CG Board Class 9 Science Chapter 12 Solutions: Work and Energy (कार्य एवं ऊर्जा) – 2026-27

CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Science Chapter 12 Solutions: Work and Energy (कार्य एवं ऊर्जा) in Bilingual Format (English & Hindi Medium).


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Part 1

In-Text Questions & Solutions

Set 1 (Pages 172

174): Scientific Concept of Work

Question 1

In which of the following situations is work…


📝 Question:

** In which of the following situations is work being done according to science? State reasons for your answer

💡 Solution:

In which of the following situations is work being done according to science? State reasons for your answer

  1. You pushed a very large stone, but it did not move.
  2. You reached the second floor of a building by climbing stairs.
  3. A traveller pulls his luggage to some distance on a platform.
  4. Stopping a moving bicycle.

No work done. Force was applied, but the displacement of the stone was zero (S = 0).

Work is done. The person applies force against gravity to displace their body vertically upwards to the second floor.

Work is done. The traveller applies a pulling force on the luggage, causing displacement along the platform in the direction of applied force.

Work is done (Negative work). The braking force is applied opposite to the direction of motion to displace the bicycle to a stop.

Question 2

A box filled with books is placed near…


📝 Question:

** A box filled with books is placed near a wall. Even after several efforts by you, the box does not move. Calculate the work done according to the equation W = FS.

💡 Solution:

A box filled with books is placed near a wall. Even after several efforts by you, the box does not move. Calculate the work done according to the equation W = FS.

  • Applied Force = F
  • Displacement (S) = 0 m
  • W = F × S = F × 0 = 0 Joules.
  • Reason: Since there is no displacement, no physical work is done on the box.

Question 3

Can you think of a situation where there…


📝 Question:

** Can you think of a situation where there is displacement in the absence of any force? Discuss.

💡 Solution:

Can you think of a situation where there is displacement in the absence of any force? Discuss.

Yes. According to Newton’s First Law of Motion, an object moving with a uniform velocity in deep outer space (where gravitational and frictional forces are completely absent) continues to displace indefinitely at constant velocity without requiring any external force (F = 0).

Set 2 (Pages 175

176): Positive, Negative Work & Vertical Movement

Question 1

A ball is thrown upwards A gravitational force…


📝 Question:

** A ball is thrown upwards. A gravitational force continuously acts on it in the downward direction.

💡 Solution:

A ball is thrown upwards. A gravitational force continuously acts on it in the downward direction.

  1. Which force acting on the ball does positive work?
  2. Which force on the ball does negative work?
  3. Support your answer with reasons.

The throwing force applied by the hand during the launch performs positive work because force and initial upward displacement are in the same direction.

The gravitational force performs negative work while the ball moves upwards.

When displacement is opposite to the direction of applied force, the work done is negative (W = -fS). Here, gravity acts downwards while displacement is upwards.

Question 2

A student picks up an object of mass…


📝 Question:

** A student picks up an object of mass m in the vertical direction to a height h and moves a distance d horizontally. How much work is done by the student in raising the object and walking horizontally?

💡 Solution:

A student picks up an object of mass m in the vertical direction to a height h and moves a distance d horizontally. How much work is done by the student in raising the object and walking horizontally?

  • Work done in lifting vertically: W1 = mgh (since applied lifting force F = mg is in the direction of vertical displacement h).
  • Work done while walking horizontally: W2 = 0 (since the uniform horizontal velocity implies zero horizontal acceleration and zero horizontal force, F = 0).
  • Total Work Done: W = W1 + W2 = mgh + 0 = mgh.

Set 3 (Page 181)

Kinetic Energy Questions

Question 1

What do you mean by kinetic energy of…


📝 Question:

** What do you mean by kinetic energy of an object?

💡 Solution:

What do you mean by kinetic energy of an object?

The energy possessed by an object due to its state of motion is called kinetic energy. Formula: Ek = 1/2mv2.

Question 2

The kinetic energy of a body of mass…


📝 Question:

** The kinetic energy of a body of mass m moving at a velocity v is 1/2mv2. If its velocity is doubled, how much will be its kinetic energy?

💡 Solution:

The kinetic energy of a body of mass m moving at a velocity v is 1/2mv2. If its velocity is doubled, how much will be its kinetic energy?

  • Initial Kinetic Energy = Ek1 = 1/2mv2.
  • New velocity = v’ = 2v.
  • New Kinetic Energy = Ek2 = 1/2m(2v)2 = 1/2m(4v2) = 4 (1/2mv2) = 4 Ek1.
  • The kinetic energy becomes 4 times its initial value.

Question 3

If the velocity and mass of an object…


📝 Question:

** If the velocity and mass of an object are twice the velocity and mass of another object, how much will be the ratio of their kinetic energies?

💡 Solution:

If the velocity and mass of an object are twice the velocity and mass of another object, how much will be the ratio of their kinetic energies?

  • Let object 1 have mass m1 = m and velocity v1 = v.
  • Ek1 = 1/2mv2.
  • For object 2: m2 = 2m and v2 = 2v.
  • Ek2 = 1/2(2m)(2v)2 = 1/2(2m)(4v2) = 8 (1/2mv2) = 8 Ek1.
  • Ratio (Ek1)/(Ek2) = 1/8 or Ek2 : Ek1 = 8 : 1.

Set 4 (Pages 183

184): Potential Energy Questions

Question 1

A cuboidal box is shown in Fig 16…


📝 Question:

** A cuboidal box is shown in Fig. 16. Its width and height are 2h and h respectively. What will be the potential energy of an object of mass m placed on the box when placed vertically vs horizontally?

💡 Solution:

A cuboidal box is shown in Fig. 16. Its width and height are 2h and h respectively. What will be the potential energy of an object of mass m placed on the box when placed vertically vs horizontally?

Height above ground = h. Gravitational Potential Energy = Ep = mgh.

Height above ground = 2h. Gravitational Potential Energy = Ep = m × g × (2h) = 2mgh.

Question 2

In the process of using a bow and…


📝 Question:

** In the process of using a bow and arrow, why do we have to pull the arrow backwards by using a thread or rubber?

💡 Solution:

In the process of using a bow and arrow, why do we have to pull the arrow backwards by using a thread or rubber?

When the string/rubber is stretched backwards, muscle work done against elasticity gets stored as elastic potential energy in the stretched string. On releasing, this potential energy converts into kinetic energy of the arrow, propelling it forward at high speed.

Question 3

Can the gravitational potential energy of an object…


📝 Question:

** Can the gravitational potential energy of an object be negative?

💡 Solution:

Can the gravitational potential energy of an object be negative?

Yes. Potential energy depends on the chosen reference zero-level. If a position below the chosen reference surface (e.g., inside a deep well or pit) is considered, the height h is negative, making Ep = -mgh negative.

Set 5 (Pages 185

188): Power & Energy Conversion

Question 1

A person of mass m jumps into a…


📝 Question:

** A person of mass m jumps into a swimming pool from a height of 10 m. Using the law of conservation of mechanical energy, calculate the velocity of the man when he is at a height of 5 m from the water surface.

💡 Solution:

A person of mass m jumps into a swimming pool from a height of 10 m. Using the law of conservation of mechanical energy, calculate the velocity of the man when he is at a height of 5 m from the water surface.

Question 2

Question 13


📝 Question:

**

💡 Solution:

  1. What do you mean by power?
  2. Define 1 watt power.
  3. If a bulb consumes 990 Joules of electric energy in 10 sec, how much will be its power?

The rate of doing work or rate of energy conversion is called power. Formula: P = W/t.

If an agent performs 1 Joule of work in 1 second, its power is 1 Watt (1 W = 1 J/s).

Part 2

End-of-Chapter Exercises

Question 1

Multiple Choice Questions


📝 Question:

Multiple Choice Questions

💡 Solution:

(i) If an object displaces 2 m in the direction of 10 N applied force, the work done on the object by the force is:**

  • (a) 15 J
  • (b) 20 J
  • (c) -20 J
  • (d) 5 J

(ii) A body of mass 5 kg is in a constant motion at 2 m/s. What will be its kinetic energy?

  • (a) 10 Joule
  • (b) 15 Joule
  • (c) 5 Joule
  • (d) 20 Joule

(iii) A body of 12 kg mass is placed at a certain height from the surface of earth. If its potential energy is 480 J, then what will be its height w.r.t. earth’s surface?

  • (a) 6 meter
  • (b) 9 meter
  • (c) 5 meter
  • (d) 4 meter

(iv) You burn a bulb of 100 W daily for 5 hours. What will be the energy spent in units by the bulb in one day?

  • (a) 0.4 unit
  • (b) 0.5 unit
  • (c) 0.05 unit
  • (d) 0.01 unit

Question 2

What do you understand by kinetic energy Establish…


📝 Question:

What do you understand by kinetic energy? Establish the equation for kinetic energy for an object in motion. – Hindi Question / : ?

💡 Solution:

(i) SI unit of work is …………………………**

(ii) 1 kwh is equivalent to ……………………….. joules.

(iii) The total energy of an object remains ………………………….

(iv) If the direction of applied force of an object and its displacement are opposite then its work will be ………………………………

The energy possessed by an object due to its motion is called kinetic energy.

Question 4

The energy consumed at a house is 250…


📝 Question:

The energy consumed at a house is 250 units in one month. How much will it be in Joules? – Hindi Question / : 250 ?

💡 Solution:

  • Step 1: Relation between Units and Joules / :
  • 1 unit = 1 kWh = 3.6 × 106 J (or 36 × 105 J).
  • Step 2: Conversion / :
  • Energy consumed = 250 units = 250 × 3.6 × 106 J.
  • Energy in Joules = 900 × 106 J = 9 × 108 Joules.

Question 5

a What is the energy conservation law Explain…


📝 Question:

– (a) What is the energy conservation law? Explain. – (b) The potential energy of a freely falling body drops gradually. Does it contradict the law of energy conservation? Explain with reasons. – Hindi Question / : – (a) ? – (b) ?

💡 Solution:

Energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy of an isolated system remains constant.

No, it does not contradict the law. As the body falls, its height decreases, causing potential energy (Ep = mgh) to decrease. However, its speed increases due to gravitational acceleration, causing an equal increase in kinetic energy (Ek = 1/2mv2). The sum of potential and kinetic energy (Ep + Ek) remains constant at all points.

Question 6

Explain potential energy and establish its equation Hindi…


📝 Question:

Explain potential energy and establish its equation. – **Hindi Question /

💡 Solution:

The energy stored in an object due to its position or change in shape is called its potential energy.

Question 7

a If the velocity of a particle is…


📝 Question:

– (a) If the velocity of a particle is doubled, what will be its kinetic energy? – (b) If the work done on the particle is zero, then what will be its velocity? – Hindi Question / : – (a) , ? – (b) , ?

💡 Solution:

Kinetic energy is proportional to the square of velocity (Ek ∝ v2). Doubling velocity (2v) increases kinetic energy to (2)2 = 4 times.

According to the Work-Energy Theorem (Δ Ek = W), if W = 0, the change in kinetic energy is zero, meaning velocity remains constant (unchanged).

Question 8

On applying brakes of a car which is…


📝 Question:

On applying brakes of a car which is moving at a speed v, it stops after covering a distance d. Calculate what will be the distance covered by the car after applying brakes if it was moving at a speed of 2v. – Hindi Question / : v d 2v , ?

💡 Solution:

The stopping distance will be 4d (four times the initial distance).

Question 9

A man holds a sack of rice on…


📝 Question:

A man holds a sack of rice on his head for 30 minutes and gets tired. Has he done any work? Explain your answer with reasons. – Hindi Question / : 30 ?

💡 Solution:

No physical work is done (W = 0).**

  • Reason: Although the man exerts an upward force to balance the weight of the sack, there is zero displacement (S = 0) of the sack while standing stationary.

Question 10

On applying a force of 8 Newton on…


📝 Question:

On applying a force of 8 Newton on a body, it moves in the direction of motion and displaces for 4 m. Calculate the work done. – Hindi Question / : 8 4 m

💡 Solution:

  • Given Values / :
  • Force (F) = 8 N
  • Displacement (S) = 4 m
  • Formula & Calculation / :
  • W = F × S = 8 N × 4 m = 32 Joules (J).

Question 11

A body of 10 kg mass is raised…


📝 Question:

A body of 10 kg mass is raised against the gravitational force of earth to a height of 10 m. How much work is done in this case? – Hindi Question / : 10 kg 10 m ?

💡 Solution:

  • Given Values / :
  • Mass (m) = 10 kg
  • Height (h) = 10 m
  • Acceleration due to gravity (g) = 9.8 m/s2 (or 10 m/s2)
  • W = mgh = 10 kg × 9.8 m/s2 × 10 m = 980 Joules (J).
  • (If using g = 10 m/s2: W = 10 × 10 × 10 = 1000 J).

Question 12

Two bodies of masses 10 kg and 15…


📝 Question:

Two bodies of masses 10 kg and 15 kg are raised to a height of 5 m and 2 m respectively above the surface of earth. Calculate the change in their potential energies. – Hindi Question / : 10 kg 15 kg 5 m 2 m /

💡 Solution:

  • Step 1: Potential Energy of First Body (Ep1) / :
  • m1 = 10 kg, h1 = 5 m, g = 9.8 m/s2
  • Ep1 = m1 g h1 = 10 × 9.8 × 5 = 490 Joules
  • Step 2: Potential Energy of Second Body (Ep2) / :
  • m2 = 15 kg, h2 = 2 m, g = 9.8 m/s2
  • Ep2 = m2 g h2 = 15 × 9.8 × 2 = 294 Joules
  • Step 3: Difference in Potential Energies / :
  • Δ Ep = Ep1 – Ep2 = 490 – 294 = 196 Joules (J).
  • (If using g = 10 m/s2: Ep1 = 500 J, Ep2 = 300 J, Difference = 200 J).

Question 13

A ball of mass m is released from…


📝 Question:

A ball of mass m is released from height h_A = 10 m at position A. Calculate the velocity of the ball at position D (height h_D = 2.5 m) using the law of conservation of mechanical energy. – Hindi Question / : A D

💡 Solution:

  • Step 1: Total Mechanical Energy at Position A (h_A = 10 m) / A :
  • Initial speed u = 0 ⇒ EkA = 0
  • EpA = m g h_A = m × 9.8 × 10 = 98m Joules
  • Total Energy = 98m
  • Step 2: Total Mechanical Energy at Position D (h_D = 2.5 m) / D :
  • EpD = m g h_D = m × 9.8 × 2.5 = 24.5m Joules
  • EkD = 1/2mv_D2
  • Step 3: Apply Conservation Law / :
  • EpA + EkA = EpD + EkD
  • 98m + 0 = 24.5m + 1/2mv_D2
  • 98m – 24.5m = 1/2mv_D2 ⇒ 73.5m = 1/2mv_D2
  • Cancel m: v_D2 = 73.5 × 2 = 147
  • v_D = √(147) ≈ 12.12 m/s.
  • (If using g = 10 m/s2: 100m – 25m = 1/2mv_D2 ⇒ 75 = 1/2v_D2 ⇒ v_D = √(150) ≈ 12.25 m/s).

Question 14

On applying a force of 15 N for…


📝 Question:

On applying a force of 15 N for 6 sec, a man displaces a box by 8 m. Calculate his power. – Hindi Question / : 15 N 6 8 m

💡 Solution:

  • Step 1: Calculate Work Done (W) / :
  • Force (F) = 15 N, Displacement (S) = 8 m
  • W = F × S = 15 × 8 = 120 Joules (J)
  • Step 2: Calculate Power (P) / :
  • Time (t) = 6 s
  • P = W/t = 120/6 = 20 Watts (W).

Question 15

To change the energy of a body of…


📝 Question:

To change the energy of a body of 0.5 kg by 1 Joule, we will have to raise it to what height? (g = 10 m/s2). – Hindi Question / : 0.5 kg 1 ? (g = 10 m/s2)

💡 Solution:

  • Given Values / :
  • Mass (m) = 0.5 kg
  • Potential Energy (Δ Ep) = 1 Joule
  • Acceleration due to gravity (g) = 10 m/s2
  • Formula & Step-by-Step Calculation / :
  • Ep = mgh
  • 1 = 0.5 × 10 × h
  • 1 = 5h ⇒ h = 1/5 = 0.2 meters (or 20 cm).


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