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**Work energy and power**

**Work**

Work is said to be done when we applied a force and body moves in the direction of applied force.

**Work done by a constant force**

It is measured as the dot product of force and displacement i.e. W =

W = FS cos

**Example**

A force of 5 N acts on s body such that the direction of the force with horizontal is 60 . If the body moves horizontal through a distance of 2 m .

Find the work done.

**Solution**

F = 5N , s = 2m = 60

W = FS cos = 5 = 5 2 = 5 j

**Dimension and unit of work**

Work = Fs (MLT^{-2}) [L] = [ML^{2} T^{-2}]

**Dimension and unit of work**

Work = F S = [MLT^{-2}] [L] = [ML^{2}T^{-2}__]__

**The unit of work are of two types**

__Absolute units__

** In SI** joule = Nm

** In CGS** erg = dyne cm

**Relation in between joule and erg**

1J = 11 l m

1J = 10^{5 }dyne 10^{2} cm

1J = 10^{6+1 }dyne cm

1J = 10^{7} erg

__Gravitational unit__

** In SI** kgm 1 kgm = 9.8J

** In cgs** gram cm (gcm ) 1 gcm = 980ergs

__Types of work done __

W = FS cos__Positive work done__

If < 90 then work done will be positive eg . when a string is stretened then he work done by stretching force as positive .

__Negative work done__

If > 90 cos is negative

Eg work done by frictional force always negative

–__Zero work done__

If = 90 cos 90 = 0

S = 0 w = F 0 = 0

Eg (i) when we fail to move a heavy stone how ever hard we may try work done by using zero (ii) when a coolie carrying some some load on his head moves on a horizontal plate form = 90

**Work done by variable force**

**Analytical method**

We want to calculate work done from A tO B S_{A} and S_{B} the distance of A and B w . r .to reference point P and Q are v . close the distance between then is ds

Work done from P to Q is

Dw = total work done

W

**Graphical method **

Small amount of work done in moving the body from P to Q in

Dw = fdx = Ps PQ = area of strip PQRS

Total work done in moving the body grom A to B is given by

W = = = __area of ABCD__

__Conservation force__A force is said to be conservation . of work done by or against the force depends only initial and final position of the does not depends the path followed by the body .

In fig (a) body being raised vertically upwards

= mg n = mgh

In fig (b) h = n_{1} +n_{2} + n_{3}

= w_{1} +w_{2} + w_{3} = mg (n_{1} +n_{2} + n_{3}) = mgh

= mgh ——(ii)

** In fig C** – the body being taken along a smooth inclined plane C B of height AB = h

= F CB = mg sin CB = mg

= mgh

**Properties of conservation forces**

- Work done aganist this forces does not not depnds the path followeed by the body .
- It depends only intial and final position
- Work done aganist this force in a close path is zero.

**Power**

__ __

Power of a body is defined as the role which the body can do the work .

P = = FV cos

P = unit – watt

1 H. P = 746 watt

Effecienry = 100 P_{o } – output

P_{1n } – input power

**Example**

An engine pumps up 100 of water through a height of 10 m is 5 sec . if the efficiency of the engine 60% . Calculate the [power of the engine in kw g = 10 m / s^{2}

__ __

**Solution**

= P_{o} = = = 200w

Pi = ? = 60%

100 = 100 = 3.33 kw

**Kinetic Energy**

Energy possessed by the body by virtue of motion .

Eg (i) moving cycle having K E

(ii) wind energy work on the KE of air

**Formula for KE**

**There are two methods**

- The amount of work done in stopping the moving body .
- The amount of work done in going the present velocity to the body from the state of rest

Here we consider II ^{nd} method

v^{2} = u2 + 2as

v^{2 } = o + 2as a =

= FS = mas = m S = mv^{2}

= KE = mv^{2}

^{ }

**Calculate Method**

Suppose ds – V . small displacement

Dw = = F ds cos = Fds

Dw = m ds = mvdv

= v^{2} – o]

= v^{2}

**Work energy principle**

According to this principle work done by a force on displacement a body measures the change in KE of the body .

dw = = p ds cos = Fds

dw = mads = m ds = mvdv

= v^{2} – u^{2}]

= v^{2} – u^{2}

Work done = final KE – initial KE

**Relation in between KE and linear momentum**

Linear momentum P = mv

Kinetic energy KE (E) = v^{2}

^{ }E =

P =

- P

A body cannot have minimum without having and vice versa.

- The variati0on of with
- For a given value of P MEK = cawt

Or EK

- For a given value of E
_{K}.

P

**Example**

The KE of a body in increase by 21% what is the percentage increases in the linear momentum of the body .

**Solution**

__ __

E2 = E1 , m = m

= =

% increase in linear momentum is 10%

**Example**

If the momentum of theody increase by 20% . what will be the increase in the KE of the body .

**Solution**

E_{1 } = P_{2} = =

E_{2} = = =

% increase in KE = =

**Potential Energy**

Energy possessed by the body by virtue of its position or configuration in some field.

**Gravitational Potential**

Gravitational potential energy of a body is the energy possessed by the body by virtue of its position above the surface of the earth.

1 1 = F h = mgh

PE = mgh

**Potential Energy of a Spring**

It is the energy associated with the state of compression or expenension of an elastic spring .when the spring is compressed or elongated it tends to recover its original length on account of elasticity . the force trying to bring the string back to its original position is called restoring force.

The restoring force (F) is directly proportional to the displacement .

F = – Lx (i) K – spring constant let the block be displaced further through dx then

Work done dw = – F dx = – ( Kxdx)

dw = K xdx (ii)

total work done in giving displacement x to the body can be obtained by integration of eq (ii)

= = K

= kx^{2}

**Mass Energy Equivalence**

^{ }

Mass can be transferred into energy . the mass energy equivalence relation is given by Einstein

E = mc^{2 } E – energy that appears

C – velocity of ligut

M – mass that disappears

**Principle of Conservation of Energy**

__ __

The sum of total energy of all kinds in a system remains constant at all times.

To prove this principle let us consider KE , PE and total energy of a body falling freely under gravity .

__Now at point A __

KE = 0 . h = 0 DE = – mgh , TE = KE +PE = 0 mgh

TE = mgh (i)

__ Total energy at B __

PE = mg (h – x) , KE = mv^{2} v^{2} = u^{2} +2as

Te = mg (h – x ) + m (2gx) = mgh TE = mgh (ii)

__Total energy at C__

PE = mg (o) = o , h = o

KE = mv^{2} v^{2} = u^{2} +2gh u = o 2gh

KE = m (2gh) = mgh

TE = mgh

__ __

**Collision**

“collision between two particles is said to occur when they actually strike aganist each other”

** Types of collision ** – (i) elastic collision (ii) inelastic collision

– A collision in which there is no loss of KE is called as elastic collision eg . coision between atomic and sub atomic particles__Elastic collision__

** Basic characteristics of elastic collison **–

- KE is conserve (ii) momentum is conserve (iii) T.E conserve

(iv) force in volving during elastic in each there occurs some loss of KE is called inelastic collision

– A collision in which there occurs some loss of KE is called inelastic collision .__Inelastic collision__

** Basic characteristics ** –

- Momentum is conserve (ii) total energy conserve
- KE is not conserve.

**Coefficient of restitution**

It is they ratio of relative velocity of separation after collision to the relative velocity of approach before collision.

E = for perfect elastic collision

e = 1

**Elastic Coefficient in one dimension**

Suppose two balls A and B of masses m1 and m2 respectively moving in dame direction with u_{1} and u_{2} when u_{1} > u_{2} then they called

Linear momentum before collision = linear momentum after collision

V E before collision = KE after collision

1/2 m1 u_1^2 + 1/2 m2 u_2^2 = 1/2 m1 v_1^2 + 1/2 m2 v_2^2

m2 (v_2^2 – u_2^2 ) = m1 (u_1^2 – v_1^2 ) (iii)

eq (iii) divided by (ii)

(m_2 (v_1^2-u_2^2) )/(m_2 (v_2-u_2)) = (m_1 (u_1^2-v_2^2) )/(m_1 (v_1-v_1)) = v2 + u2 = u1 + v1 = v2 – v1 = u1 – u2

(v_2- v_1)/(u_1- u_2 ) = 1 = e

**Calculation of velocities after collision**

V_{2} = u_{1} – u_{2} + v_{1 } putting in eq (i)

m_{1}v_{1} + m_{2} (u_{1} – u_{2} + v_{1}) = m_{1} u_{1} + m_{2} u_{2}

V_{1}(m_{1} + m_{2} ) = (m_{1} – m_{2}) u_{1} +2m_{2} u_{2}

V_{1} = ((m_1- m_2 ) u_1+2m_2 u_2)/(m_1+m_2 ) ————–(iv)

Similarly V_{2} = ((m_2- m_1 ) u_2+2m_1 1)/(m_1+m_2 ) ———-(v)

**Particular cases**

– m__When masses of two bodies are equal___{1 }= m_{2}= m

From eq (iv) v_{1} = = u_{2}

From eq (v) v_{2} = = u_{1}

Ie “ they inter change their velocities after collision”

– u__When target body is at rest___{2}= 0

From eq (iv) v_{1} = (vi) from eq (v) v_{2} = (vii)

m__When masses of two bodies are equal___{1}= m_{2}

From eq (vi) v_{1} = 0 from eq (vii) v_{2} = = u_{1}

ie “body A comes at rest and body B starts moving with velocity of A”

m__When body B at rest is very heavy___{2}m_{1}

m_{1}_{ }0

from eq (vi) v_{1} = – u_{1} from eq (vii) v_{2} = 0

hence when a light body A collides against a heavy body at rest . A rebounds with its velocity.

–__When body B at rest has negligible mass__

m_{1}_{ }m_{2} m_{2} 0

from eq (vi) v_{1} = u_{1} from (vii) v_{2} = = 2u_{1}

the body A keeps on moving with the same velocity and body B starts moving with twice of velocity of A .

**elastic collision in two dimension**

suppose m1 and m2 are the masses of two bodies moving with velocity u_{1} and u_{2 }in direction .

if u_{1} > u_{2} then they collides after collision let the body A moves with velocity v_{1} at angle _{1} with x axis and body B moves with velocity v_{2} at angle _{2} with x axis K E is conserve .

let KE before collision = total KE after collision