[latexpage]

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]    = [ML2   T-2]

Dimension and unit of work   

Work = F  S    = [MLT-2]       [L]    = [ML2T-2]

The unit of work  are of two types

In SI               joule = Nm 

In CGS           erg       = dyne cm

Relation in between joule and erg

1J   =   11 l  m

1J    = 105 dyne 102 cm

1J   = 106+1   dyne cm

1J    = 107 erg

In SI     kgm                          1 kgm    = 9.8J

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

 

Types of 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  .

If  > 90     cos  is negative

Eg   work done by frictional force always negative

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   SA    and SB 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 = n1  +n2  + n3

=  w1 +w2 + w3   = mg  (n1  +n2  + n3)  = 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
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        Po  – output

P1n  – 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 / s2

 

Solution

=                            Po    =     =   = 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

Here we consider II nd method

v2 = u2  + 2as

v2  =  o + 2as         a =

=   FS  = mas   =  m     S        =   mv2

= KE  =   mv2

 

Calculate Method

Suppose ds   –   V . small displacement

Dw   =    =  F ds  cos     = Fds

Dw  = m   ds     = mvdv

= v2 – o]

=   v2

 

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

= v2 – u2]

=   v2 – u2

 

Work done  = final KE  – initial KE

Relation in between KE and linear momentum

Linear momentum P = mv

Kinetic energy KE (E) = v2

                  E =

P =

A body cannot have minimum without having and vice versa.

Or    EK

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

E1   =                 P2 =   =

E2 =     =      =

% 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

=  kx2

Mass Energy Equivalence

 

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

E = mc2                   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 =    mv2                   v2 = u2 +2as

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

Total   energy at C

PE   = mg (o)   = o    , h = o

KE =    mv2        v2 = u2 +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

Basic characteristics  of elastic collison

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

Basic characteristics  –

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 u1 and u2  when u1 > u2  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

V2 = u1 – u2 + v1   putting in eq   (i)

m1v1 + m2 (u1 – u2 + v1)  = m1 u1 + m2 u2

V1(m1 + m2 ) = (m1 – m2) u1 +2m2 u2

V1 = ((m_1- m_2 ) u_1+2m_2 u_2)/(m_1+m_2 )        ————–(iv)

Similarly       V2 =    ((m_2- m_1 ) u_2+2m_1 1)/(m_1+m_2 )        ———-(v)

 

Particular cases

From  eq (iv)   v1  =    = u2

From eq  (v)   v2 =    = u1

Ie “ they inter change their velocities after collision”

From eq (iv)  v1  =     (vi)   from eq (v)    v2 =          (vii)

From  eq  (vi)    v1 = 0    from eq (vii)     v2 =   = u1

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

m1 0

from eq (vi)  v1 = –  u1    from  eq  (vii)   v2 = 0

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

m1 m2                m2   0

from eq (vi)  v1  =    u1   from (vii) v2  =    = 2u1

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 u1 and u2 in direction .

if u1 > u2  then they collides after collision let the body A moves with velocity v1 at angle 1 with x axis and body B moves  with velocity v2 at angle 2 with x axis K E is conserve .

let KE before collision  = total KE after collision