Work, Energy and Power

SCALAR PRODUCT (DOT PRODUCT) OF VECTORS

**Definition:** The scalar product (or dot product) of two vectors **A** and **B** is a scalar quantity defined as:

**A · B = AB cos θ**

where:

A and B are magnitudes of vectors

θ is the angle between them

The result is always a scalar (no direction)

**Geometric Interpretation:**

B cos θ is the projection of vector **B** onto **A**

A cos θ is the projection of vector **A** onto **B**

**A · B** equals the product of one vector's magnitude and the other's component along it

**Properties of Scalar Product:**

1. **Commutative:** A · B = B · A

2. **Distributive:** A · (B + C) = A · B + A · C

3. **Scalar Multiplication:** A · (λB) = λ(A · B), where λ is a real number

**For Unit Vectors (î, ĵ, k̂):**

î · î = ĵ · ĵ = k̂ · k̂ = 1

î · ĵ = ĵ · k̂ = k̂ · î = 0

**In Component Form:**

If **A** = Aₓî + Aᵧĵ + Aᵤk̂ and **B** = Bₓî + Bᵧĵ + Bᵤk̂, then:

**A · B = AₓBₓ + AᵧBᵧ + AᵤBᵤ**

**Important Outcomes:**

1. **Magnitude of a vector:** |**A**| = √(A · A) = √(Aₓ² + Aᵧ² + Aᵤ²)

2. **Perpendicular vectors:** If A · B = 0, then vectors are perpendicular

3. **Finding angle:** cos θ = (A · B)/(AB)

**Example:** Find angle between **F** = 3î + 4ĵ - 5k̂ and **d** = 5î + 4ĵ + 3k̂

**F · d** = 3(5) + 4(4) + (-5)(3) = 15 + 16 - 15 = 16 units

|**F**| = √(9 + 16 + 25) = √50 units

|**d**| = √(25 + 16 + 9) = √50 units

cos θ = 16/(√50 × √50) = 16/50 = 0.32

θ = cos⁻¹(0.32) ≈ 71.3°

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NOTIONS OF WORK AND KINETIC ENERGY: THE WORK-ENERGY THEOREM

**Derivation from Kinematics:**

Starting with kinematic equation: **v² - u² = 2as**

Multiply both sides by m/2:

(m/2)v² - (m/2)u² = mas

(m/2)v² - (m/2)u² = Fs (using F = ma)

**In Three Dimensions (Vector Form):**

v² - u² = 2**a** · **d**

Multiplying by m/2:

**(1/2)mv² - (1/2)mu² = m(**a** · **d**) = **F** · **d****

**Statement of Work-Energy Theorem:**

The change in kinetic energy of a particle equals the work done on it by the net force:

**Kf - Ki = W**

or **ΔK = W**

**Physical Meaning:** Work done by net force changes the kinetic energy of an object. Positive work increases KE; negative work decreases it.

**Exam Example:** A raindrop of mass 1.00 g falls from height 1.00 km and hits ground with speed 50.0 m/s.

Change in KE: ΔK = (1/2)mv² - 0 = (1/2)(10⁻³)(50)² = 1.25 J

Work by gravity: Wg = mgh = 10⁻³ × 10 × 10³ = 10 J

Work by air resistance: Wr = ΔK - Wg = 1.25 - 10 = -8.75 J (negative, opposes motion)

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WORK

**Definition:** Work done by a force **F** over a displacement **d** is:

**W = F · d = Fd cos θ**

where θ is angle between force and displacement.

**Also expressed as:**

W = (F cos θ) × d = component of force along displacement × displacement

W = F × d_component = force × component of displacement along force

**Dimensions:** [ML²T⁻²]

**SI Unit:** **Joule (J)** = 1 N·m = 1 kg·m²/s²

**Alternative Units:**

1 erg = 10⁻⁷ J

1 calorie = 4.186 J

1 kWh = 3.6 × 10⁶ J

1 eV = 1.6 × 10⁻¹⁹ J

**When is NO Work Done?**

1. **Zero displacement:** Weightlifter holding 150 kg mass on shoulder for 30 s does zero work (no displacement)

2. **Zero force:** Block on frictionless surface moving with constant velocity (no net horizontal force)

3. **Force ⊥ Displacement:** θ = 90°, cos 90° = 0, so W = 0

Moon in circular orbit: gravitational force (radial) ⊥ displacement (tangential)

Normal force on object moving on horizontal surface

**Sign of Work:**

**Positive work (0° ≤ θ < 90°):** Force has component along displacement, increases energy

**Negative work (90° < θ ≤ 180°):** Force opposes displacement, decreases energy

**Friction always does negative work** (θ = 180°, cos 180° = -1)

**Example:** Cyclist skids to stop in 10 m. Stopping force = 200 N opposite to motion.

W = Fd cos 180° = 200 × 10 × (-1) = -2000 J

Negative work by friction brings cycle to halt.

**Important Note (Newton's Third Law):** Work done by force of A on B is NOT equal and opposite to work by B on A. If B doesn't move, work done on B is zero (even if equal/opposite force acts). In above example, road does -2000 J work on cycle, but cycle does 0 J work on road (road doesn't move).

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KINETIC ENERGY

**Definition:** Kinetic energy of object with mass m and velocity v is:

**K = (1/2)mv² = (1/2)m(**v** · **v**)**

**Key Points:**

**Scalar quantity** (always positive or zero)

**Always depends on magnitude of velocity** (not direction)

Measures work an object can do by virtue of its motion

Can be thought of as "energy of motion"

**Physical Significance:**

Kinetic energy represents the capacity of a moving object to do work on surroundings. A fast-flowing stream does work grinding corn; sailing ships use wind's kinetic energy.

**Typical Kinetic Energy Values:**

Electron in Cu wire at room temperature: ~10⁻²⁰ J

Raindrop (1 mm radius) falling: ~10⁻⁶ J

Cricket ball (100 g) at 10 m/s: ~5 J

Car (1000 kg) at 20 m/s: ~2 × 10⁵ J

Bullet (50 g) at 200 m/s: ~1000 J

Jumbo jet at takeoff: ~10¹⁰ J

**Example:** Bullet of mass 50 g fired at 200 m/s through plywood, emerges with 10% initial kinetic energy.

Initial KE: Ki = (1/2)(0.05)(200)² = 1000 J

Final KE: Kf = 0.1 × 1000 = 100 J

Final velocity: vf = √(2Kf/m) = √(2 × 100/0.05) = √4000 = 63.2 m/s

Note: Speed reduced by ~68% but energy reduced by 90% (quadratic relationship).

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WORK DONE BY A VARIABLE FORCE

**Problem:** Most real forces vary with position. How to calculate work?

**Method: Integration Approach**

Divide displacement into small intervals Δx where F(x) is approximately constant:

**ΔW = F(x)Δx** (small element of work)

Total work: **W ≈ ΣF(x)Δx** (sum from xi to xf)

Taking limit as Δx → 0:

**W = ∫[xi to xf] F(x) dx**

**Graphical Interpretation:**

Plot F vs x

Work = area under the curve between xi and xf

Above x-axis = positive work

Below x-axis = negative work

**For Multiple Dimensions:**

**W = ∫ **F** · d**r****

where d**r** is infinitesimal displacement vector.

**Example:** Woman pushes trunk 20 m on rough platform.

First 10 m: F = 100 N (constant)

Next 10 m: F decreases linearly from 100 N to 50 N

Friction force: f = -50 N (constant, opposing motion)

Work by applied force:

Rectangle (0-10 m): W₁ = 100 × 10 = 1000 J

Trapezium (10-20 m): W₂ = (1/2)(100 + 50) × 10 = 750 J

**Total: W = 1000 + 750 = 1750 J**

Work by friction:

W = -50 × 20 = -1000 J (negative, full 20 m)

Net work = 1750 - 1000 = 750 J

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WORK-ENERGY THEOREM FOR VARIABLE FORCE

**Derivation:**

Starting from Newton's Second Law:

dK/dt = d/dt(1/2 mv²) = m(dv/dt) × v = F × v = F(dx/dt)

Therefore: **dK = F dx**

Integrating from xi to xf:

**∫[Ki to Kf] dK = ∫[xi to xf] F dx**

**Kf - Ki = ∫[xi to xf] F(x) dx = W**

**Therefore: ΔK = W**

**or Kf - Ki = W**

This is the **Work-Energy Theorem** for variable force.

**Important Notes:**

1. **Integral form of Newton's 2nd Law:** WE theorem integrates F = ma over distance, losing explicit time information

2. **Scalar form:** Works in one dimension; for 3D, use **F** · d**r** (directionality information compressed into scalar)

3. **Applies to any force:** Constant, variable, conservative, or non-conservative

**Example:** Block (m = 1 kg) enters rough patch (x = 0.1 to 2.01 m) with vi = 2 m/s.

Retarding force: Fr = -k/x = -0.5/x (where k = 0.5 J)

Using work-energy theorem:

Kf = Ki + ∫[0.1 to 2.01] (-0.5/x) dx

Kf = (1/2)(1)(2)² - 0.5[ln x] from 0.1 to 2.01

Kf = 2 - 0.5 ln(2.01/0.1) = 2 - 0.5 ln(20.1)

Kf = 2 - 0.5(3.0) = 2 - 1.5 = **0.5 J**

Final speed: vf = √(2Kf/m) = √(2 × 0.5/1) = **1 m/s**

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CONCEPT OF POTENTIAL ENERGY

**Definition:** Potential energy is the "stored energy" possessed by an object by virtue of its position or configuration. It represents the work done against a conservative force to place the object in that position.

**Gravitational Potential Energy:**

Near Earth's surface (h << RE), gravitational force F = -mg (upward positive).

Work done by external agent against gravity to raise object from ground to height h:

**Wext = mgh**

This stored energy is **gravitational potential energy:**

**V(h) = mgh**

Taking ground (h = 0) as reference, V = 0.

**Relationship Between Force and Potential Energy:**

**F = -dV/dh**

Negative sign: Force opposes increase in potential energy.

For gravity: F = -d(mgh)/dh = -mg ✓

**Conservation During Free Fall:**

When object of mass m is released from height h:

Initial energy: K = 0, V = mgh

At ground: K = (1/2)mv², V = 0

From kinematics: v² = 2gh

Therefore: (1/2)mv² = mgh ✓

**Potential energy converts to kinetic energy.**

**Conditions for Defining Potential Energy:**

Not all forces have associated potential energy. **Conservative forces** have:

**F(x) = -dV(x)/dx**

Equivalently: **∫ F(x)dx = -(Vf - Vi) = -ΔV**

**Work depends only on initial and final positions,** not path taken.

**Conservative Forces:**

Gravity (near Earth)

Elastic spring force

Electric force (for fixed charges)

**Non-conservative Forces:**

Friction

Air resistance

Tension (variable)

**Physical Meaning:** For conservative forces, work done gets "stored" as potential energy. When constraints removed, manifests as kinetic energy.

**Example (Inclined Plane):** Object released from rest at top of smooth inclined plane of height h (any angle θ).

At bottom: v = √(2gh)

This is **independent of angle.** All gravitational PE converts to KE.

Initial: K = 0, V = mgh (taking bottom as reference)

Final: K = (1/2)mv² = mgh, V = 0

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CONSERVATION OF MECHANICAL ENERGY

**Definition:** Mechanical energy E is sum of kinetic and potential energy:

**E = K + V = (1/2)mv² + V(x)**

**Law of Conservation of Mechanical Energy:**

For a system under action of **conservative forces only:**

**E = K + V = constant**

**or Ei = Ef**

**or Ki + Vi = Kf + Vf**

**Proof:**

From work-energy theorem: ΔK = W

For conservative force: W = -ΔV

Therefore: ΔK = -ΔV

or: ΔK + ΔV = 0

or: **d(K + V)/dx = 0**

This means **K + V = constant**

**Important Points:**

1. **Applies only for conservative forces:** If friction/air resistance present, mechanical energy decreases

2. **Isolates the system:** No external work done on the system by non-conservative forces

3. **Energy transforms:** K ↔ V, but total remains constant

**Graphical Representation:**

For object in gravitational field, plotting K and V vs height:

As h increases: V increases linearly, K decreases

Graphs intersect at total energy line

At turning point: K = 0, V = E (maximum)

At ground: K = E, V = 0 (if V(0) = 0)

**Example 1 (Vertical Motion):** Ball thrown upward with initial speed v₀.

Initial (at ground, h = 0): K = (1/2)mv₀², V = 0, E = (1/2)mv₀²

At maximum height h: K = 0, V = mgh, E = mgh

By conservation: (1/2)mv₀² = mgh

Therefore: **h = v₀²/(2g)**

At any intermediate height h':

K + V = (1/2)mv² + mgh' = (1/2)mv₀²

Therefore: **v = √(v₀² - 2gh')** ✓

**Example 2 (Frictionless Incline):** Block slides down smooth incline of height h, initial velocity = 0.

E = K + V = (1/2)mv² + mgh' = mgh (constant)

At bottom (h' = 0): (1/2)mv² = mgh, so **v = √(2gh)**

Independent of angle (unlike time)!

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POTENTIAL ENERGY OF A SPRING

**Hooke's Law:** Spring force proportional to extension/compression:

**F = -kx**

where:

k = spring constant (N/m)

x = displacement from natural length (m)

Negative sign: restoring force (opposes displacement)

**Derivation of Spring PE:**

Work done by external agent against spring force to extend by x:

dW = -F·dx = -(-kx)dx = kx dx

Integrating from 0 to x:

**V(x) = ∫[0 to x] kx dx = (1/2)kx²**

**Spring Potential Energy: V = (1/2)kx²**

**Verification:**

F = -dV/dx = -d/dx(1/2 kx²) = -kx ✓

**Characteristic of Spring PE:**

1. Always positive (always V ≥ 0)

2. Maximum at maximum compression or extension

3. Zero at natural length (reference point)

4. Parabolic function of displacement

**Energy Conservation in Spring System:**

For mass m attached to spring, initial velocity v₀, initial extension x₀:

**E = (1/2)mv² + (1/2)kx² = constant**

At turning point: v = 0, x = xmax

**(1/2)mv₀² + (1/2)kx₀² = (1/2)kxmax²**

**Example:** Spring (k = 100 N/m) compressed 10 cm releases block (m = 2 kg).

Initial energy: E = (1/2)(100)(0.1)² = 0.5 J

On frictionless surface, when spring returns to natural length:

All PE converts to KE

(1/2)(2)v² = 0.5

v = √(0.5) = **0.707 m/s**

If friction force f = 2 N opposes motion over 10 cm:

Work by friction = -2 × 0.1 = -0.2 J

Final KE = 0.5 - 0.2 = 0.3 J

v = √(0.3) = **0.548 m/s**

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POWER

**Definition:** Power is the rate of doing work; measure of how quickly work is done.

**P = W/t** (average power)

**P = dW/dt** (instantaneous power)

**From W = F · d, and v = d/dt:**

**P = F · v** (instantaneous power)

**or P = Fv cos θ** (scalar form)

**Dimensions:** [ML²T⁻³]

**SI Unit:** **Watt (W)** = J/s = kg·m²/s³

**Other Units:**

1 kW = 10³ W

1 MW = 10⁶ W

1 horse power (hp) = 746 W

1 kWh = 3.6 × 10⁶ J (energy unit, not power)

**Physical Meaning:** Power measures how fast energy is being transferred. Same work done in less time = higher power.

**Average vs Instantaneous Power:**

**Average:** Pave = ΔW/Δt (useful for overall energy budget)

**Instantaneous:** P = dW/dt (at any particular moment)

**For Constant Force:**

P = Fv cos θ

When F parallel to v (θ = 0°): **P = Fv**

When F perpendicular to v (θ = 90°): **P = 0**

**Example 1 (Constant Velocity):** Car (mass 1000 kg) driven at 20 m/s. If friction force = 2000 N.

Applied force = 2000 N (for constant velocity)

Power required: P = Fv = 2000 × 20 = **40,000 W = 40 kW**

**Example 2 (Accelerating Car):** Car accelerates from rest to 20 m/s in 10 s.

Average power: Pave = ΔK/Δt = (1/2)mv²/t = (1/2)(1000)(400)/10 = **20 kW**

**Example 3 (Climbing Stairs):** Person (60 kg) climbs stairs to height 5 m in 10 s.

Work done: W = mgh = 60 × 10 × 5 = 3000 J

Power: P = W/t = 3000/10 = **300 W**

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COLLISIONS

**Definition:** Collision is an event where two or more bodies exert large forces on each other in a very short time.

**Key Assumptions:**

1. External forces negligible compared to collision forces

2. Time duration very short

3. Momentum conserved

**Types of Collisions:**

**1. Elastic Collision:**

Kinetic energy conserved

Momentum conserved

Objects may stick together or separate

**Before:** K₁ + K₂ = Ki (total)

**After:** K₁' + K₂' = Kf (total), where Kf = Ki

**2. Inelastic Collision:**

Kinetic energy NOT conserved (lost to heat, sound, deformation)

Momentum conserved

**Before:** K₁ + K₂ = Ki

**After:** K₁' + K₂' = Kf, where Kf < Ki

**3. Perfectly Inelastic (Perfectly Plastic) Collision:**

Maximum KE loss

Objects stick together

**After collision:** both move with same velocity

**m₁v₁ + m₂v₂ = (m₁ + m₂)v**

**Law of Conservation of Momentum (applies to all collisions):**

**Total momentum before = Total momentum after**

**m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂**

where u = velocity before, v = velocity after

**One-Dimensional Elastic Collision (Head-on):**

Before: m₁ moving at u₁, m₂ moving at u₂

After: m₁ at v₁, m₂ at v₂

Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ... (1)

Conservation of KE: (1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂² ... (2)

**Solving (1) and (2):**

**v₁ = [(m₁ - m₂)u₁ + 2m₂u₂]/(m₁ + m₂)**

**v₂ = [(m₂ - m₁)u₂ + 2m₁u₁]/(m₁ + m₂)**

**Special Cases:**

**Case 1: m₁ = m₂ = m, u₂ = 0 (equal masses, one at rest)**

v₁ = 0

v₂ = u₁

**Velocities exchange completely**

**Case 2: m₂ >> m₁, u₂ = 0 (light ball hits heavy stationary wall)**

v₁ ≈ -u₁

v₂ ≈ 0

**Light ball rebounds; wall doesn't move**

**Case 3: m₁ >> m₂, u₂ = 0 (heavy object hits light ball)**

v₁ ≈ u₁

v₂ ≈ 2u₁

**Heavy object continues; light ball flies off at ~2u₁**

**Perfectly Inelastic Collision Example:**

Car (1000 kg) at 10 m/s collides head-on with stationary truck (2000 kg).

Before: p = 1000 × 10 + 2000 × 0 = 10,000 kg·m/s

After: (1000 + 2000)v = 10,000

v = 10,000/3000 = **3.33 m/s** (combined velocity)

KE lost: ΔK = (1/2)(1000)(10)² - (1/2)(3000)(3.33)²

= 50,000 - 16,667 = **33,333 J** (lost to deformation, heat)

**Coefficient of Restitution (e):**

**e = (relative velocity of separation)/(relative velocity of approach)**

**e = (v₂ - v₁)/(u₁ - u₂)**

e = 1: perfectly elastic

0 < e < 1: inelastic

e = 0: perfectly inelastic

For elastic collision: KE = (1/2)e²m₁m₂(u₁ - u₂)²/(m₁ + m₂)

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IMPORTANT FORMULAE SUMMARY

| Quantity | Formula | SI Unit | Notes |

|----------|---------|---------|-------|

| Scalar Product | **A · B** = AB cos θ | - | Component form: AₓBₓ + AᵧBᵧ + AᵤBᵤ |

| Work | W = **F** · **d** = Fd cos θ | J | Negative if force opposes displacement |

| Kinetic Energy | K = (1/2)mv² | J | Always positive; depends on v² |

| Work-Energy Theorem | ΔK = W | J | Change in KE = work by net force |

| Variable Force Work | W = ∫F(x)dx | J | Integration over displacement |

| Gravitational PE | V(h) = mgh | J | Reference at h = 0 |

| Spring PE | V(x) = (1/2)kx² | J | Reference at natural length |

| Total Mechanical Energy | E = K + V | J | Constant for conservative forces |

| Power (Average) | P = W/t | W | Rate of energy transfer |

| Power (Instantaneous) | P = **F** · **v** = Fv cos θ | W | Can vary with time |

| Momentum Conservation | m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ | kg·m/s | All collisions |

| Elastic Collision (1D) | v₁ = [(m₁-m₂)u₁ + 2m₂u₂]/(m₁+m₂) | m/s | KE conserved |

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KEY EXAM POINTS

1. **Work is scalar, depends on force component along displacement** — pushing perpendicular to motion does no work

2. **Kinetic energy always (1/2)mv², never (1/2)mu²** — depends on magnitude only

3. **Conservative forces (gravity, spring) have associated PE** — friction does not

4. **Mechanical energy conserved only for conservative forces** — friction → mechanical energy lost

5. **Power = rate of doing work** — same work in less time = more power

6. **Momentum conserved in all collisions** — KE only in elastic collisions

7. **Negative work decreases KE** — friction opposes motion, removes energy

8. **Spring PE = (1/2)kx², always positive** — zero at natural length only

9. **F = -dV/dx** — force is negative gradient of PE

10. **Scalar product gives projection information** — crucial for component calculations