WORK
Work is said to be done
when a force applied on the
body displaces the body
through a certain distance
WORK DONE BY
CONSTANT FORCE
Fsin
WORK DONE BY
CONSERVATIVE & NON
CONSERVATIVE FORCE
WORK DONE IN PULLING
THE CHAIN
L
B
(1/n)
3
WA
S
W=Fcos x S
=FScos =F.S
WA
(Path 1)= WA
B
(Path 2)=
B
(Path 3)
B
(for conservative force)
NATURE OF WORK DONE
1) Positive work (0o< <90o)
F
Direction
of motion
S
WA
WA
B
B
(Path 1)= WA
B
Direction
of motion
S
2) Zero work
Work done becomes 0 for
three conditions
1.Force is perpendicular to
displacement
2.if there is no displacement
3.if there is no force acting
on the body
(Path 3)
W= F.ds = Fds cos
in terms of rectagular
components
W= Fxdx+
Fydy+ Fzdz
Graphical representation of
work done
Note :
Work done for a complete cycle
for a conservative force is zero
WORK DONE BY DIFFERENT
FORCES
B
B
B
l
h1
A
A
h
h2
A
x
=mgh
W3=mgh1+0+mgh2+0+mgh3+0+mgh4
Work done by spring force
magnitude of spring force, Fs=-kx
x=0
x>0
Fs
x<0 Fs
s
Fext
force
xi
xf
= F.dx
xi
xi
dx
xf
Displacement
W= Area under F-x graph
S
xf
Ws= Fs.dx =- kxdx
=- 1 K xf2-xi2
2
xi
WORK
ENERGY&
POWER
Fext
V
p2
m constant, E
K
P
K
P constant,& E
1
m
NUETRAL
If particle displaced from
equillibrium position, force
acting will try to bring the
particle back to the
equilibrium position
If particle displaced from
equillibrium position, force
acting on it tries to displace it
further away from equillibrium
position
If particle is slightly
displaced from equillibrium,
then it does not experience
a force or continues to be
in equillibrium
Potential energy is minimum
at stable equilibrium
Potential energy is maximum
at unstable equilibrium
Potential energy is
constant
F= -dU =0
dx
F= -dU =0
dx
F= -dU =0
dx
d 2U
=negative
dx2
d 2U
=positive
dx2
d2U
=0
dx2
P
m
P constant,& E
P
- Defined only for conservative force
- Energy possessd by a body by
virtue of its position/configuration
- Can either be positive, negative or
zero according to point of reference
- Force always acts from higher
potential to lower potential
Identifying forces with the help of
potential energy
1) Force opposing the motion:On increasing x, if U increases
ENERGY
Relation between different units
h
l
v2
UNSTABLE
POTENTIAL ENERGY
Capacity of doing work
Scalar quantity
Dimension ML2T-2
A
W1=mgh = mgh
W2=mgxl sin =mgxl
m constant, E
STABLE
1eV=1.6 10-19Joules
1kWh=3.6 106Joules
1calorie= 4.18Joules
1 Joule=107erg
2) Force supporting the motion:On increasing x, if U decreases
dU
=negative
dx
(AB portion of graph)
Work Energy Theorm
C
U(x)
dU
=0
dx
A
B
B,C points on the graph
Change in kinetic energy of a
body is equal to network done
on the body

K2-K1= F.dr
D
x
Types of Potential Energy
-Elastic Potential Energy
-Electric Potential Energy
-Gravitational Potential Energy
•Types of equilibrium
If net force acting on a particle
is zero it is said to be in equillibrium
E (total
energy)
For an isolated system or body in the
presence of only conservative forces,the
sum of kinetic and potential energies at
any point remains constant throughout
the motion
K.E+P.E=constant
POWER
Rate at which body does work
W
Average power(Pav)=
t
Instanteneous power

3) Zero force:On increasing x, if U doesnot change
Kinetic Energy
Energy possessed by virtue of
motion
K.E= 1 mv2
2
Always positive
Depends on frame of reference

dU
=positive
dx
(BC portion of graph)
CONSERVATION OF ENERGY
P.E=max
P.E=max
K.E=min
K.E=min
P.E=min
P.E
K.E=max
K.E
work
)
θ
tan θ = d W
dt
=Pinst

 
(Pinst)= d W = F.ds = F.v
dt
dt
Relation between units:
1 watt=1joule/sec =107 erg/sec
1 HP=746watt,1MW =106 watt
1 KW=103 watt
If work done by two bodies
1
is same then power
time
Unit of power multiplied by time
always gives work
1 KWh=3.6 106 Joules
Slope of work-time curve gives
instanteneous power
+
W= dW
portion back on the table
MgL
W=
2n2
h4
h3
dW=F.dx
xf
M
Mass of chain
Work done in pulling the hanging
(for non conservative force)
F=Fx i+Fy j+FZ k
ds=dx i+dy j+dz k
part of length hanging
+
dW=F.ds
Total length
th
+
WORK DONE BY VARIABLE
FORCE
K
K
(Path 2)=
2) Negative work (90o< <180o
F
L/n
Variation of graph of kinetic Energy
1
A
Fcos
Linear momentum:- P= 2m K
Conservative: work done doesnot
depend on path followed
Non-conservative: work depends
on the path followed
2
F
RELATION OF KINETIC
ENERGY WITH OTHER QUANTITIES
time
Area under power-time graph
gives work done
P= d W
dt
Power
=> dW=Pdt
t2
W= Pdt
t1
t1
dt
t2
time
W=Area under P-t graph
Position and velocity in
terms of power (P=constant)
[
1/2
Line of impact
Line passing through common normal to surfaces
in contact during impact
1/2
[
1) Velocity,V=[2Pt
m
2) Position,S= [ 8P
9m
Collision is the event in which impulsive force acts
between two or more bodies which results in change
of their velocities.
t3/2
B
Power delivered by an elevator
a=0, T=(M+m) g
  
U1
Common normal
(Line of impact)
Line of motion of ball A
(M+m)g
[
V2
Power,P= d W = d m gh+ 2
dt
dt
h=height of water level
d m => mass flow rate of pump
dt
[
velocity of the water outlet
Power required to just lift
water, V=0
dm
P=gh
dt
Efficiency of pump
(
Output Power
Input Power
=
Velocity of separation along the line of impact
Velocity of approach along the line of impact
Relative velocity along the line of impact after collision
V1= U1
m1-m2
m1+m2
Oblique
K.E before and after collision is not same
Head-on collision / One dimensional collision
Initial velocities of the bodies are along the line of impact.
Before collision
m2
U2
m1
m2
m2
After collision
2m2U2
m1+m2
2m1U1
2) If massive projectile collides with a light target
i.e. m1>>>m2 ,then v1=u1,v2=-u2+2u1
3) If a light projectile collides with a very heavy
target, m1<<m2 ,then v1=-u1+2u2,v2=u2
Energy transfer from projectile to target
1) Fractional decrease in kinetic energy of projectile
(If target is at rest)
4m1m2
K
=
K
(m1-m2)2+4m1m2
Greater the difference in masses, less will be
transfer of K.E and vice versa
After collision
V1
=
V2
Impact porameter b=0
Transfer of K.E will be maximum when difference
in masses is maximum
If m2=nm1
K = 4n
K
(1+n)2
1+e
+
m1+m2
(1+e) m1U1
V2 =
m1+m2
Loss in kinetic energy
+
(m1-em2)U1
m1+m2
(m2-em1)U2
K=
m1+m2
m1m2
1
2
m1+m2
(1-e2) (U1-U2)2
h1 = e2ho
h1
h2
Vo
V2
1) Projectile and target having same mass m1=m2 ,then
v1=u2,v2=u1, the velocities get interchanged.
Inelastic collision
(1+e) m2U2
1-e
ho
V2= U2
+
m1+m2
m1+m2
Head-on/
one dimensional Special cases:
K.E before and after collision is same
m1
+
Relative velocity of approach
Height after 1st rebound
Velocity after collision
m2-m1
U1
V2
V1
m1
m2
Relative velocity of seperation
Rebounding of ball
U2
Before collision
Based on direction of
colliding bodies
Perfectly
inelastic
V1
After collision
m1
Classification
Inelastic
U2
U1
Conditions
1.For elastic collision: e=1
2. For inelastic collision: e<1
3. For perfectly inelastic collision: e=0
Perfectly
elastic
U1-U2
Perfectly elastic Head-on collission
Perfectly elastic collision
WORK
ENERGY&
POWER
m2
V2-V1
Ratio of velocities
V2
- Particle collision is glancing
- Directions of motion after collision are not
along initial line of motion
- Impact parameter 0<b<(r1+r2) where r1,r2 are radii
of colliding bodies
Relative velocity along the line of impact before collision
Based on conservation of
kinetic energy
e =
Velocity after collision V1 =
m2
Coefficient of restitution (e)
e =
b
m1
Before collision
A
Power of a water drawing pump
μ=
Line of motion
of ball B
Inelastic collision
V1
Oblique collision
m1
(constant)
(
If initial velocities of the bodies are not along the line
of impact.
T
P=T.V
=TV
V
Power, P=(M+m)gV
V
COLLISION
V1
to
2t1
V2
h3
2t2
V3
2t3
h4
V4
t
2t4
Total height covered by the ball before it stops bouncing
H =ho
1+e2
1-e2
Total time taken by the ball until ot stops bouncing
T =
Perfectly inelastic collision
1+e
1-e
Colliding bodies stick together
After collision are moving in the same
m1
U1
m2
Loss in kinetic energy
m1
U2
k=
1
2
m1m2
m1+m2
m2
V
V =
2ho
g
m1U1+m2U2
m1+m2
(U1-U2)2
Colliding bodies are moving in the opposite direction
V =
m1U1-m2U2
m1+m2
, Change in kinetic enargy K= 1
2
m1m2
m1+m2
(U1-U2)2