Level: Level

1. √

1. Work

Definition

Work is said to be done when a force acts on an object and the object is displaced in the direction of the applied force.

Mathematical Expression

$W = F \times s$

W=F×s

where

• $W$W = Work done

• $F$F = Force applied

• $s$s = Displacement of the object

Condition for Work

1. Force must act on the object.

2. Object must be displaced.

3. Displacement must have a component in the direction of the force.

Positive Work

If the force and displacement are in the same direction, work done is positive.
Example: Work done by a person in lifting an object upward.

$W = F \times s \times \cos 0° = F \times s$

W=F×s×cos0°=F×s

Negative Work

If the force and displacement are in opposite directions, work done is negative.
Example: Work done by friction on a moving body.

$W = F \times s \times \cos 180° = -F \times s$

W=F×s×cos180°=−F×s

Zero Work

When displacement is zero or perpendicular to the force,

$W = 0$

W=0.
Example: Work done by centripetal force in circular motion.

Unit of Work

• SI Unit: Joule (J)

• 1 Joule = Work done when 1 N force displaces a body by 1 m in the direction of force.

  $1\text{ J} = 1\text{ N} \times 1\text{ m}$1 J=1 N×1 m

2. Energy

Definition

Energy is the capacity to do work.

Unit of Energy

Same as work — Joule (J).

Forms of Energy

1. Mechanical Energy

   • Kinetic Energy

   • Potential Energy

2. Heat Energy

3. Light Energy

4. Chemical Energy

5. Electrical Energy

6. Nuclear Energy

3. Kinetic Energy (KE)

Definition

Energy possessed by a body due to its motion is called kinetic energy.

Expression

$KE = \frac{1}{2}mv^2$

KE=21​mv2

where

• $m$m = Mass of body (kg)

• $v$v = Velocity (m/s)

Derivation

Let a force

$F$

F act on an object causing displacement

$s$

s and final velocity

$v$

v,
Work done,

$W = F \times s$

W=F×s

From equation of motion:

$v^2 – u^2 = 2as$

v2−u2=2as

$\Rightarrow s = \frac{v^2 – u^2}{2a}$

⇒s=2av2−u2​

Substitute,

$F = ma$

F=ma

$W = ma \times \frac{v^2 – u^2}{2a} = \frac{1}{2}m(v^2 – u^2)$

W=ma×2av2−u2​=21​m(v2−u2)

If the body starts from rest (

$u = 0$

u=0),

$W = \frac{1}{2}mv^2$

W=21​mv2

Hence,

$KE = \frac{1}{2}mv^2$

KE=21​mv2

4. Potential Energy (PE)

Definition

Energy possessed by a body due to its position or configuration is called potential energy.

Example

• Water stored in a dam

• A stretched bow

Expression for Gravitational Potential Energy

$PE = mgh$

PE=mgh

where

• $m$m = Mass (kg)

• $g$g = Acceleration due to gravity (9.8 m/s²)

• $h$h = Height (m)

5. Mechanical Energy

The sum of kinetic and potential energy of a body is called mechanical energy.

$E_{\text{total}} = KE + PE$

Etotal​=KE+PE

6. Law of Conservation of Energy

Statement

Energy can neither be created nor destroyed. It can only be transformed from one form to another, but the total energy remains constant.

Example:

In the case of a freely falling body:

At height

$h$

h:

$E_{\text{total}} = mgh$

Etotal​=mgh (Potential energy maximum)

At ground:

$E_{\text{total}} = \frac{1}{2}mv^2$

Etotal​=21​mv2 (Kinetic energy maximum)

Total mechanical energy remains constant throughout the motion.

7. Power

Definition

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

Formula

$P = \frac{W}{t}$

P=tW​

where

• $P$P = Power

• $W$W = Work done

• $t$t = Time

Unit of Power

• SI Unit: Watt (W)

• 1 Watt = 1 Joule per second (1 W = 1 J/s)

Larger Units

• 1 kilowatt (kW) = 1000 W

• 1 megawatt (MW) = 10⁶ W

8. Commercial Unit of Energy

Kilowatt-hour (kWh)

Energy consumed by an appliance of power 1 kW in 1 hour.

$1\text{ kWh} = 1\text{ kW} \times 1\text{ h} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}$

1 kWh=1 kW×1 h=1000 W×3600 s=3.6×106 J

9. Example Numerical

Example 1

A force of

$10\text{ N}$

10 N moves an object through a distance of

$2\text{ m}$

2 m in the direction of force.
Find the work done.

$W = F \times s = 10 \times 2 = 20\text{ J}$

W=F×s=10×2=20 J

Example 2

Calculate the kinetic energy of a body of mass

$5\text{ kg}$

5 kg moving with a velocity of

$4\text{ m/s}$

4 m/s.

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times 4^2 = 40\text{ J}$

KE=21​mv2=21​×5×42=40 J

Example 3

Find the potential energy of a body of mass

$2\text{ kg}$

2 kg raised to a height of

$5\text{ m}$

5 m.

$PE = mgh = 2 \times 9.8 \times 5 = 98\text{ J}$

PE=mgh=2×9.8×5=98 J

✅ Key Points to Remember

• $1\text{ J} = 1\text{ N·m}$1 J=1 N\cdotpm

• $KE = \frac{1}{2}mv^2$KE=21​mv2

• $PE = mgh$PE=mgh

• $P = \frac{W}{t}$P=tW​

• $1\text{ kWh} = 3.6 \times 10^6\text{ J}$1 kWh=3.6×106 J

• Energy is conserved in all physical processes.