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ICSE • Class 8 • Science • Ch 4
Estimated Time: 45 Mins
Study Progress: In Progress

Energy

In ICSE Class 8 Science (Physics), "Energy" provides an authoritative, mathematically rigorous study guide investigating work, kinetic and potential mechanical energy, the Law of Conservation of Energy, and energy dissipation. This comprehensive chapter explores Concept of Work (Scientific definition: work is done when a force produces displacement along the direction of the force: $W = F \times s$; Conditions for work: 1. Force must act, 2. Displacement must take place, 3. Displacement must not be perpendicular to force; Work done against gravity: $W = mgh$; SI unit: Joule [J]; CGS unit: Erg where $1\text{ J} = 10^7\text{ ergs}$), Concept of Power (Rate of doing work: $P = \frac{W}{t}$; SI unit: Watt [W = J/s]; Commercial units: kilowatt [kW], megawatt [MW], horsepower [hp where $1\text{ hp} = 746\text{ W}$]), Mechanical Energy (Energy possessed by a body due to its state of rest, position, configuration, or motion), Potential Energy ($U$: energy stored by virtue of position or configuration; 1. Gravitational Potential Energy: $U = mgh$, 2. Elastic Potential Energy: stretched bow, coiled spring), Kinetic Energy ($K$: energy possessed by a body by virtue of its state of motion; Derivation: $K = \frac{1}{2}mv^2$; Relationship between momentum and kinetic energy: $K = \frac{p^2}{2m}$), The Law of Conservation of Energy (Energy can neither be created nor destroyed; it can only transform from one form to another, with total energy remaining constant: $E_{\text{total}} = K + U = \text{constant}$; Step-by-step mathematical proof of conservation for a freely falling body under gravity), Energy Transformations in Common Devices, and Energy Degradation / Dissipation aligned with the 2026–27 CISCE ICSE curriculum.

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How Can a Person Holding a 50-Kilogram Stone on Their Head for Five Hours Do Exactly ZERO Joules of Work?

Imagine a porter standing completely stationary under the scorching sun at a railway station, balancing a crushing $50\text{-kilogram}$ steel trunk on his head for three exhausting hours. His muscles ache, sweat pours down his face, and he collapses from fatigue. He asks for his wages. But a physicist walks up and says: "Scientifically, you did ZERO Joules of work!" How could that be? In everyday language, "work" means physical exertion. But in PHYSICS, Work is mathematically defined as Force multiplied by Displacement in the direction of the force ($W = F \times s$)! Because the trunk never moved even a millimeter ($s = 0$), the physical work done on the trunk was EXACTLY ZERO! What is the secret connection between kinetic energy ($K = \frac{1}{2}mv^2$) and momentum ($p = mv$)? Why does a roller coaster at the top of a loop trade pure gravitational potential energy for kinetic speed? Let's master energy.

Why This Chapter Matters

Energy conservation is the supreme fundamental invariant of the physical universe: powering hydroelectric turbines, nuclear power plants, sports biomechanics, automotive braking regenerators, and orbital satellite mechanics. Mastering work, power, and energy equations is essential for ICSE physics.

Before You Begin (Prerequisites)

  • Kinematics equations of motion ($v^2 - u^2 = 2as$).
  • Newton's second law ($F = ma$).
  • Concept of gravity and mass.

What You Will Learn (Core Objectives)

  • Define work, state its SI unit (Joule), and calculate $W = F \times s$ and $W = mgh$.
  • Identify conditions under which work done is zero.
  • Define power, state its units (Watt, horsepower), and calculate $P = \frac{W}{t}$.
  • Derive and calculate Gravitational Potential Energy ($U = mgh$) and Kinetic Energy ($K = \frac{1}{2}mv^2$).
  • Prove the Law of Conservation of Energy mathematically for a freely falling body.
  • Trace energy transformation chains in hydroelectric stations, electric motors, and microphones.

Chapter Roadmap & Progression

1 1. Work: Definition, Units & Zero W...
2 2. Power: Rate of Energy Consumptio...
3 3. Potential & Kinetic Mechanical E...
4 4. Law of Conservation of Energy (F...

Complete Concept Guide (100% Curriculum Coverage)

1. Work: Definition, Units & Zero Work Conditions

Understand
A. What is Work?

Work is said to be done when an applied force produces displacement in the body in the direction of the force:

$$\mathbf{W = F \times s}$$
  • SI Unit: Joule ($\text{J}$): $1\text{ J} = 1\text{ N} \times 1\text{ m}$.
  • CGS Unit: Erg: $1\text{ Erg} = 1\text{ dyne} \times 1\text{ cm}$.
  • $$\mathbf{1\text{ J} = 10^5\text{ dynes} \times 10^2\text{ cm} = 10^7\text{ Ergs}}$$
  • Work Against Gravity: Lifting a mass $m$ through vertical height $h$: $$\mathbf{W = F_g \times h = mgh}$$
B. When is Work Done Equal to Zero ($W = 0$)?
  1. When displacement is zero ($s = 0$, e.g., pushing a concrete wall).
  2. When force is perpendicular to displacement ($\theta = 90^{\circ}$, e.g., a coolie carrying luggage on a horizontal road against vertical gravity, or the Earth orbiting the Sun).

2. Power: Rate of Energy Consumption

Power
A. Definition & Units:

The rate of doing work (or rate of energy consumption) is defined as Power:

$$\mathbf{P = \frac{\text{Work Done}}{\text{Time Taken}} = \frac{W}{t} = \frac{F \times s}{t} = F \times v}$$
  • SI Unit: Watt ($\text{W}$): $1\text{ W} = 1\text{ J/s}$.
  • Horsepower (hp): Commercial unit: $$\mathbf{1\text{ hp} = 746\text{ Watts}}$$
  • Kilowatt ($1\text{ kW} = 1000\text{ W}$); Megawatt ($1\text{ MW} = 10^6\text{ W}$).

3. Potential & Kinetic Mechanical Energy

Mechanical Energy
A. Potential Energy ($U$):

The energy stored in a body by virtue of its position above the ground or its altered configuration:

$$\mathbf{U = mgh \quad (\text{Gravitational Potential Energy})}$$
B. Kinetic Energy ($K$):

The energy possessed by a body by virtue of its velocity / motion:

$$\mathbf{K = \frac{1}{2} m v^2}$$

Relation to Momentum ($p = mv$):

$$K = \frac{1}{2} m \left( \frac{p}{m} \right)^2 = \mathbf{\frac{p^2}{2m} \quad \Longleftrightarrow \quad p = \sqrt{2mK}}$$

4. Law of Conservation of Energy (Freely Falling Body)

Conservation of Energy
Mathematical Proof for a Body of Mass $m$ Dropped from Height $H$:
  1. At Top Point $A$ ($h = H, v = 0$): $$U_A = mgH, \quad K_A = 0 \implies \mathbf{E_A = mgH}$$
  2. At Intermediate Point $B$ (Fallen by distance $x$, height $= H - x$): $$v_B^2 = 0 + 2gx = 2gx$$ $$K_B = \frac{1}{2} m v_B^2 = \frac{1}{2} m (2gx) = mgx$$ $$U_B = mg(H - x) = mgH - mgx$$ $$\mathbf{E_B = K_B + U_B = mgx + (mgH - mgx) = mgH}$$
  3. At Ground Point $C$ (Height $= 0$, fallen distance $= H$): $$v_C^2 = 2gH \implies K_C = \frac{1}{2} m (2gH) = mgH, \quad U_C = 0$$ $$\mathbf{E_C = mgH}$$

$$\mathbf{E_A = E_B = E_C = mgH = \text{Constant!}}$$

Key Formulas, Reactions & Definitions

Work Formula
$$W = F \times s = mgh \quad [1\text{ J} = 10^7\text{ Ergs}]$$
Force times displacement in direction of force.
Kinetic & Potential Energy
$$K = \frac{1}{2}mv^2 = \frac{p^2}{2m}, \quad U = mgh$$
Mechanical energy components.
Power Relation
$$P = \frac{W}{t} = F \cdot v \quad [1\text{ hp} = 746\text{ W}]$$
Rate of doing work.

Physics: Freely Falling Body Energy Conservation

Energy: Work, Power & Freely Falling Conservation Law WORK & POWER MECHANICS Work (W) = F × s [1 J = 107 Ergs] Power (P) = W / t [1 hp = 746 Watts] • When is Work ZERO? 1. Displacement s = 0 (Pushing wall) 2. Force ⊥ Displacement (Coolie walking horizontally) • Mechanical Energy: Potential U = mgh • Kinetic K = ½mv2 = p2 / 2m • Energy can neither be created nor destroyed CONSERVATION OF ENERGY (FREE FALL) Point A: U = mgH, K = 0 ⇒ Total = mgH Point B: U = mg(H-x), K = mgx ⇒ mgH Point C: U = 0, K = mgH ⇒ Total = mgH E_total = K + U = mgH = Constant! Potential energy transforms perfectly into Kinetic energy W = F×s • P = W/t • K = ½mv^2 • U = mgh • 1 hp = 746 W • ENERGY IS ALWAYS CONSERVED

Chapter Summary & 10 Key Takeaways

Takeaway 1
Work is done when a force causes displacement in the direction of the force: W = F * s.
Takeaway 2
SI unit of work is Joule (J); CGS unit is Erg (1 J = 10^7 Ergs).
Takeaway 3
Work is zero if displacement is zero, or if force is perpendicular to displacement.
Takeaway 4
Power is the rate of doing work: P = W / t, measured in Watts (1 hp = 746 W).
Takeaway 5
Gravitational potential energy is stored energy: U = mgh.
Takeaway 6
Kinetic energy is energy of motion: K = 1/2 * m * v^2 = p^2 / (2m).
Takeaway 7
Law of conservation of energy states that total energy of an isolated system is constant.
Takeaway 8
In free fall, potential energy decreases as kinetic energy increases, maintaining total E = mgH.
Takeaway 9
Energy transformation in hydroelectric plants: Potential -> Kinetic -> Mechanical -> Electrical.
Takeaway 10
Energy dissipation refers to conversion of useful energy into non-recoverable thermal and sound energy.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A body of mass $10\text{ kg}$ is lifted vertically to a height of $5\text{ meters}$. Calculate: (a) the work done against gravity, (b) the gravitational potential energy gained. (Take $g = 9.8\text{ m/s}^2$).
Reveal Answer & Explanation
Answer:

• (a) Work done against gravity:

$$W = mgh = 10\text{ kg} \times 9.8\text{ m/s}^2 \times 5\text{ m} = \mathbf{490\text{ Joules}}$$



• (b) Gravitational Potential Energy gained:

$$U = W = mgh = \mathbf{490\text{ Joules}}$$

.


$W = U = mgh = 10 \times 9.8 \times 5 = 490\text{ J}$.
2
A vehicle of mass $800\text{ kg}$ is moving at a uniform speed of $54\text{ km/h}$. Calculate its kinetic energy.
Reveal Answer & Explanation
Answer: Step 1: Convert speed to SI units ($\text{m/s}$):
$$v = 54\text{ km/h} = 54 \times \frac{5}{18} = 3 \times 5 = \mathbf{15\text{ m/s}}$$
Step 2: Apply the Kinetic Energy Formula ($K = \frac{1}{2}mv^2$):
$$K = \frac{1}{2} \times 800 \times (15)^2 = 400 \times 225 = \mathbf{90,000\text{ Joules} \quad (\text{or } 90\text{ kJ})}$$.
$v = 15\text{ m/s}$. $K = \frac{1}{2}(800)(15^2) = 400 \times 225 = 90,000\text{ J}$.
3
Why is the physical work done by a coolie carrying a heavy suitcase on his head while walking along a horizontal railway platform considered ZERO in physics?
Reveal Answer & Explanation
Answer:

• Work done is mathematically given by: $W = F \cdot s \cdot \cos\theta$.
• The coolie exerts a vertical upward force $F$ to support the luggage against the downward pull of gravity.
• His displacement $s$ is along the horizontal platform.
• The angle $\theta$ between the upward force and horizontal displacement is $90^{\circ}$.
• Since $\cos 90^{\circ} = 0$:

$$W = F \times s \times 0 = \mathbf{0}$$

.
Therefore, the work done on the luggage against gravity is strictly zero.


Force is vertical while displacement is horizontal ($ heta = 90^{\circ}$, $\cos 90^{\circ} = 0$).
4
An electric water pump of power $1.5\text{ kW}$ lifts $200\text{ liters}$ of water to an overhead tank at a height of $15\text{ meters}$ in $20\text{ seconds}$. Calculate: (a) work done by the pump, (b) its efficiency. (Take $g = 10\text{ m/s}^2$).
Reveal Answer & Explanation
Answer:

Step 1: Mass of $200\text{ liters}$ of water $= 200\text{ kg}$.
• (a) Useful Work Done ($W_{\text{out}}$):

$$W = mgh = 200 \times 10 \times 15 = \mathbf{30,000\text{ Joules}}$$


• (b) Total Energy Input ($W_{\text{in}}$):

$$W_{\text{in}} = \text{Power} \times \text{Time} = 1500\text{ W} \times 20\text{ s} = \mathbf{30,000\text{ Joules}}$$


• Efficiency ($\eta$):

$$\eta = \frac{\text{Useful Output}}{\text{Total Input}} \times 100\% = \frac{30,000}{30,000} \times 100\% = \mathbf{100\%}$$

.


$W = 200 \times 10 \times 15 = 30,000\text{ J}$. Input energy $= 1500 \times 20 = 30,000\text{ J}$. Efficiency is $100\%$.
5
Derive the algebraic relationship between Kinetic Energy ($K$) and Linear Momentum ($p$).
Reveal Answer & Explanation
Answer: • Linear momentum is defined as: $p = mv \implies v = \frac{p}{m}$.
• Kinetic energy is given by: $K = \frac{1}{2} mv^2$.
• Substitute $v = \frac{p}{m}$ into the kinetic energy equation:
$$K = \frac{1}{2} m \left( \frac{p}{m} \right)^2 = \frac{1}{2} m \frac{p^2}{m^2} = \mathbf{\frac{p^2}{2m}}$$
• Inverting for momentum gives: $\mathbf{p = \sqrt{2mK}}$.
Substitute $v = p/m$ into $K = rac{1}{2}mv^2$ to obtain $K = p^2 / (2m)$.
6
State the Law of Conservation of Energy and describe the energy transformations taking place in an electric iron and an electric bell.
Reveal Answer & Explanation
Answer:

• The Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only transform from one form into another, with the total energy of an isolated system remaining strictly constant.
• Electric Iron: Electrical Energy $\to$ Heat (Thermal) Energy (Joule heating).
• Electric Bell: Electrical Energy $\to$ Magnetic Energy $\to$ Mechanical Energy $\to$ Sound Energy.


Electric iron: electrical to heat. Electric bell: electrical to magnetic to mechanical to sound.
7
What is Horsepower and how is it related to the SI unit of power?
Reveal Answer & Explanation
Answer:

• Horsepower (hp) is a traditional imperial unit of power defined by James Watt to compare the output of steam engines with draft horses.
• It is related to the SI unit (Watt) by the conversion factor:

$$\mathbf{1\text{ Horsepower (hp)} = 746\text{ Watts}}$$

.


$1\text{ hp} = 746\text{ W}$.
8
A ball of mass $2\text{ kg}$ is dropped from a height of $20\text{ meters}$. Find its velocity and kinetic energy just before striking the ground. (Take $g = 10\text{ m/s}^2$).
Reveal Answer & Explanation
Answer: Step 1: By Conservation of Energy, Kinetic Energy at ground equals initial Potential Energy at height $H$:
$$K_{\text{bottom}} = U_{\text{top}} = mgh = 2 \times 10 \times 20 = \mathbf{400\text{ Joules}}$$
Step 2: Find velocity using $K = \frac{1}{2}mv^2$:
$$400 = \frac{1}{2} \times 2 \times v^2 \implies v^2 = 400 \implies v = \sqrt{400} = \mathbf{20\text{ m/s}}$$.
*(Velocity is $20\text{ m/s}$ and kinetic energy is $400\text{ J}$)*.
$K = mgh = 2 \times 10 \times 20 = 400\text{ J}$. $v = \sqrt{2gh} = \sqrt{2(10)(20)} = 20\text{ m/s}$.
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