EXAMSPRINT

WORK AND ENERGY

Physics - Chapter 10

Introduction

Work, energy, and power are fundamental concepts in physics that help us understand and interpret many natural phenomena. All living beings need energy for various life processes, and machines require energy for their functioning.

Energy is the capacity to do work. Work and energy are closely related concepts with the same unit - Joule (J).

10.1 Work

The scientific definition of work differs from our everyday understanding of the term.

Scientific Definition of Work

Work is said to be done when:

  1. A force acts on an object, AND
  2. The object is displaced in the direction of the force
If either condition is not satisfied, no work is done scientifically.

Examples of No Work Done:

Examples of Work Done:

10.1.1 Work Done by a Constant Force

Object Force (F) Displacement (s)
Work Done = Force × Displacement
W = F × s

Unit of Work:

The SI unit of work is Joule (J) or Newton-metre (N⋅m)

1 Joule = 1 Newton × 1 metre
1 J = 1 N⋅m
examsprint.pages.dev
13
1

Positive and Negative Work

Positive Work F displacement Negative Work F displacement Force and displacement in same direction → Positive Work Force and displacement in opposite direction → Negative Work
Work is positive when force and displacement are in the same direction.
Work is negative when force and displacement are in opposite directions.

Example 1:

Problem: A force of 7 N acts on an object. The displacement is 8 m in the direction of force. Calculate work done.

Solution:
Given: F = 7 N, s = 8 m
Work done = F × s = 7 N × 8 m = 56 J

Example 2:

Problem: A porter lifts luggage of 15 kg from ground to his head 1.5 m above ground. Calculate work done.

Solution:
Given: m = 15 kg, h = 1.5 m, g = 10 m/s²
Force required = mg = 15 × 10 = 150 N
Work done = F × s = 150 N × 1.5 m = 225 J

10.2 Energy

Energy is the capacity to do work. An object that possesses energy can exert force on another object and transfer energy to it.

Energy = Capacity to do work
Unit: Joule (J) = Same as work

Forms of Energy

examsprint.pages.dev
2

10.2.1 Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion.

Moving Ball Moving Car Flying Arrow Objects with Kinetic Energy

Formula for Kinetic Energy

Ek = ½mv²

Where: m = mass of object, v = velocity of object

Example 3:

Problem: An object of mass 15 kg moves with velocity 4 m/s. Find its kinetic energy.

Solution:
Given: m = 15 kg, v = 4 m/s
Ek = ½mv² = ½ × 15 × (4)² = ½ × 15 × 16 = 120 J

Kinetic energy depends on both mass and velocity. When velocity doubles, kinetic energy becomes four times larger.

10.2.2 Potential Energy

Potential energy is the energy possessed by an object due to its position or configuration.

Stretched Spring h Ground Object at Height Compressed Ball

Types of Potential Energy:

examsprint.pages.dev
3

10.2.3 Gravitational Potential Energy

Ground Level (Reference) h₁ h₂ h₃ PE = mgh₁ PE = mgh₂ PE = mgh₃ Higher the height → Greater the Potential Energy
Gravitational Potential Energy = mgh
Ep = mgh

Where: m = mass, g = acceleration due to gravity (9.8 m/s²), h = height

Example 4:

Problem: Find energy possessed by 10 kg object at height 6 m. (g = 9.8 m/s²)

Solution:
Given: m = 10 kg, h = 6 m, g = 9.8 m/s²
Ep = mgh = 10 × 9.8 × 6 = 588 J

Potential energy depends on the reference level chosen. The work done by gravity depends only on the difference in heights, not the path taken.

10.2.4 Interconversion of Energy

Energy can be converted from one form to another but cannot be created or destroyed.

Examples of Energy Conversion:

examsprint.pages.dev
4

10.3 Law of Conservation of Energy

Energy can neither be created nor destroyed.
It can only be converted from one form to another.
Total Energy = Constant

Free Fall Example

At top: PE = mgh, KE = 0 Total Energy = mgh Midway: PE = mgh/2, KE = mgh/2 Total Energy = mgh Near bottom: PE ≈ 0, KE ≈ mgh Total Energy = mgh At ground: PE = 0, KE = mgh Total Energy = mgh FREE FALL
At any point during free fall:
PE + KE = mgh + ½mv² = Constant
Mechanical Energy = Kinetic Energy + Potential Energy = Constant (in absence of friction)

Energy Transformation Examples

Process Energy Transformation
Pendulum swing PE ⇌ KE
Burning fuel Chemical → Heat + Light
Electric motor Electrical → Mechanical
Solar panel Light → Electrical
Hydroelectric dam PE → KE → Electrical
examsprint.pages.dev
5

10.4 Power

Power is the rate of doing work or rate of energy transfer.

Power = Work Done / Time Taken
P = W/t

Unit of Power

SI Unit: Watt (W)
1 Watt = 1 Joule/second
1 W = 1 J/s

Larger unit: Kilowatt (kW)
1 kW = 1000 W
Person A Time: 10s Power: High Person B Time: 20s Power: Low Same Work, Different Time = Different Power

Example 5:

Problem: Two girls of weight 400 N each climb 8 m height. Girl A takes 20 s, Girl B takes 50 s. Find power of each.

Solution:
Work done by each = mgh = 400 × 8 = 3200 J

Power of Girl A = W/t = 3200/20 = 160 W
Power of Girl B = W/t = 3200/50 = 64 W

Higher power means work is done faster. Power indicates how quickly energy is transferred or consumed.
examsprint.pages.dev
6

Summary of Key Formulas

Concept Formula Unit
Work Done W = F × s Joule (J)
Kinetic Energy Ek = ½mv² Joule (J)
Potential Energy Ep = mgh Joule (J)
Power P = W/t Watt (W)
Conservation of Energy Etotal = PE + KE = Constant Joule (J)

Important Relationships

Work-Energy Theorem: Work done on an object equals change in its kinetic energy.
W = ΔKE = ½mv² - ½mu²
Units Conversion:
• 1 kJ = 1000 J
• 1 kW = 1000 W
• 1 kWh = 3.6 × 10⁶ J (commercial unit of energy)

Key Points to Remember

  1. Work requires both force and displacement in the direction of force
  2. Energy is the capacity to do work
  3. Kinetic energy increases with square of velocity
  4. Potential energy depends on height above reference level
  5. Total mechanical energy remains constant (without friction)
  6. Power measures how fast work is done
  7. Energy can be converted but never created or destroyed

Common Misconceptions

examsprint.pages.dev
7

🏆 PRACTICE QUESTIONS & ANSWERS

Q1. A force of 10 N displaces an object by 2 m at an angle of 60° to the force. Calculate the work done.
Answer:
Given: F = 10 N, s = 2 m, θ = 60°
Work done = F × s × cos θ = 10 × 2 × cos 60° = 10 × 2 × 0.5 = 10 J
Q2. An object of mass 5 kg is moving with velocity 10 m/s. Find its kinetic energy. If velocity becomes 20 m/s, what will be new kinetic energy?
Answer:
Initial: Ek1 = ½mv² = ½ × 5 × (10)² = 250 J
Final: Ek2 = ½mv² = ½ × 5 × (20)² = 1000 J
When velocity doubles, kinetic energy becomes 4 times.
Q3. A 2 kg object is at rest at height 10 m. It falls freely. Find its kinetic energy when it reaches height 4 m above ground. (g = 10 m/s²)
Answer:
Initial PE = mgh = 2 × 10 × 10 = 200 J
PE at 4 m height = 2 × 10 × 4 = 80 J
By conservation: PE + KE = 200 J
Therefore: KE = 200 - 80 = 120 J
Q4. A machine does 5000 J of work in 25 seconds. Calculate its power in watts and kilowatts.
Answer:
Given: W = 5000 J, t = 25 s
Power = W/t = 5000/25 = 200 W
Power in kW = 200/1000 = 0.2 kW
Q5. A car of mass 1000 kg moving at 20 m/s is brought to rest by applying brakes. Calculate the work done by braking force.
Answer:
Initial KE = ½mv² = ½ × 1000 × (20)² = 200,000 J
Final KE = 0 J (car at rest)
Work done by brakes = Final KE - Initial KE = 0 - 200,000 = -200,000 J
(Negative because braking force opposes motion)
examsprint.pages.dev
8

Detailed Examples & Problem Solving

Worked Example 1: Energy Conversion in Pendulum

Problem: A pendulum bob of mass 0.5 kg is raised to height 0.2 m and released. Find its velocity at the lowest point.

Solution:
At highest point: PE = mgh = 0.5 × 10 × 0.2 = 1 J, KE = 0
At lowest point: PE = 0, KE = ½mv²
By conservation: 1 = ½ × 0.5 × v²
v² = 4, therefore v = 2 m/s

Worked Example 2: Work Against Gravity

Problem: Calculate work done in raising 20 kg object from ground to 5 m height by taking two different paths: (a) Straight up (b) Along an inclined plane of length 10 m.

Solution:
In both cases: W = mgh = 20 × 10 × 5 = 1000 J
Note: Work done against gravity depends only on vertical height, not on path taken.

Worked Example 3: Power Calculation

Problem: A pump lifts 200 kg water to height 10 m in 20 seconds. Find power of pump.

Solution:
Work done = mgh = 200 × 10 × 10 = 20,000 J
Power = W/t = 20,000/20 = 1000 W = 1 kW

Problem Solving Strategy

  1. Identify what is given and what is to be found
  2. Choose appropriate formula based on the concept
  3. Convert units if necessary
  4. Substitute values in the formula
  5. Calculate and write answer with proper unit
  6. Check if the answer is reasonable

Energy in Daily Life

SUN Primary Source Fossil Fuels H₂O Hydro Wind Energy Sources
examsprint.pages.dev
9

Advanced Concepts

Work Done at an Angle

When force acts at an angle θ to the displacement:

W = F × s × cos θ
s F θ F cos θ Force at Angle θ

Special Cases:

Elastic Potential Energy

For a spring compressed or stretched by distance x:

Eelastic = ½kx²

Where k is the spring constant

Normal Length Compressed PE = ½kx² Stretched PE = ½kx²
examsprint.pages.dev
10

Energy Transformation Examples

Hydroelectric Power Plant Water at Height PE = mgh Turbine KE → Mechanical GEN Generator Mechanical → Electrical Electricity
Energy Transformation:
Gravitational PE → Kinetic Energy → Mechanical Energy → Electrical Energy

Real World Applications

Application Energy Type Work Done
Lifting weights Gravitational PE Against gravity
Car braking KE to Heat Negative work
Winding a clock Elastic PE Storing energy
Solar panels Light to Electrical Energy conversion
Photosynthesis Light to Chemical Energy storage

Important Constants

examsprint.pages.dev
11

Conceptual Understanding

Why Energy Cannot be Created or Destroyed?

The law of conservation of energy is a fundamental principle of nature. Energy transformations occur continuously around us, but the total energy of an isolated system remains constant. This principle governs all physical and chemical processes.

In real situations, some mechanical energy converts to heat due to friction. This doesn't violate conservation law - the total energy (including heat) remains constant.

Efficiency and Energy Loss

No machine is 100% efficient. Some energy always converts to unwanted forms like heat, sound, or vibration.

Efficiency = (Useful Energy Output / Total Energy Input) × 100%

Common Exam Questions Types

  1. Direct formula application: Calculate work, energy, or power using given values
  2. Energy conservation: Apply conservation law to find unknown quantities
  3. Conceptual questions: Identify when work is/isn't done
  4. Energy transformation: Trace energy changes in given scenarios
  5. Power comparison: Compare power of different agents doing same work

Tips for Problem Solving

Quick Reference Formulas

Quantity Formula Special Cases
Work W = F⋅s⋅cos θ W = Fs (θ = 0°), W = 0 (θ = 90°)
Kinetic Energy KE = ½mv² KE ∝ v² (quadratic relationship)
Potential Energy PE = mgh PE ∝ h (linear relationship)
Power P = W/t P = F⋅v (when F∥v)
Conservation Einitial = Efinal KE₁ + PE₁ = KE₂ + PE₂
examsprint.pages.dev
12

Chapter Summary

Work: Product of force and displacement in direction of force. Unit: Joule (J)
Energy: Capacity to do work. Exists in various forms and can be interconverted. Unit: Joule (J)
Kinetic Energy: Energy due to motion. Ek = ½mv²
Potential Energy: Energy due to position or configuration. Ep = mgh
Power: Rate of doing work or energy transfer. P = W/t. Unit: Watt (W)
Conservation of Energy: Energy cannot be created or destroyed, only transformed from one form to another.

Final Tips for Success

Study Smart, Succeed Faster!

Master these concepts through practice and understanding

examsprint.pages.dev