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WORK POWER AND ENERGY – III

Work, Power, and Energy: Conservation of Energy, Types of Energy, Mass-Energy Equivalence & Power Mechanics

Welcome to Part 3 of our comprehensive physics series on Work, Power, and Energy. In this guide, we dive deep into the Law of Conservation of Mechanical Energy with mathematical derivations and diagrams, explore various forms of energy, examine Einstein’s Mass-Energy Equivalence, master the mechanics of Power, and test your understanding with 10 conceptual questions and 10 step-by-step solved numerical problems.

Table of Contents

  1. 1. Conservation of Mechanical Energy (Statement, Diagram & Mathematical Proof)
  2. 2. Law of Conservation of Energy (Explanation & Real-World Examples)
  3. 3. Different Types of Energy and Their Definitions
  4. 4. Einstein’s Mass-Energy Equivalence
  5. 5. Power: Definition, Formulas, Units, Dimensions & Nature
  6. 6. Units of Power & Inter-Conversion Relations
  7. 7. Conceptual Questions & Solved Numerical Examples

1. Statement and Explanation of Conservation of Mechanical Energy

Statement:

The Law of Conservation of Mechanical Energy states that if only conservative forces act upon a system, the total mechanical energy $E$ (sum of Kinetic Energy $K$ and Potential Energy $U$) remains constant over time.

$$E = K + U = \text{Constant}$$

Mathematical Proof: Motion under Gravity (Freely Falling Body)

Consider a body of mass $m$ dropped from rest from a height $H$ above the ground under the influence of gravity alone.

Step 1: At Position A (At initial height $H$)

• Initial velocity: $v_A = 0$

• Kinetic Energy ($K_A$):

$$K_A = \frac{1}{2} m v_A^2 = 0$$

• Potential Energy ($U_A$):

$$U_A = mgH$$

• Total Mechanical Energy ($E_A$):

$$E_A = K_A + U_A = 0 + mgH = mgH \quad \text{— (Equation 1)}$$

Step 2: At Position B (After falling a distance $x$, height above ground $= H – x$)

• Velocity at point B using $v^2 = u^2 + 2gs$:

$$v_B^2 = 0 + 2gx = 2gx$$

• Kinetic Energy ($K_B$):

$$K_B = \frac{1}{2} m v_B^2 = \frac{1}{2} m (2gx) = mgx$$

• Potential Energy ($U_B$):

$$U_B = mg(H – x) = mgH – mgx$$

• Total Mechanical Energy ($E_B$):

$$E_B = K_B + U_B = mgx + (mgH – mgx) = mgH \quad \text{— (Equation 2)}$$

Step 3: At Position C (Just before hitting ground, height $= 0$)

• Velocity at point C using $v^2 = u^2 + 2gH$:

$$v_C^2 = 0 + 2gH = 2gH$$

• Kinetic Energy ($K_C$):

$$K_C = \frac{1}{2} m v_C^2 = \frac{1}{2} m (2gH) = mgH$$

• Potential Energy ($U_C$):

$$U_C = mg(0) = 0$$

• Total Mechanical Energy ($E_C$):

$$E_C = K_C + U_C = mgH + 0 = mgH \quad \text{— (Equation 3)}$$

From Equations (1), (2), and (3):
$$E_A = E_B = E_C = mgH = \text{Constant}$$Thus, total mechanical energy remains conserved throughout the motion.

Ground Level (U = 0) H Point A: h = H, v = 0 E = K (0) + U (mgH) = mgH Point B: h = H – x E = K (mgx) + U (mg(H-x)) = mgH Point C: h = 0 E = K (mgH) + U (0) = mgH

Figure 1: Conservation of Mechanical Energy for a freely falling mass at different positions.

2. Law of Conservation of Energy: Explanation with Examples

The General Law of Conservation of Energy is a fundamental law of physics. It states that: Energy can neither be created nor destroyed; it can only transform from one form into another. The total energy of an isolated universe remains constant.

Real-World Energy Transformations:

1. Hydroelectric Power Plant: Potential energy of water stored in dams $\rightarrow$ Kinetic energy of flowing water $\rightarrow$ Rotational mechanical energy of turbines $\rightarrow$ Electrical energy from generators.

2. Electric Bulb: Electrical energy $\rightarrow$ Light energy and Thermal (heat) energy.

3. Microbial / Cellular Respiration: Chemical energy stored in food molecules (glucose) $\rightarrow$ Thermal energy and Mechanical work in muscle contraction.

4. Automobile Engines: Chemical energy of petrol/diesel $\rightarrow$ Thermal energy through combustion $\rightarrow$ Mechanical kinetic energy driving the wheels.

3. Different Types of Energies and Their Definitions

Energy manifests in numerous forms depending on the atomic, molecular, macro-mechanical, or field states of a system:

1. Mechanical Energy

The total energy possessed by a body due to its motion or position. It is the sum of Kinetic Energy ($K = \frac{1}{2}mv^2$) and Potential Energy ($U = mgh$ or $U = \frac{1}{2}kx^2$).

2. Thermal (Heat) Energy

The internal energy possessed by a system due to the random kinetic energy and vibration of its constituent atoms and molecules.

3. Chemical Energy

The energy stored within chemical bonds holding atoms together. Released or absorbed during chemical reactions (e.g., combustion, batteries).

4. Electrical Energy

The energy associated with the movement of electric charges (electrons) or stored within electric potential fields.

5. Radiant / Light Energy

The energy carried by electromagnetic waves (photons) capable of traveling through vacuum (e.g., solar radiation, X-rays, gamma rays).

6. Nuclear Energy

The binding energy stored inside atomic nuclei holding protons and neutrons together. Released during nuclear fission (splitting nuclei) or nuclear fusion (combining nuclei).

7. Sound Energy

Mechanical wave energy propagated through longitudinal pressure oscillations in a material medium (solid, liquid, or gas).

4. Mass-Energy Equivalence

Albert Einstein demonstrated through his Special Theory of Relativity (1905) that mass and energy are not separate quantities, but different manifestations of the same fundamental entity.

Einstein’s Equation:

$$E = m c^2$$

Where:

• $\mathbf{E}$ = Equivalent Energy (in Joules)

• $\mathbf{m}$ = Mass defect / converted mass (in kilograms)

• $\mathbf{c}$ = Speed of light in vacuum $\mathbf{(3 \times 10^8\text{ m/s})}$

Key Significance:

1. Even an extremely tiny mass $m$ corresponds to a huge quantity of energy because $c^2 = 9 \times 10^{16}\text{ m}^2/\text{s}^2$.

2. Energy Equivalent of $1\text{ kg}$ Mass:

$$E = (1\text{ kg}) \times (3 \times 10^8\text{ m/s})^2 = 9 \times 10^{16}\text{ Joules}$$

3. Atomic Mass Unit ($1\text{ amu}$ or $1\text{ u}$): Equivalent to $\mathbf{931.5\text{ MeV}}$ of energy.

5. Definition of Power: Formula, Units, Dimensional Formula & Nature

Definition of Power:

Power ($P$) is defined as the time rate of doing work or the rate of transfer/consumption of energy.

Mathematical Formulas:

• Average Power ($P_{\text{avg}}$): Total work done divided by total time elapsed.

$$P_{\text{avg}} = \frac{\Delta W}{\Delta t} = \frac{W}{t}$$

• Instantaneous Power ($P$): The instantaneous rate of work done as $\Delta t \to 0$.

$$P = \frac{dW}{dt} = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \vec{v} = F v \cos\theta$$

where $\vec{F}$ is the applied force vector, $\vec{v}$ is velocity vector, and $\theta$ is the angle between them.

Physical Nature, Units & Dimensions:

• Scalar or Vector? Power is a SCALAR quantity. It is defined by the dot product of two vector quantities ($\vec{F} \cdot \vec{v}$), resulting in a scalar magnitude with no direction.

• SI Unit: Watt ($\text{W}$), named after James Watt. $\mathbf{1\text{ W} = 1\text{ Joule/second (J/s)} = 1\text{ kg}\cdot\text{m}^2/\text{s}^3}$.

• CGS Unit: $\text{erg/s}$ ($\mathbf{1\text{ W} = 10^7\text{ erg/s}}$).

• Dimensional Formula: $\mathbf{[\text{M}^1 \text{L}^2 \text{T}^{-3}]}$.

6. Different Types of Units of Power & Their Conversions

Power is measured in various engineering, industrial, and electrical units depending on the practical context:

Common Units & Conversion Factors:

1. Kilowatt ($\text{kW}$): $\mathbf{1\text{ kW} = 10^3\text{ W} = 1000\text{ W}}$

2. Megawatt ($\text{MW}$): $\mathbf{1\text{ MW} = 10^6\text{ W} = 1000\text{ kW}}$

3. Gigawatt ($\text{GW}$): $\mathbf{1\text{ GW} = 10^9\text{ W} = 1000\text{ MW}}$

4. Horsepower (Imperial – $\text{hp}$): $\mathbf{1\text{ hp} = 746\text{ Watts} = 0.746\text{ kW}}$

5. Metric Horsepower ($\text{PS}$ or $\text{CV}$): $\mathbf{1\text{ metric hp} = 735.5\text{ Watts}}$

6. $\text{erg/s}$ to Watt: $\mathbf{1\text{ erg/s} = 10^{-7}\text{ W}}$

Important Distinction: Power vs. Energy Units

Kilowatt-hour ($\text{kWh}$) is a commercial unit of ENERGY (not power).
$$\mathbf{1\text{ kWh} = 1\text{ kW} \times 1\text{ hour} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ Joules}}$$

7. Conceptual Questions & Solved Numerical Examples

Part A: Conceptual Understanding Questions

Q1. Can mechanical energy be conserved in the presence of friction? Explain.

No. Friction is a non-conservative dissipative force. It converts part of the mechanical energy into thermal (heat) and sound energy, causing total mechanical energy ($K + U$) to decrease.

Q2. Is power a scalar or a vector quantity? Why?

Power is a scalar quantity. Mathematically, instantaneous power is the dot product (scalar product) of force and velocity vectors ($\vec{F} \cdot \vec{v}$), which results in a pure scalar magnitude.

Q3. Distinguish clearly between kilowatt ($\text{kW}$) and kilowatt-hour ($\text{kWh}$).

Kilowatt ($\text{kW}$) is a unit of Power (rate of energy consumption), whereas Kilowatt-hour ($\text{kWh}$) is a commercial unit of Energy ($1\text{ kWh} = 3.6 \times 10^6\text{ J}$).

Q4. Does a light body and a heavy body having equal kinetic energy have the same momentum?

No. Momentum $p = \sqrt{2mK}$. Since momentum is proportional to $\sqrt{m}$ for constant kinetic energy, the heavier body has greater momentum.

Q5. What work is done by centripetal force in uniform circular motion, and what is the instantaneous power delivered?

Work done is zero and instantaneous power delivered is zero because centripetal force is perpendicular to velocity ($\theta = 90^\circ \implies P = F v \cos 90^\circ = 0$).

Q6. State Einstein’s mass-energy relation and give its physical meaning.

$E = m c^2$. It implies that mass can be destroyed to release an equivalent amount of energy, and energy can be converted into mass.

Q7. Can kinetic energy of a system be negative? Explain.

No. Kinetic energy $K = \frac{1}{2}mv^2$. Since mass $m$ is positive and $v^2 \ge 0$, kinetic energy can never be negative.

Q8. A water pump rates $2\text{ kW}$. What does this rating indicate physically?

It means the pump is capable of doing $2000\text{ Joules}$ of electrical work or energy conversion every second.

Q9. What is the dimensional formula of power? Derive it from basic units.

$$\text{Power} = \frac{\text{Work}}{\text{Time}} = \frac{[\text{M}^1 \text{L}^2 \text{T}^{-2}]}{[\text{T}^1]} = \mathbf{[\text{M}^1 \text{L}^2 \text{T}^{-3}]}$$

Q10. How does doubling the velocity of a moving object affect its power required to maintain motion against constant drag?

Since $P = F \cdot v$, if resisting force $F$ is constant, doubling velocity ($v \to 2v$) doubles the required power ($P \to 2P$).

Part B: Solved Numerical Problems

Problem 1: Mechanical Energy Conservation in a Pendulum

Question: A simple pendulum bob of mass $0.5\text{ kg}$ is pulled aside to a vertical height of $0.8\text{ m}$ and released from rest. Calculate its speed at the lowest point. (Take $g = 9.8\text{ m/s}^2$)

Given: $m = 0.5\text{ kg}$, $h = 0.8\text{ m}$, $g = 9.8\text{ m/s}^2$

Formula: $mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$

Calculation:

$$v = \sqrt{2 \times 9.8 \times 0.8} = \sqrt{15.68} \approx 3.96\text{ m/s}$$

Final Answer: Speed at lowest point $= 3.96\text{ m/s}$

Problem 2: Mass-Energy Conversion ($1\text{ mg}$ Mass)

Question: Calculate the energy released in Joules and Kilowatt-hours ($\text{kWh}$) when $1\text{ mg}$ of mass is completely converted into energy.

Given: $m = 1\text{ mg} = 1 \times 10^{-6}\text{ kg}$, $c = 3 \times 10^8\text{ m/s}$

Step 1 (Joules):

$$E = m c^2 = (10^{-6}\text{ kg}) \times (3 \times 10^8\text{ m/s})^2 = 9 \times 10^{10}\text{ Joules}$$

Step 2 ($\text{kWh}$ conversion):

$$E_{\text{kWh}} = \frac{9 \times 10^{10}}{3.6 \times 10^6} = 2.5 \times 10^4\text{ kWh} = 25,000\text{ kWh}$$

Final Answer: $9 \times 10^{10}\text{ J}$ or $25,000\text{ kWh}$

Problem 3: Power of an Overhead Crane

Question: An electric crane lifts a load of $2000\text{ kg}$ through a vertical height of $15\text{ m}$ in $20\text{ seconds}$. Calculate the power of the crane in Kilowatts and Horsepower. (Take $g = 9.8\text{ m/s}^2$)

Given: $m = 2000\text{ kg}$, $h = 15\text{ m}$, $t = 20\text{ s}$, $g = 9.8\text{ m/s}^2$

Step 1 (Power in Watts):

$$P = \frac{mgh}{t} = \frac{2000 \times 9.8 \times 15}{20} = 14,700\text{ W} = 14.7\text{ kW}$$

Step 2 (Horsepower):

$$P_{\text{hp}} = \frac{14,700}{746} \approx 19.7\text{ hp}$$

Final Answer: $14.7\text{ kW}$ or $19.7\text{ hp}$

Problem 4: Instantaneous Power of a Constant Force

Question: A constant force $\vec{F} = (3\hat{i} + 4\hat{j} + 5\hat{k})\text{ N}$ acts on a body causing a velocity $\vec{v} = (2\hat{i} + 3\hat{j} – 4\hat{k})\text{ m/s}$. Calculate instantaneous power delivered.

Formula: $P = \vec{F} \cdot \vec{v} = F_x v_x + F_y v_y + F_z v_z$

Calculation:

$$P = (3 \times 2) + (4 \times 3) + (5 \times -4) = 6 + 12 – 20 = -2\text{ Watts}$$

Final Answer: $-2\text{ Watts}$ (indicates force opposes motion)

Problem 5: Water Pump Rate Calculation

Question: How many liters of water per minute can a $5\text{ kW}$ pump raise from a well $40\text{ m}$ deep? (Take $g = 10\text{ m/s}^2$ and mass of $1\text{ liter}$ water $= 1\text{ kg}$)

Given: Power $P = 5000\text{ W}$, $h = 40\text{ m}$, $t = 60\text{ s}$, $g = 10\text{ m/s}^2$

Formula: $P = \frac{mgh}{t} \implies m = \frac{P \cdot t}{g \cdot h}$

Calculation:

$$m = \frac{5000 \times 60}{10 \times 40} = \frac{300,000}{400} = 750\text{ kg}$$

Final Answer: $750\text{ Liters per minute}$

Problem 6: Car Engine Power on Inclined Road

Question: An automobile of mass $1000\text{ kg}$ moves up an incline of $1\text{ in } 20$ ($\sin\theta = 0.05$) at a constant speed of $72\text{ km/h}$. If total friction resistance is $300\text{ N}$, find the power of the engine in $\text{kW}$. (Take $g = 9.8\text{ m/s}^2$)

Given: $m = 1000\text{ kg}$, $v = 72\text{ km/h} = 20\text{ m/s}$, $\sin\theta = 0.05$, $f = 300\text{ N}$

Step 1 (Total Resistance Force):

$$F_{\text{total}} = mg\sin\theta + f = (1000 \times 9.8 \times 0.05) + 300 = 490 + 300 = 790\text{ N}$$

Step 2 (Power):

$$P = F_{\text{total}} \times v = 790 \times 20 = 15,800\text{ W} = 15.8\text{ kW}$$

Final Answer: $15.8\text{ kW}$

Problem 7: Conversions of Commercial Energy Units

Question: An electric heater of power $1500\text{ W}$ runs continuously for $8\text{ hours}$ daily. Find the electrical energy consumed per day in $\text{kWh}$ and in Joules.

Given: $P = 1.5\text{ kW}$, $t = 8\text{ hours}$

Step 1 (In $\text{kWh}$):

$$E_{\text{kWh}} = P(\text{kW}) \times t(\text{hours}) = 1.5 \times 8 = 12\text{ kWh (Units)}$$

Step 2 (In Joules):

$$E_{\text{Joules}} = 12 \times 3.6 \times 10^6 = 4.32 \times 10^7\text{ Joules}$$

Final Answer: $12\text{ kWh}$ or $4.32 \times 10^7\text{ Joules}$

Problem 8: Compressed Spring Conservation of Energy

Question: A spring of spring constant $k = 400\text{ N/m}$ is compressed by $0.1\text{ m}$. When released, it launches a mass of $0.2\text{ kg}$ horizontally along a smooth surface. Find the launch speed of mass.

Given: $k = 400\text{ N/m}$, $x = 0.1\text{ m}$, $m = 0.2\text{ kg}$

Conservation Equation: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2 \implies v = x\sqrt{\frac{k}{m}}$

Calculation:

$$v = 0.1 \times \sqrt{\frac{400}{0.2}} = 0.1 \times \sqrt{2000} = 0.1 \times 44.72 \approx 4.47\text{ m/s}$$

Final Answer: $4.47\text{ m/s}$

Problem 9: Horsepower to CGS Units Conversion

Question: Convert $5\text{ Horsepower}$ into CGS unit of power ($\text{erg/s}$).

Step 1 (Convert hp to Watts):

$$P = 5 \times 746\text{ W} = 3730\text{ Watts}$$

Step 2 (Convert Watts to $\text{erg/s}$):

Since $1\text{ W} = 10^7\text{ erg/s}$:

$$P = 3730 \times 10^7 = 3.73 \times 10^{10}\text{ erg/s}$$

Final Answer: $3.73 \times 10^{10}\text{ erg/s}$

Problem 10: Energy Equivalent of 1 Atomic Mass Unit ($1\text{ u}$)

Question: Calculate the energy equivalent of $1\text{ u}$ ($1.66 \times 10^{-27}\text{ kg}$) in Mega-electron-volts ($\text{MeV}$). (Take $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$)

Given: $m = 1.66 \times 10^{-27}\text{ kg}$, $c = 3 \times 10^8\text{ m/s}$

Step 1 (Energy in Joules):

$$E = (1.66 \times 10^{-27}) \times (3 \times 10^8)^2 = 1.494 \times 10^{-10}\text{ Joules}$$

Step 2 (Energy in $\text{eV}$ and $\text{MeV}$):

$$E_{\text{eV}} = \frac{1.494 \times 10^{-10}}{1.6 \times 10^{-19}} = 9.3375 \times 10^8\text{ eV}$$

$$E_{\text{MeV}} = \frac{9.3375 \times 10^8}{10^6} \approx 931.5\text{ MeV}$$

Final Answer: $931.5\text{ MeV}$