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

(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

(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.

• 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.

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 (E_p = mgh) to decrease. However, its speed increases due to gravitational acceleration, causing an equal increase in kinetic energy (E_k = 1/2mv²). The sum of potential and kinetic energy (E_p + E_k) remains constant at all points.

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

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

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

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

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.

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

• 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).

• Step 1: Potential Energy of First Body (E_p1) / :
• m₁ = 10 kg, h₁ = 5 m, g = 9.8 m/s²
• E_p1 = m₁ g h₁ = 10 × 9.8 × 5 = 490 Joules
• Step 2: Potential Energy of Second Body (E_p2) / :
• m₂ = 15 kg, h₂ = 2 m, g = 9.8 m/s²
• E_p2 = m₂ g h₂ = 15 × 9.8 × 2 = 294 Joules
• Step 3: Difference in Potential Energies / :
• Δ E_p = E_p1 – E_p2 = 490 – 294 = 196 Joules (J).
• (If using g = 10 m/s²: E_p1 = 500 J, E_p2 = 300 J, Difference = 200 J).

• Step 1: Total Mechanical Energy at Position A (h_A = 10 m) / A :
• Initial speed u = 0 ⇒ E_kA = 0
• E_pA = 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 :
• E_pD = m g h_D = m × 9.8 × 2.5 = 24.5m Joules
• E_kD = 1/2mv_D²
• Step 3: Apply Conservation Law / :
• E_pA + E_kA = E_pD + E_kD
• 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).

• 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).

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