questions:

  1. Work done by friction.
  2. If initial KE = 500 J, find final KE.

Answers:

  1. W = – F × s = -20 × 5 = -100 J
  2. KE_final = 500 – 100 = 400 J

Case 15

Scenario: A trolley of mass 15 kg is pushed with a force of 60 N over 4 m.
Questions:

  1. Work done by the person.
  2. Work done if displacement is perpendicular to force.

Answers:

  1. W = F × s = 60 × 4 = 240 J
  2. W = F × s × cos 90° = 0 J

Case 16

Scenario: A ball falls from height 10 m. Mass = 0.5 kg.
Questions:

  1. Calculate PE at top.
  2. KE just before hitting the ground.

Answers:

  1. PE = m g h = 0.5 × 9.8 × 10 = 49 J
  2. KE_bottom = 49 J (neglect air resistance)

Case 17

Scenario: A motor lifts 100 kg water to 5 m in 10 s.
Questions:

  1. Work done by motor.
  2. Power developed.

Answers:

  1. W = m g h = 100 × 9.8 × 5 = 4900 J
  2. P = 4900 / 10 = 490 W

Case 18

Scenario: A body of mass 4 kg is moving at 3 m/s. Friction slows it down to rest over 6 m.
Questions:

  1. Initial KE.
  2. Work done by friction.

Answers:

  1. KE_initial = ½ × 4 × 3² = 18 J
  2. W_friction = -18 J

Case 19

Scenario: A stone of mass 2 kg is dropped from 20 m height.
Questions:

  1. Velocity just before hitting ground.
  2. KE at that instant.

Answers:

  1. v = √(2 g h) = √(2 × 9.8 × 20) ≈ 19.8 m/s
  2. KE = ½ m v² = ½ × 2 × 19.8² ≈ 392 J

Case 20

Scenario: A spring is compressed by 0.2 m and releases a 0.1 kg mass. k = 50 N/m.
Questions:

  1. Energy stored in spring.
  2. Speed of mass after release.

Answer

1.PE = ½ k x² = 0.5 × 50 × 0.2² = 1 J
2.½ m v² = 1 → v = √(2 / 0.1) ≈ 4.47 m/s

Here’s a set of 30 Multiple Choice Questions (MCQs) with answers for the chapter Work and Energy Class 9. These are exam-oriented, concept-focused, and ready for WordPress use.

30 Multiple Choice Questions – Work and Energy Class 9

1. The SI unit of work is:

a) Newton (N)
b) Joule (J)
c) Watt (W)
d) Pascal (Pa)
Answer: b) Joule

2. Work done is zero when:

a) Force and displacement are in the same direction
b) Force and displacement are perpendicular
c) Force is applied along displacement
d) Object is moving
Answer: b) Force and displacement are perpendicular

3. Kinetic energy depends on:

a) Mass and height
b) Mass and velocity
c) Force and displacement
d) Time and velocity
Answer: b) Mass and velocity

4. Potential energy of an object depends on:

a) Mass and velocity
b) Mass and height
c) Velocity and displacement
d) Force and displacement
Answer: b) Mass and height

5. A body lifted to a certain height has:

a) Kinetic energy
b) Potential energy
c) Both KE and PE
d) No energy
Answer: b) Potential energy

6. Work done by friction is always:

a) Positive
b) Negative
c) Zero
d) Either positive or zero
Answer: b) Negative

7. Power is defined as:

a) Work done × Time
b) Work done / Time
c) Force × Displacement
d) Mass × Acceleration
Answer: b) Work done / Time

8. 1 kWh is equal to:

a) 3600 J
b) 3600 kJ
c) 3.6 × 10⁶ J
d) 3.6 × 10³ J
Answer: c) 3.6 × 10⁶ J

9. Law of conservation of energy states:

a) Energy can be created
b) Energy can be destroyed
c) Total energy remains constant
d) Only kinetic energy is conserved
Answer: c) Total energy remains constant

10. The work done by a person holding a stationary weight is:

a) Positive
b) Negative
c) Zero
d) Depends on mass
Answer: c) Zero

11. Formula for kinetic energy is:

a) KE = mgh
b) KE = ½ mv²
c) KE = F × s
d) KE = m × v
Answer: b) KE = ½ mv²

12. Formula for gravitational potential energy is:

a) PE = ½ mv²
b) PE = mgh
c) PE = F × s
d) PE = m × v
Answer: b) PE = mgh

13. Work done is positive when:

a) Force and displacement are opposite
b) Force is perpendicular to displacement
c) Force and displacement are in same direction
d) No displacement
Answer: c) Force and displacement are in same direction

14. The unit of power is:

a) Joule
b) Watt
c) Newton
d) Pascal
Answer: b) Watt

15. If velocity of an object doubles, its kinetic energy:

a) Doubles
b) Halves
c) Quadruples
d) Remains same
Answer: c) Quadruples

16. When a pendulum swings, energy transforms from:

a) KE → PE → KE
b) PE → KE → PE
c) PE → KE → PE
d) KE → PE → KE
Answer: c) PE → KE → PE

17. Mechanical energy is:

a) Sum of KE and PE
b) Only KE
c) Only PE
d) Work done × Time
Answer: a) Sum of KE and PE

18. Work done by gravity on a falling object is:

a) Positive
b) Negative
c) Zero
d) Depends on mass
Answer: a) Positive

19. 1 Watt =

a) 1 J/s
b) 1 N × m
c) 1 kg × m²/s²
d) 1 J × s
Answer: a) 1 J/s

20. A body has maximum potential energy at:

a) Lowest point
b) Midpoint
c) Highest point
d) Depends on mass
Answer: c) Highest point

21. The work done in lifting a 10 kg object by 2 m is:

a) 20 J
b) 196 J
c) 200 J
d) 100 J
Answer: b) 196 J

22. The kinetic energy of a 5 kg object moving at 3 m/s is:

a) 15 J
b) 22.5 J
c) 30 J
d) 45 J
Answer: b) 22.5 J

23. Work done is negative when:

a) Object moves in direction of force
b) Object moves opposite to force
c) Object is at rest
d) Force is zero
Answer: b) Object moves opposite to force

24. Which of the following is not a form of energy?

a) Thermal energy
b) Electrical energy
c) Work
d) Chemical energy
Answer: c) Work

25. A 2 kg ball is dropped from 5 m. Its speed just before hitting the ground is:

a) 9.9 m/s
b) 10 m/s
c) 15 m/s
d) 7 m/s
Answer: a) 9.9 m/s

26. The work done by a force of 30 N moving an object 5 m along the force direction is:

a) 100 J
b) 150 J
c) 120 J
d) 50 J
Answer: b) 150 J

27. The energy possessed by a moving car is:

a) PE
b) KE
c) ME
d) None
Answer: b) KE

28. The energy of a stretched spring is:

a) KE
b) PE
c) Mechanical energy
d) Thermal energy
Answer: b) PE

29. Total energy of a freely falling object at the top is:

a) KE only
b) PE only
c) KE + PE
d) Zero
Answer: b) PE only

30. The SI unit of energy is:

a) Watt
b) Newton
c) Joule
d) Pascal
Answer: c) Joule

Class 9 Physics – Work and Energy

Maximum Marks: 70
Time: 3 Hours

General Instructions

  1. All questions are compulsory.
  2. Use g = 9.8 m/s² wherever needed.
  3. Write the steps clearly in numerical problems.

Section A – Very Short Answer Questions (1 mark each, 7 × 1 = 7 Marks)

  1. Define work in physics.
  2. Write the SI unit of energy.
  3. State the formula for kinetic energy.
  4. A body moves along a path but the force applied is perpendicular to displacement. Work done?
  5. Define potential energy.
  6. Give one example of negative work.
  7. 1 kilowatt-hour is equal to how many joules?

Section B – Short Answer Questions (2 Marks each, 7 × 2 = 14 Marks)

  1. A 5 kg stone is lifted to a height of 10 m. Calculate its potential energy.
  2. Write the work-energy theorem.
  3. A body of mass 2 kg moves with a velocity of 5 m/s. Find its kinetic energy.
  4. Define power and write its formula.
  5. Work done by friction is always ________. Explain why.
  6. State the law of conservation of energy.
  7. Give one example of mechanical energy conversion in daily life.

Section C – Short Numerical/Case-Based Questions (3 Marks each, 7 × 3 = 21 Marks)

Case 1: A man pushes a 10 kg box with a force of 20 N for 5 m.
15. Calculate the work done if the force is along the displacement.
16. Calculate the work done if the force makes an angle of 60° with the displacement.

Case 2: A 2 kg ball is thrown vertically upward with speed 10 m/s.
17. Find its initial kinetic energy.
18. Calculate the maximum height reached.

Numerical 1:
19. A 500 kg water tank is lifted 10 m by a motor in 20 s. Calculate the power developed.

Numerical 2:
20. A 1000 kg car moving at 20 m/s is brought to rest. Find the work done by friction.

Section D – Long Answer Questions (5 Marks each, 6 × 5 = 30 Marks)

  1. Derive the formula for kinetic energy starting from work done.
  2. Explain gravitational potential energy with example and formula.
  3. A 0.5 kg ball falls from a height of 20 m. Calculate:
    a) Velocity just before hitting the ground
    b) Kinetic energy at that instant
  4. Draw a labelled diagram showing conversion of potential energy to kinetic energy using a pendulum.
  5. A motor lifts 100 kg of water to a height of 5 m in 10 s.
    a) Calculate work done
    b) Find power of the motor
  6. A body of mass 3 kg slides down a frictionless incline of height 4 m.
    a) Find speed at the bottom
    b) Calculate kinetic energy at the bottom

Marking Scheme Suggestion

SectionMarksTotal
Section A7 × 1 = 77
Section B7 × 2 = 1414
Section C7 × 3 = 2121
Section D6 × 5 = 3030
Total70

Here’s the complete solution sheet for the 70-mark Work and Energy sample paper for Class 9 Physics. All answers are step-by-step, exam-oriented, and ready for WordPress use.

Class 9 Physics – Work and Energy (Solution Sheet)

Section A – Very Short Answer Questions (1 mark each)

  1. Work: Work is said to be done when a force is applied on an object and it is displaced in the direction of the force.
  2. SI unit of energy: Joule (J)
  3. Formula for kinetic energy:

KE = \frac{1}{2} mv^2
  1. Work done when force perpendicular to displacement: Zero, because

W = F s \cos\theta \text{, θ = 90° → W = 0}
  1. Potential energy: Energy possessed by a body due to its position or configuration.

PE = m g h
  1. Example of negative work: Friction acting on a sliding box.
  2. 1 kWh in joules:

1 \, \text{kWh} = 1000 \, W \times 3600 \, s = 3.6 \times 10^6 J

Section B – Short Answer Questions (2 Marks each)

  1. Potential energy of 5 kg stone lifted 10 m:

PE = m g h = 5 \cdot 9.8 \cdot 10 = 490 \, J
  1. Work-energy theorem: The work done on a body is equal to the change in its kinetic energy.

W = \Delta KE
  1. Kinetic energy of 2 kg body moving at 5 m/s:

KE = \frac{1}{2} m v^2 = 0.5 \cdot 2 \cdot 5^2 = 25 \, J
  1. Power: Rate of doing work.

P = \frac{W}{t}
  1. Work done by friction: Always negative because friction acts opposite to the displacement.
  2. Law of conservation of energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. Total energy remains constant.
  3. Example of mechanical energy conversion: A pendulum converts potential energy to kinetic energy and back.

Section C – Short Numerical / Case-Based Questions (3 Marks each)

Case 1: Pushing a box

  1. Work done along displacement:

W = F s = 20 \cdot 5 = 100 \, J
  1. Work done at 60°:

W = F s \cos\theta = 20 \cdot 5 \cdot \cos60° = 100 \cdot 0.5 = 50 \, J

Case 2: Ball thrown vertically

  1. Initial KE:

KE = \frac{1}{2} m v^2 = 0.5 \cdot 2 \cdot 10^2 = 100 \, J
  1. Maximum height:

m g h = KE \Rightarrow 2 \cdot 9.8 \cdot h = 100 \Rightarrow h = \frac{100}{19.6} \approx 5.10 \, m

Numerical 1: Motor lifting water

  1. Work done:

W = m g h = 500 \cdot 9.8 \cdot 10 = 49000 \, J


P = \frac{W}{t} = \frac{49000}{20} = 2450 \, W

Numerical 2: Car brought to rest

  1. Initial KE:

KE = \frac{1}{2} m v^2 = 0.5 \cdot 1000 \cdot 20^2 = 200,000 \, J


W = - KE = -200,000 \, J

Section D – Long Answer Questions (5 Marks each)

  1. Derive formula for KE:
  • Work done W = F × s
  • From Newton’s second law, F = ma
  • Using kinematics, s = (v² – u²)/2a

W = F s = ma \frac{v^2 - u^2}{2a} = \frac{1}{2} m (v^2 - u^2)
  1. Gravitational potential energy:

PE = m g h
  • Example: Stone lifted to height h has PE = mgh.
  1. Falling ball

a) Velocity just before hitting ground:


v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 20} \approx 19.8 \, m/s

b) Kinetic energy:


KE = ½ m v^2 = 0.5 \cdot 0.5 \cdot 19.8^2 \approx 98 \, J
  1. Pendulum diagram:

Description:

  • Highest point: PE = max, KE = 0
  • Middle: PE decreases, KE increases
  • Lowest point: KE = max, PE = min

(Diagram: simple pendulum showing swing from highest to lowest point with arrows indicating PE ↔ KE)

  1. Motor lifting water

a) Work done:


W = m g h = 100 \cdot 9.8 \cdot 5 = 4900 \, J

b) Power:


P = W/t = 4900/10 = 490 \, W
  1. Sliding body down frictionless incline

a) Speed at bottom:


v = \sqrt{2 g h} = \sqrt{2 \cdot 9.8 \cdot 4} = \sqrt{78.4} \approx 8.86 \, m/s

b) Kinetic energy at bottom:


KE = ½ m v^2 = 0.5 \cdot 3 \cdot 8.86^2 \approx 117.5 \, J
Related Story

Leave a Reply

Your email address will not be published. Required fields are marked *