# Class 9 Work and Energy Notes

Notes Important Questions**Work and Energy Notes**

## Work:-

Work is said to be done if a force is applied on an object & it displaces in the direction of force applied.

- S.I. unit:-
**Joule (J)** - It is a
**scalar**quantity. - Work done = Force x displacement (
**W = FS**)

**1 Joule :-** The work done on an object is said to be 1 joule when a force of 1 N displaces it by 1 m along the line of action of the force.

#### Positive, Negative and Zero Work Done:-

**W = F S Cos θ**

## Energy:-

The capacity of doing work is called energy.

- S.I. unit :-
**Joule****(J)**

### Kinetic energy:-

The energy possessed by an object by virtue of its motion is known as kinetic energy. **Ex:- **Energy in moving fan.

#### Expression of kinetic energy:-

Consider an object having mass ‘m’ moving with uniform acceleration ‘a’. Let its initial velocity be ‘u’ & final velocity be ‘v’.

From **2 ^{nd } **law of motion, F = ma

W = FS

or, W = ma(v^{2}-u^{2})/2a **[**Since,** v ^{2}-u^{2} = 2aS **So,

**S = (v**

^{2}-u^{2})/2a]or, W = m(v^{2}-u^{2})/2 = ½ m(v^{2}-u^{2})

If u = 0 then, W = ½ m(v^{2}-0^{2}) = ½ mv^{2}

So, **E _{k} = ½**

**mv**

^{2}Work done= Change in kinetic energy

or, W = E_{kf} -E_{ki}

or,** W = ½** **mv ^{2} – ½**

**mu**

^{2}### Potential energy:-

The energy possessed by an object by virtue of its position or configuration is known as its potential energy. **Ex:-** Water stored in a dam.

#### Expression of potential energy:-

Consider an object having mass ‘m’ at a height ‘h’.

From **2 ^{nd} **law of motion,

F = ma = mg (Where, g is acceleration due to gravity.)

Now, W = FS

or, W = mgh [here,** S = h**]

So, **E _{p}= mgh**

#### Law of conservation of energy:-

It states that energy can neither be created nor be destroyed. It can only be converted from one form to another.

### Mechanical energy:-

The sum of kinetic energy & potential energy of an object is known as mechanical energy.

**Mechanical energy = Potential energy +Kinetic energy = constant**

### Mathematical proof of conservation of mechanical energy for a freely falling object:-

Consider an object having mass ‘m’ at height ‘h’.

At point A, velocity = 0, acceleration = g and height = h

K.E. = **½** mv^{2} = **½** m.0^{2} = 0

P.E. = mgh

M.E. = K.E.+P.E. = 0 + mgh = mgh

**At point B, **

Using, v^{2}-u^{2} = 2aS

or, v^{2} = 0^{2}+2gx [here,u=0, a=g & S=x]

or, v^{2} = 2gx

K.E.= **½** mv^{2} = **½** m.2gx = mgx

P.E.= mg(h-x) = mgh – mgx

So, M.E.=K.E.+P.E. = mgx + mgh – mgx = mgh

**At point C,**

Using, v^{2}-u^{2} = 2aS

or, v^{2}= u^{2}+2aS = 0^{2}+2gh = 2gh

K.E.= **½** mv^{2} = **½** m(2gh) = mgh

and P.E.= mgh = mg.0 = 0

so, M.E.= K.E.+P.E. = mgh+0 = mgh

Clearly, the total mechanical energy for a freely falling object is ‘mgh’ i.e., constant at every point.

## Power:-

The rate of doing work is called power.

- Power= work/time
**(P=W/t)** - S.I. unit :-
**J/S**or**Watt (W)**

**1 watt of power:-** The power of an agent is said to be 1 watt if it does 1 J work in 1 sec.

- 1 kW = 1000 W = 1000 J/s
- Commercial unit of energy :- kilowatt hour (
**kW h**)

## Conversion of commercial unit of energy to S.I. unit of energy:-

1 kwh = 1000 W x 3600 sec = 3.6 x10^{6} W sec = 3.6 x10^{6} J

or, **1 unit = 1 kwh = 3.6 x 10 ^{6} J**

## Comments

## Ayush prabhakar

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