Work is said to be done when we applied a force and body moves in the direction of applied force.
It is measured as the dot product of force and displacement i.e. W =
W = FS cos
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.
F = 5N , s = 2m = 60
W = FS cos = 5 = 5 2 = 5 j
Work = Fs (MLT-2) [L] = [ML2 T-2]
Work = F S = [MLT-2] [L] = [ML2T-2]
- Absolute units
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
- Gravitational unit
In SI kgm 1 kgm = 9.8J
In cgs gram cm (gcm ) 1 gcm = 980ergs
Types of work done
- Positive work done W = FS cos
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
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
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
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
- 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 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
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
= Po = = = 200w
Pi = ? = 60%
100 = 100 = 3.33 kw
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
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
v2 = u2 + 2as
v2 = o + 2as a =
= FS = mas = m S = mv2
= KE = mv2
Suppose ds – V . small displacement
Dw = = F ds cos = Fds
Dw = m ds = mvdv
= v2 – o]
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
Linear momentum P = mv
Kinetic energy KE (E) = v2
A body cannot have minimum without having and vice versa.
- The variati0on of with
- For a given value of P MEK = cawt
- For a given value of EK .
The KE of a body in increase by 21% what is the percentage increases in the linear momentum of the body .
E2 = E1 , m = m
% increase in linear momentum is 10%
If the momentum of theody increase by 20% . what will be the increase in the KE of the body .
E1 = P2 = =
E2 = = =
% increase in KE = =
Energy possessed by the body by virtue of its position or configuration in some field.
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
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
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
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 between two particles is said to occur when they actually strike aganist each other”
Types of collision – (i) elastic collision (ii) inelastic collision
- Elastic collision – A collision in which there is no loss of KE is called as elastic collision eg . coision between atomic and sub atomic particles
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
- Inelastic collision – A collision in which there occurs some loss of KE is called inelastic collision .
Basic characteristics –
- Momentum is conserve (ii) total energy conserve
- KE is not conserve.
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
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
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)
- When masses of two bodies are equal – m1 = m2 = m
From eq (iv) v1 = = u2
From eq (v) v2 = = u1
Ie “ they inter change their velocities after collision”
- When target body is at rest – u2 = 0
From eq (iv) v1 = (vi) from eq (v) v2 = (vii)
- When masses of two bodies are equal m1 = m2
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”
- When body B at rest is very heavy m2 m1
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.
- When body B at rest has negligible mass –
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 .
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