# Chapter: Work, Energy, and Power ⚡

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## **Work (W)**

[cite_start]Work is said to be done when a force applied to a body causes it to displace in the direction of the force. [cite: 2, 4, 5]

[cite_start]Alternatively, work can be defined as the **dot product** of the force vector ($\vec{F}$) and the displacement vector ($\vec{s}$). [cite: 8, 9]


[cite_start]If $\theta$ is the angle between the applied force and the direction of displacement, then the work done is calculated as: [cite: 10, 11, 12]
[cite_start]$$ W = \vec{F} \cdot \vec{s} = Fs\cos\theta $$ [cite: 15, 17]

### **Cases for Work Done**

**Case 1: Positive Work (Maximum)**
[cite_start]When the applied force is parallel to the direction of displacement, the angle $\theta = 0^\circ$. [cite: 18]
* [cite_start]$W = Fs\cos(0^\circ)$ [cite: 21]
* [cite_start]$W = Fs$ (This is the maximum possible work). [cite: 21]

**Case 2: Zero Work (Minimum)**
[cite_start]When the applied force is perpendicular to the direction of displacement, the angle $\theta = 90^\circ$. [cite: 23]
* [cite_start]$W = Fs\cos(90^\circ)$ [cite: 23]
* [cite_start]$W = 0$ (This is the minimum possible work). [cite: 23]
* [cite_start]**Note:** The gravitational force on the Moon as it orbits the Earth does no work because the force is always perpendicular to the Moon's direction of motion. [cite: 23] [cite_start]Similarly, the work done by centripetal force in any circular motion is zero. [cite: 370]

### **Work Done by a Variable Force**

[cite_start]When the force applied is not constant, the work done can be calculated by finding the area under the force-displacement graph. [cite: 23, 40]

**Derivation:**
1.  Suppose a body displaces from position $x_1$ to $x_2$ under a variable force. [cite_start]We can divide the path into a large number of very small displacements, $dx$. [cite: 24, 25]
2.  [cite_start]Over each small displacement $dx$, the force 'F' can be considered nearly constant. [cite: 25, 26]
3.  [cite_start]The small amount of work done ($\Delta W$) over this small displacement is the area of a small strip on the graph: [cite: 28, 29]
    [cite_start]$\Delta W = F \cdot dx$ [cite: 28]
4.  The total work done is the sum of the work done over all these small intervals. [cite_start]By making the strips infinitesimally small, this sum becomes an integral: [cite: 31, 33, 34, 35]
    [cite_start]$$ W = \int_{x_1}^{x_2} F \cdot dx $$ [cite: 36]
[cite_start]This integral represents the total area under the force-displacement curve from $x_1$ to $x_2$. [cite: 40, 41]

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## **Power (P)**

[cite_start]Power is defined as the rate at which work is done. [cite: 43, 45]
* [cite_start]**Formula**: $P = \frac{W}{t}$ [cite: 48]
* [cite_start]Since $W = Fs$, we can write $P = \frac{Fs}{t}$. [cite: 49]
* As velocity $v = \frac{s}{t}$, power can also be expressed as:
    [cite_start]$P = F \times v$ [cite: 51]
* [cite_start]**Unit**: The SI unit of power is the **Watt (W)**, which is equivalent to Joules per second (J/s) or Newton-meters per second (Nm/s). [cite: 52]

**Important Conversions:**
* [cite_start]1 Horsepower (HP) = 746 Watts [cite: 54]
* [cite_start]1 kilowatt (kW) = 1000 Watts [cite: 57]
* [cite_start]1 megawatt (MW) = $10^6$ Watts (Note: source has a typo of $10^8$) [cite: 58]

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## **Energy (E)**

[cite_start]Energy is the capacity of a body to do work. [cite: 61, 62, 63] [cite_start]Its SI unit is the **Joule (J)**. [cite: 63]

### **Mechanical Energy**
[cite_start]Mechanical energy is the energy possessed by a body due to its position, configuration (structure), or motion. [cite: 64, 68] [cite_start]It is the sum of a body's potential energy and kinetic energy. [cite: 69]
[cite_start]$$ ME = PE + KE $$ [cite: 71]

### **Potential Energy (PE)**
[cite_start]Potential energy is the energy stored in a body due to its position or configuration. [cite: 73, 75]
* [cite_start]**Examples**: Energy stored in a stretched rubber band, or the energy of a body placed at a certain height above the ground. [cite: 77, 79]

**Derivation of PE = mgh:**
* [cite_start]Consider a body of mass 'm' lifted to a vertical height 'h' against the force of gravity. [cite: 85, 86]
* The force required to lift the body is equal to its weight, $F = mg$.
* [cite_start]The work done on the body is: $W = \text{Force} \times \text{Displacement} = (mg) \times h = mgh$. [cite: 91]
* [cite_start]This work done against gravity is stored in the body as its gravitational potential energy. [cite: 98, 99]
    [cite_start]$$ PE = mgh $$ [cite: 101]

### **Kinetic Energy (KE)**
[cite_start]Kinetic energy is the energy possessed by a body due to its motion. [cite: 102, 103] [cite_start]It is equal to the amount of work done to bring a body from rest to its current velocity. [cite: 105]

**Derivation of KE = ½mv²:**
* [cite_start]Consider a body of mass 'm' initially at rest ($u=0$). [cite: 111, 116] [cite_start]A constant force 'F' is applied, causing it to accelerate and reach a velocity 'v' after traveling a distance 's'. [cite: 117, 119, 120]
* From the equation of motion, $v^2 = u^2 + 2as$. [cite_start]Since $u=0$, we have $v^2 = 2as$. [cite: 123, 125]
* [cite_start]This gives the acceleration, $a = \frac{v^2}{2s}$. [cite: 127]
* [cite_start]The work done on the body is $W = Fs$. [cite: 133]
* Using Newton's second law, $F=ma$:
    [cite_start]$W = (ma)s = m \left(\frac{v^2}{2s}\right)s$ [cite: 134, 135]
* [cite_start]This simplifies to $W = \frac{1}{2}mv^2$. [cite: 136]
* [cite_start]This work done is stored in the moving body as its kinetic energy. [cite: 141, 142]
    [cite_start]$$ KE = \frac{1}{2}mv^2 $$ [cite: 144]

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## **Principle of Conservation of Energy**

[cite_start]This fundamental principle states that **"Energy can neither be created nor be destroyed, but it can be transferred from one form to another."** [cite: 147, 148, 150] The total energy in an isolated system remains constant.

### **Proof for a Freely Falling Body**
[cite_start]Consider a body of mass 'm' at a height 'h' above the ground, falling freely under gravity. [cite: 152, 157, 159, 161]

* **At Point A (Top, height h):**
    * [cite_start]Initial velocity, $u = 0$. [cite: 166]
    * [cite_start]$PE = mgh$. [cite: 169]
    * [cite_start]$KE = \frac{1}{2}mu^2 = 0$. [cite: 170]
    * [cite_start]**Total Energy (TE) = $PE + KE = mgh + 0 = mgh$**. [cite: 171, 172]

* **At Point B (After falling distance x, height h-x):**
    * The new height is $h' = h-x$. [cite_start]So, $PE = mg(h-x) = mgh - mgx$. [cite: 175, 178]
    * [cite_start]The velocity $v'$ after falling distance x is found using $v'^2 = u^2 + 2gx \implies v'^2 = 2gx$. [cite: 176, 177]
    * [cite_start]$KE = \frac{1}{2}mv'^2 = \frac{1}{2}m(2gx) = mgx$. [cite: 178]
    * [cite_start]**Total Energy (TE) = $PE + KE = (mgh - mgx) + mgx = mgh$**. [cite: 179, 180]

* **At Point C (Just before hitting the ground, height 0):**
    * The height is $h'' = 0$. [cite_start]So, $PE = 0$. [cite: 184, 186]
    * [cite_start]The final velocity $v''$ after falling height h is found using $v''^2 = u^2 + 2gh \implies v''^2 = 2gh$. [cite: 185]
    * [cite_start]$KE = \frac{1}{2}mv''^2 = \frac{1}{2}m(2gh) = mgh$. [cite: 186]
    * [cite_start]**Total Energy (TE) = $PE + KE = 0 + mgh = mgh$**. [cite: 186]

[cite_start]Since the total energy at points A, B, and C is the same ($mgh$), the principle of conservation of energy is verified. [cite: 188]

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## **Types of Forces**

### **Conservative Force**
[cite_start]A force is said to be conservative if the work done by it in moving a body between two points is **independent of the path followed**. [cite: 192, 196] [cite_start]The work done by a conservative force in a closed loop (returning to the start) is always zero. [cite: 196]
* [cite_start]**Examples**: Gravitational force, electrostatic force. [cite: 200]

### **Non-Conservative Force**
[cite_start]A force is non-conservative if the work done by it **depends on the path taken**. [cite: 202]
* [cite_start]**Example**: Frictional forces. [cite: 205, 206]

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## **Collisions**

[cite_start]A collision is a mutual interaction between particles for a short duration. [cite: 207, 209]

### **Types of Collisions**

**1. Elastic Collision**
[cite_start]An elastic collision is one in which both the **total linear momentum** and the **total kinetic energy** of the system are conserved. [cite: 212, 214, 216] [cite_start]This typically occurs between atomic or subatomic particles. [cite: 215]
* [cite_start]$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (Momentum is conserved) [cite: 217]
* [cite_start]$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ (KE is conserved) [cite: 218]

**2. Inelastic Collision**
[cite_start]An inelastic collision is one in which the **total linear momentum is conserved**, but the **total kinetic energy is not conserved**. [cite: 213, 219, 221] Some kinetic energy is usually lost as heat, sound, or deformation. [cite_start]This is common in collisions between macroscopic (heavy) objects. [cite: 220]
* [cite_start]$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (Momentum is conserved) [cite: 222]
* [cite_start]$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 \neq \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ (KE is not conserved) [cite: 223]

### **Derivation: Final Velocities in a 1D Elastic Collision**
For a one-dimensional elastic collision between masses $m_1$ and $m_2$:
* [cite_start]From conservation of momentum: $m_1(u_1 - v_1) = m_2(v_2 - u_2)$ --- (i) [cite: 244]
* [cite_start]From conservation of kinetic energy: $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$ --- (ii) [cite: 245]
* Dividing (ii) by (i) gives: $u_1 + v_1 = v_2 + u_2$, which rearranges to:
    [cite_start]$u_1 - u_2 = v_2 - v_1$ [cite: 250]
    [cite_start]This important result means the **relative velocity of approach is equal to the relative velocity of separation**. [cite: 251]
* By solving these equations simultaneously, we can find the final velocities:
    [cite_start]$$ v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2 $$ [cite: 272, 273]
    [cite_start]$$ v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2 $$ [cite: 257]
* **Special Case (Equal Masses):** If $m_1 = m_2$, then $v_1 = u_2$ and $v_2 = u_1$. [cite_start]The bodies simply exchange their velocities. [cite: 274, 278, 283]

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## **Relationship Between Kinetic Energy and Momentum**

We can derive a direct relationship between kinetic energy (E) and linear momentum (p).
* [cite_start]$p = mv$ [cite: 292]
* [cite_start]$E = \frac{1}{2}mv^2$ [cite: 293]
* Multiplying the energy equation by $m/m$:
    [cite_start]$E = \frac{1}{2} \frac{m^2v^2}{m} = \frac{(mv)^2}{2m}$ [cite: 296, 299]
* This gives the final relation:
    [cite_start]$$ E = \frac{p^2}{2m} \quad \text{or} \quad p = \sqrt{2mE} $$ [cite: 301]

**Comparisons:**
1.  [cite_start]**If two bodies have the same momentum ($p_1=p_2$):** [cite: 316]
    Then $\frac{E_1}{E_2} = \frac{m_2}{m_1}$, which means **$E \propto \frac{1}{m}$**. [cite_start]The lighter body will have greater kinetic energy. [cite: 320, 321, 357]
2.  [cite_start]**If two bodies have the same kinetic energy ($E_1=E_2$):** [cite: 324]
    Then $\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}$, which means **$p \propto \sqrt{m}$**. [cite_start]The heavier body will have greater momentum. [cite: 328, 329, 437, 439]

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## **Short Answer Questions & Concepts**

**Q1: A light and heavy body have the same momentum. Which has greater KE?**
[cite_start]The lighter body will have greater kinetic energy. [cite: 357] [cite_start]Since $KE = p^2/(2m)$ and p is the same for both, KE is inversely proportional to mass ($KE \propto 1/m$). [cite: 367]

**Q2: How does the KE of a body change if its momentum is halved?**
[cite_start]The kinetic energy becomes one-fourth of its original value. [cite: 426, 427] [cite_start]Since $KE \propto p^2$, if the new momentum is $p' = p/2$, the new kinetic energy is $KE' = (p/2)^2 / (2m) = \frac{1}{4}(p^2/2m) = \frac{1}{4}KE$. [cite: 425]

**Q3: How does the KE of an object change if its momentum is doubled?**
[cite_start]The kinetic energy becomes four times its original value. [cite: 462] [cite_start]If $p' = 2p$, then $KE' = (2p)^2 / (2m) = 4(p^2/2m) = 4KE$. [cite: 459, 460]

**Q4: A stationary mass explodes into a heavy and a light fragment. Which has greater KE?**
[cite_start]The lighter fragment has greater kinetic energy. [cite: 400] [cite_start]By conservation of momentum, the two fragments have equal and opposite momenta ($p_1 = -p_2$). [cite: 341, 392] [cite_start]Since $KE = p^2/(2m)$, and p is the same, the fragment with the smaller mass ('m') must have the greater KE. [cite: 400]

**Q5: What happens to the KE of a bullet when it embeds in a block of wood?**
[cite_start]The kinetic energy of the bullet decreases significantly. [cite: 465] [cite_start]The initial KE is converted into other forms of energy, such as the work done to create a hole in the wood, heat, and sound energy. [cite: 466, 468, 471, 473]

**Q6: Does a man carrying a bucket on a level road do any work on the bucket?**
[cite_start]No, he does no mechanical work on the bucket. [cite: 407, 415] [cite_start]The upward force he applies to hold the bucket is perpendicular to the horizontal direction of displacement ($\theta=90^\circ$). [cite: 412] [cite_start]Therefore, $W = Fs\cos(90^\circ) = 0$. [cite: 412]