#### Class IX Work Power and Energy notes(Physics)

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**Work:** In our daily life anything that makes us tired is known as work. For example, reading, writing, painting, walking, etc.

In physics work (W) is said to be done, when a force (F) acts on the body and point of application of the force is displaced (s) in the direction of force.

Work done = force x displacement

W = F x s

(i) If the body is displaced in the same direction of force, Work done is positive

(ii) If the displacement is against a force, the work is done against the force. Work done is negative

(iii) If the displacement is perpendicular to the direction of the force, work done is zero.

**Unit of work**

Unit of work is joule (J). One joule of work is said to be done when a force of 1 Newton acting on a body displacing it by a distance of 1 m.

Larger units of work are

- i) kilojoules (1000 joule) ii) megajoule (10 lakh joule)

**ENERGY –** The energy of the body is defined as its capacity to do work

Unit of energy – Energy is measured in terms of work. Unit of energy is also joule. One joule of energy is required to do one joule of work

__Different forms of energy__

- Mechanical Energy

The energy used to displace a body or to change the position of the body or to deform the body is known as mechanical energy.

**Mechanical energy is of two types** (i) Potential energy (ii) Kinetic energy.

**POTENTIAL ENERGY**

The energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors like state of strain is called potential energy.

Example :The work done to lift a body above the ground level gives the potential energy of the body. Eg. Weightlifting.

Water stored in a reservoir has a large amount of potential energy due to which it can drive a water turbine when allowed to fall down. This is the principle of production of hydroelectric energy.

**Expression for potential energy of a body above the ground level**

Consider an object of mass m. It is raised through a height “h” meter from the ground.

By applying force F, The object gains energy to do the work done (w) on it.

Work done = force x displacement

w = F x h (Since F= m a , a = g , F = mg)

w = m g h

**KINETIC ENERGY- ** Energy possessed by an object due to its motion is called kinetic energy.

The kinetic energy of an object increases with its speed. Kinetic energy of an object moving with a velocity is equal to the work done on it to make it acquire that velocity

Example – Kinetic energy of a hammer is used to drive a nail into the wall. A bullet fired from a gun can penetrate into a target due to its kinetic energy.

Expression for kinetic energy:

Let a body (ball) of mass m is moving with an initial velocity v. If it is brought to rest by applying a retarding (opposing) force F, then it comes to rest by a displacement S.

Let, E_{k} = work done against the force used to stop it.

E_{k} = F x S —- (i)

But retarding force F = ma—–(ii)

Let initial velocity u = v, final velocity v = 0

From III equation of motion

v^{2} = u^{2} + 2aS

Applying, 0 = v^{2} – 2aS ( a is retardation)

2aS = v^{2}

Displacement, S = v^{2}/2a —- (iii)

Substituting (ii) and (iii) in (i)

E_{k} = ma x v^{2}/2a

E_{k} = 1/2 mv^{2}

**LAW OF CONSERVATION OF ENERGY**

Energy can neither be created nor destroyed, but it is transformed from one form to another. Alternatively, whenever energy gets transformed, the total energy remains unchanged.

Proof – Freely falling body

Consider a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.

At A,

Kinetic energy E_{k} = 0

Potential energy E_{p} = mgh

Total energy E = E_{p} + E_{k}

= mgh + 0

E = mgh

During the fall, the body is at a position B. The body has moved a distance x from A.

At B,

Velocity v^{2} = u^{2} + 2as

Applying, v^{2} = 0 + 2ax = 2ax

E_{k} = 1/2 mv^{2}

=1/2 m x 2gx

= mgx

Potential energy

E p = mg (h – x)

Total energy E = E_{p} + E_{k}

= mg (h-x) + mgx

= mgh – mgx + mgx

E = mgh

If the body reaches the position C.

At C,

Potential energy E _{p} = 0

Velocity of the body C is v^{2} = u^{2} + 2as

u = 0, a = g, s = h

Applying v^{2} = 0 + 2gh = 2gh

Kinetic energy Ek = 1/2 mv^{2} = 1/2x m x 2gh E_{k} = mgh

Total energy at C

E = E _{p} + E _{k}

E = 0 + m g h

E = m g h

Thus sum of potential and kinetic energy of freely falling body at all points remains same.

Power : Power is defined as the rate of doing work or work done per unit time

Power = work done/time taken

P = w / t

**UNIT OF POWER**

The unit of power is J/S known as watt, its symbol is W.

1 watt = 1 joule/1 second

1 W = 1 J S ^{-1}

1 kilowatt = 1000 watts

1 kW = 1000 W

1 kW = 1000 J /s.

Commercial unit of energy is kilowatt hour

__Average power__

Work done by a person or agent may be different at different intervals of time. Therefore the concept of average power is useful.

We obtain average power by dividing the total energy consumed by the total time taken.

Example 1. How much energy will be used when a hundred watt bulb is used for 10 hours?

Energy = 100 watt x 10 hour

= 1000 w h = 1kw h

I k w h is known as 1 unit.

One kilowatt hour means thousand watts of power is consumed in one hour.

1 kWh = 1 kW x 1 h

= 1000 W x 60 x 60 s

= 1000 Js^{-1} x 3600 s

= 3.6 x 10^{6} J

1 unit = 1 kilowatt hour = 3.6×10^{6} J

Example 1: An electric bulb of 60 W is used for 6 h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.

Solution: Power of electric bulb = 60 W = 0.06 kW. Time used, *t *= 6 h

Energy = power × time taken = 0.06 kW × 6 h = 0.36 kW h = 0.36 ‘units’.

The energy consumed by the bulb is 0.36 ‘units’.

**Click here NCERT Solution Work and Energy Class 9th**

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