Login / Register
Follow Us
0
0
Register Free!
Register Free!
Online Mock Tests & Exam Preparation | Target Exams
Register Free Login
माध्यम चुनें / Select Medium:
Eng (English) Hindi (हिन्दी)
ICSE • कक्षा 8 • Science • अध्याय 4
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

ऊर्जा (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.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें
प्रस्तावना अध्याय अवधारणाएँ सूत्र एवं अभिक्रियाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

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.

यह अध्याय क्यों महत्वपूर्ण है

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.

अध्ययन से पूर्व (आवश्यक ज्ञान)

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

इस अध्याय के लक्ष्य

  • 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.

अध्याय रूपरेखा एवं प्रगति

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...

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

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 $= H$): $$v_C^2 = 2gH \implies K_C = \frac{1}{2} m (2gH) = mgH, \quad U_C = 0$$ $$\mathbf{E_C = mgH}$$

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

महत्वपूर्ण रासायनिक सूत्र, अभिक्रियाएँ एवं समीकरण

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

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

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

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

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 $g = 9.8\text{ m/s}^2$).
उत्तर एवं व्याख्या देखें
उत्तर:

• (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.
उत्तर एवं व्याख्या देखें
उत्तर: 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?
उत्तर एवं व्याख्या देखें
उत्तर:

• 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 $g = 10\text{ m/s}^2$).
उत्तर एवं व्याख्या देखें
उत्तर:

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 ($p$).
उत्तर एवं व्याख्या देखें
उत्तर: • 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.
उत्तर एवं व्याख्या देखें
उत्तर:

• 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?
उत्तर एवं व्याख्या देखें
उत्तर:

• 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 $g = 10\text{ m/s}^2$).
उत्तर एवं व्याख्या देखें
उत्तर: 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 $400\text{ J}$)*.
$K = mgh = 2 \times 10 \times 20 = 400\text{ J}$. $v = \sqrt{2gh} = \sqrt{2(10)(20)} = 20\text{ m/s}$.
अध्याय का अध्ययन पूर्ण हुआ?
अभ्यास के लिए तैयार?

ऑनलाइन CBT टेस्ट देकर तैयारी का मूल्यांकन करें

ICSE परीक्षा पैटर्न पर आधारित बहुविकल्पीय प्रश्नों का ऑनलाइन टेस्ट दें। तुरंत परिणाम, समय विश्लेषण और प्रत्येक प्रश्न का विस्तृत हल प्राप्त करें।

फाउंडेशन टेस्ट
15 MCQs • मूल अवधारणाएँ
एडवांस्ड टेस्ट
15 MCQs • HOTS और ट्रैप्स

कक्षा 8 Science के सभी अध्याय

अध्याय 1: पदार्थ (गतिज सिद्धांत) (Matter (Kinetic Theory)) अध्याय 2: भौतिक राशियाँ और मापन (Physical Quantities and Measurement) अध्याय 3: बल तथा दाब (Force and Pressure) अध्याय 4: ऊर्जा (Energy) अध्याय 5: प्रकाश ऊर्जा (Light Energy) अध्याय 6: गर्मी का हस्तांतरण (Heat Transfer) अध्याय 7: ध्वनि (Sound) अध्याय 8: विद्युत (Electricity) अध्याय 9: पदार्थ और उसकी संरचना (Matter and Composition) अध्याय 10: भौतिक एवं रासायनिक परिवर्तन (Physical and Chemical Changes) अध्याय 11: तत्व, यौगिक और मिश्रण (Elements, Compounds and Mixtures) अध्याय 12: परमाणु संरचना (Atomic Structure) अध्याय 13: रसायन विज्ञान की भाषा (Language of Chemistry) अध्याय 14: रासायनिक प्रतिक्रिएं (Chemical Reactions) अध्याय 15: हाइड्रोजन (Hydrogen) अध्याय 16: जल (Water) अध्याय 17: कार्बन और उसके यौगिक (Carbon and Its Compounds) अध्याय 18: पौधों में भोजन एवं खनिजों का परिवहन (Transport of Food and Minerals in Plants) अध्याय 19: पौधों और जानवरों में प्रजनन (Reproduction in Plants and Animals) अध्याय 20: पारिस्थितिकी प्रणालियों (Ecosystems) अध्याय 21: मानव शरीर - अंतःस्रावी, परिसंचरण और तंत्रिका तंत्र (Human Body - Endocrine, Circulatory and Nervous Systems) अध्याय 22: स्वास्थ्य एवं स्वच्छता (Health and Hygiene) अध्याय 23: खाद्य उत्पादन (Food Production)

AI अध्ययन मित्र

त्वरित शंका समाधान

ऊर्जा (Energy) में कोई संदेह या प्रश्न है? हमारे AI अध्ययन मित्र से तुरंत समझें।

STAY UPDATED

Get the Latest Updates

New test series, syllabus updates, exam notifications and more.

Syllabus
Updates
Exam
Alerts
Study
Resources