LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

WORK ENERGY POWER

In everyday language, we often use the terms work, energy and power. A teacher teaching a class,
a student preparing for examination, a farmer ploughing the field, all are said to be working. A person
who can put in long hours of work is said to have large stamina or more energy. In karate or boxing
we talk of powerful punches, that are delivered at greater speed for small time durations. But all
these terms have specific scientific meanings, which are to be learnt in detail in this chapter.
WORK
Any agent which can deliver a force can perform a work also. In science, work done by a force may
be zero even if the force is acting. Suppose a man is pushing a wall and in that case the person is
applying a force. But scientifically the work done by that man is zero. Also, when a man with a load
on his head is walking on a straight horizontal road, then also work done by that man in displacing
that load is zero. Scientifically a force is said to perform a work only when the following two conditions
are satisfied.
(i) The body on which the force is acting must undergo a displacement.
(ii) The direction of this displacement must not be perpendicular to the direction of that force.

So work is said to be by a force acting on a body,only if that body undergo a

displacement under theinf luence of that force such that the direction of
displacement is not perpendicular to the direction of that force.

It is to be noted that, even if the force is not directly responsible for the motion of the body, it can
perform work on the body.
Consider a body placed on a horizontal surface. Let a force F is applied on the body as shown. Let

the body undergo a displacement s along the surface such that the direction of s makes an angle 

with the direction of F

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 
F cos  is the component of force F in the direction of s
Work done by the force is

W = (F cos  ) s w  FScos 
OR
 
w  FS
So the work done by a force is defined as the product of component of the force in the direction of
displacement and the magnitude of the displacement. So in general work done by a force can be
defined as the dot product of force vector and displacement vector.
Calculation of work done by a force can be carried out in three different ways.
(i) w = Fs cos 

(ii) w = s x (fcos  )
w = value of displacement × component of force in the direction of displacement
(iii) w = F x (s cos  ) = value of force x component of displacement in the direction of force
In terms of rectangular components, if

F  Fx i  Fy j  Fz k and

s  x i  y j  zk

Then work done by the force

 
w  F s(Fxi  Fy j  Fzk)
  (xi  y j  zk)

w  xFx yFy zFz

Dimensions of work = ML2 T 2 ; Unit of work done

In SI system ; unit of work is Joule (J)
In cgs system ; unit of work is erg
1 Joule = 107 ergs
Practical unit of work is Kilogram - metre (kg-m)
1 Kgm = 9.8 J
NATURE OF WORK DONE
Depending on the angle (  ) between the force vector and displacement vector, work done by a force
can be positive, negative or zero.
a. Positive work
If  is acute (0     / 2) , then cos  is positive and work done will be also positive.

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

eg :
(i) When a body falls freely under the action of gravity, weight mg acts downwards and body also displaces
down wards. so  = 00 and hence work done by weight will be positive.
(ii) When a lawn roller is pulled by applying a force along the handle at an acute angle, work done by
applied force is positive.
(iii) When a gas filled in a cylinder fitted with a movable piston is allowed to expand, work done by
the gas is positive.
(iv) When a spring is stretched, work done by the stretching force is positive.
b. Negative work

 
If  is obtuse       , cos  will be negative and hence work done will be negative.
2 
eg: (i) When a body is thrown vertically upwards, weight of body is acting vertically downwards but the
displacement of body is vertically upwards. So  = 1800 and hence work done by weight will be
negative.
(ii) When a body is moved over a rough horizontal surface, the force of friction opposes the motion.
So angle between friction and displacement is 1800 so work done by friction is negative.
(iii) When breaks are applied on a moving vehicle, work done by breaking force is negative.
c. Zero work
Work done by a force will be zero in two cases.
(i) If body is not moving even after the application of a force, then displacement of the body will be zero
and hence work done will be zero.
(ii) If the direction of displacement is perpendicular to the direction of force, then  = 900 and cos  = 0.
then work done by the force will be zero.
eg :
(i) when we push hard against a wall, the force we exert on the wall does no work because displacement
is zero. However in this process our muscles are contracting and relaxing alternately and internal
energy is being used up. That is why we get tired.
(ii) When a person carrying some load on his head moves on a horizontal platform. Then  = 900 and
hence work done by him is zero.
(iii) When a particle moves in a circular orbit, centripetal force always acts towards the centre of circle
along the radius and the displacement is tangential to the circular orbit.

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 
So angle between F andds is 900. So work done by centripetal force is always zero.

If a satellite is revolving around the earth in a circular orbit, gravitational force on it from earth acts as
the centripetal force. So work done by the gravitational force is zero on the satellite.
(iv) Tension in the string of a simple pendulum is always perpendicular to the displacement of bob.

So  = 900 hence work done by tension is zero.

WORK DONE BY A CONSTANT FORCE
If the value and direction of a force remains same throughout it is a constant force. Then work done
by it to make a body to displace through s is;
 
w  F  S  FS cos 
WORK DONE BY A VARIABLE FORCE
If the value or direction or both of a force varies either with position or time, it is a variable force.
eg: Restoring force developed in a spring.

Consider a variable force F (x) acting in the x direction. Under the influence of this force, let a body
displaces in the x direction from an initial position x1 to a final position x2 as shown.

To find the work done by this force, let us consider an elementary displacement dx in the path. Then
small work done to displace the body through the element.
 
dw = F  dx

To find the total work done to displace the body from A to B, the path length AB is split in to a number
of such elements, then find dw for each element and take their sum. Instead, we can integrate dw
within limits x1 to x2

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

x2  

w   F  dx
x1

The above situation is used when force vector and displacement vector are one dimensional.

Now let us consider a variable force in three dimensions. F  Fx i  Fy j  Fz k

Let this force displaces a body from a position p(x1, y1, z1) to Q (x2, y2, z2)

Consider a small element dr in the path, given by

dr = dx i + dy j + dz k
Then small work done to displace the body through this element.
 
dw = F  dr = (Fx i  Fy j  Fz k)
  (dx i  dy j  dz k)
 = Fx dx + Fy dy + Fz dz

Then total workdone to displace the body from P to Q;

x 2 ,y 2 ,z 2

w 
x1 ,y1 ,z1
Fx dx  Fy dy  Fzdz

x2 y2 z2

w   Fx dx   Fy dy   Fz dz
x1 y1 z1

WORK DONE BY A SPRING FORCE

Consider a spring of force constant k with one end fixed and the other end connected to a block of
mass M. Let the spring is in its natural length position. Let the spring is elongated or compressed
through a distance x as shown.

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If the spring is released, the system will return. It means that a restoring force acts on the spring given
by
F = -kx
This is a variable force work done by spring force is given by

1
w k  x initial2  x final2 
2 

x initial  initial elongation or compression on the spring

x final  final elongation or compression on the spring
WORK DONE FROM FORCE - DISPLACEMENT GRAPH
Consider a graph draw between a force F acting on a body and displacement s the body has
undergone from position A to position B.

To find the workdone by the force consider a small element as shown over which the displacement is
ds and force is almost constant at F0. Then, element is like a rectangular strip.
Area of element
dA = F0 x ds = work done by force for a small displacement ds

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Then to find total workdone, split the area of region below the graph into such small elements, find
area of each element and then take their sum.
s2

w   Fds  area of the region between graph AB and displacement axis.

s1

Hence workdone can be calculated as the area under force displacement graph.

Work done  area of theregion between the force displacement graph and displacement axis

While taking the area, sign of force and displacement must also be considered. If Fs product is
positive work done must be taken as positive and if the product is negative, workdone must be taken
as negative.
WORK DONE BY STATIC FRICTION
Static friction is the frictional force developed at the contact surface between two bodies when there
is no relative motion between them at their contact surface.
eg
(1) : Consider a block placed on a rough horizontal surface acted upon by an
external force F as shown.

Let the body continues to be at rest even if F is acting. This is because F is nullified by friction
fs and is static. Since displacement of body is zero, work done by static friction in this case is
zero.
eg (2): Consider a block A placed over a smooth horizontal floor. Another block B is placed over A and
the cotact surface between A and B is rough. Let an external force F acts on the lower body as
shown so that both the blocks move together with a common acceleration.

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Since both move with a common acceleration, there is no relative motion or slipping between
A and B. So friction at their contact surface is static (fs). Since F on A is towards right, fs will act
towards left on A. So the direction of fs on B is towards right. So fx behaves like an applied force
on B and makes it move along with A with same acceleration a. Let s be the displacement of
the system after time t. Then static friction perform work on both A and B. Work done by static
friction on A
wA = fs s cos 180 = -fs s
Work done by static friction on B;
wB = fs s cos 0 = + fs s
 Total work done by static friction on the system = wA + wB = 0
Points
(i) Work done by static friction can be positive, negative or zero.
(ii) Static friction perform work on both bodies forming the contact surface
(iii) Total work done by the static friction on the system forming the contact surface is zero.
WORK DONE BY KINETIC FRICTION
Kinetic friction is the frictional force developed at the contact surface between two bodies when there
is a relative motion between them at their contact surface.
eg : Consider a block sliding on a fixed rough horizontal surface as shown. then friction is kinetic,
opposite to its direction of motion.

Let s be the displacement . Then work done by kinetic friction = fkS cos 180 = -fkS
So here workdone by kinetic friction is negative.
eg : A block of mass m is projected with a velocity Vo on a plank of mass M such that the block slides
through a displacement over the plank.

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So friction is kinetic.
(fk) between them
It acts towards left on M as it slides towards right. So fk acts towards right on m due to which it also
starts moving towards right. Let S1 and S2 be the displacements of m and M until the slipping
between them stops. So S1>S2.
Work done by kinetic friction on M ; W 1 = fkS1 cos 180 = -fkS1
Work done by kinetic friction on M ; W 2 - fk S2 cos 0 = fk S2
Total work done by Kinetic friction = W 1 + W 2

[Since S1 > S2 ; w 1  W2 ] = (-)

Points
(i) Work done by kinetic friction can be positive or negative.
(ii) Total workdone by kinetic friction on a system forming the contact surface is always negative.
CONSERVATIVE FORCES
Depending on the work done, forces are generally classified into two types
(i) Conservative forces
(ii) Non conservative forces
Let a body is displaced from an initial point A to final point B under the influence of a force F, through
different paths as shown.

Let W 1, W 2 and W 3 respectively represents the workdone by F along paths (1), (2) and (3).
If W 1 = W 2 = W 3, then the force F is said to be conservative.

A force is said to be conservativeif work done by or against the force in moving a body depends only
the initial and final positions of the body and is completely independent of thepath through which
the body is displaced.

eg : Gravitational force, spring force, electric force between charges, magnetic force between magnetic
poles etc.
It is to be noted that work done by an external agent against a conservative force is also independent
of the path and depends only on initial and final points.

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If we consider mountain roads, they do not go straight up but wind up gradually in hair pin mode.
Since engine of vehicles do work against gravitational force (conservative), work done will be same
along both the paths. Since w = Fs cos  , if s is large F will be small for same amount of work. So in
hair pin mode, S will be very large and hence engine force required will be small.
Now let us check the conservative nature of gravitational force. Let a body of mas m is taken from
the ground to a platform at a height h, through different paths as shown
Properties
(i) Work done by a conservative force or work done by an external agent against a conservative
force to displace a body from one point to another depends only on the initial and final points and is
completely independent of the path through which the body is displaced.
(ii) Work done by a conservative force to move a body around a closed loop is always zero.
(iii) If only conservative forces are performing work on a body, then total mechanical energy (KE + PE) of
that body remains a constant. That is why the name conservative is used.
(iv) If an external agent performs a work against a conservative force to displace a body, then that
energy does not lost but is stored as potential energy on that body.
(v) A conservative force on a body is always directed towards a fixed point.
NON CONSERVATIVE FORCES
Properties of a non conservative force is exactly opposite to those of conservative forces.

A force is said to be non conservative, if work done or against the force in moving a
body from one position to another, depends on the path followed between these
two positions.

eg : frictional force, viscous force etc.

Properties
(i) Work done by a non conservative force or by an external agent against a non conservative force
depends on the path through which body is moved.
(ii) Work done by a non conservative force to move a body around a closed loop is non zero.
(iii) If non conservative forces are performing work on a moving body, its total mechanical energy (KE +
PE) will not be conserved.
(iv) If an external agent performs a work against a non conservative force. it is lost as heat or other
energy forms. That is why the non conservative forces are known as dissipative forces or damping
forces.
(v) Non conservative forces on a body are not always directed to a fixed point.
It is to be noted that all the forces which are not conservative are non conservative. There are many
forces which are neither purely conservative or purely nonconservative.
ENERGY
Energy of a body is defined as the capacity or ability of the body to do the work. When a body is
capable of doing more work, it is said to pusess more energy. Like work, energy is also a scalar
quantity.
There are many forms of energy like mechanical energy, heat energy, chemical energy, light energy,

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sound energy, electrical energy, nuclear energy etc.

Unit of energy is also joule (SI)or ergs (cgs). There are many practical units of energy. Calorie for
heat energy; I calorie = 4.18 J electron volt for nuclear energy; 1 eV = 1.6  10-19 J
Kilowatt hour for electrical energy; 1 KWh = 3.6  106 J
Mechanical energy of a body is defined as the ability of the body to do mechanical work. It is equal
to the sum of kinetic energy and potential energy. ME = KE + PE
KINETIC ENERGY
Kinetic energy of a body is defined as the energy posessed by the body by virtue of its motion.
eg :
(i) A bullet fired from a gun can pierce through a target on account of Kinetic energy of the bullet
(ii) Wind mills work on the Kinetic energy of air.
(iii) A nail is driven into a wooden block on account of KE of the hammer striking the nail.
EQUATION FOR KE
Consider a body of mass m placed at rest. Let a constant force F is applied on it so that it starts
moving with a constant acceleration a. Let it gains a velocity v when it is moved through a displacement
s.

F
a using V2 = U2 + 2as
m

F m v 2
V2 = U + 2 s  S 
m 2F

mv 2
Work done by the force W = Fs cos  = F   cos 0
2F

1
 mv 2
2
This work done appears as translational or linear KE

1
KE  mv 2
2

1
(i) The expression mv 2 holds even when the force applied varies. That is expression is valid irrespective
2
of how the body acquires velocity v.

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(ii) Kinetic energy of a body is always positive.

(iii) Kinetic energy of a body depends upon the frame of reference.
RELATION BETWEEN KE AND MOMENTUM

1 m
KE; E = mv 2 
2 m

m2 v 2
= mv = p, linear momentum
2m

p2
E
2m

KE  momentum2
POTENTIAL ENERGY
The potential energy of a body is defined as the energy possessed by the body by virtue of its position
or configuration in some field.
Two important types of potential energy are
(i) Gravitational Potential energy
(ii) Elastic Potential energy
Potential energy of a body refers to an energy stored in a system which gives that system an ability to
do mechanical work. For that this PE is converted into KE.
For example, when we wind the spring of our watch, PE is stored in the spring on account of
configuration of the turns of the spring. As the spring unwinds, it works to move the hands of the
watch. Thus wound spring has the potential to do the work.
Two important points are need to be considered while considering PE

Point 1 : Potential energy is associated only with conservative forces

Point 2 : If an external agent performs work against a conservative force to displace a body, without an acceleration, .
the entire workdone by the agent appearas change in potential energy of that system or body

What is presented as point 2 above is the method by which we can provide PE to a body. The
process is done without an acceleration because, in such a case, speed will not change and hence
KE will also not change. So entire work done will appear as PE change only.
eg: Let a body of mass m is bring under the influence of gravity, which is conservative. Let the body is
moved to a height h by an external agent very slowly without an acceleration. then the agent is
performing work against gravitational force which is conservative and this work has a value mgh.
This mgh is stored as PE on that body. If released, this mgh will convert back to KE.
GRAVITATIONAL POTENTIAL ENERGY
Gravitational potential energy of a body is the energy posessed by the body by virtue of its position
with respect to earth. This is kind of interaction PE developed due to the gravitational ineraction
between the body and earth. It means that, if a body is placed at the surface of earth, its gravitational
PE is not zero, because gravitational force exist between that body and earth.

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But for practical calculations of PE, we will usually assume that PE of a body placed at the surface
of earth is zero. Consider a body of mas m placed on the ground, where PE is assumed to be zero.

Let an external agent moves this body to a height h without an acceleration by doing work against
gravitational force, which is conservative. So this workdone will store as gravitational PE on the body.
Work done by external agent
wext = Fext s cos 
[Fext is the force given by external agent to overcome mg.
= mgh cos 0 = mgh
so Fext = mg (upwards) ]
This work done appears as change in PE, given as
 u = mgh
Here Uinitial = 0, Ufinal = mgh
Ufinal - Uinitial = mgh
Let Ufinal = U
U - 0 = mgh
U = mgh
This equation is obtained on the assumption that PE is zero at the ground level. Such an assumption
taken is known as zero reference. Any position or level can be taken as zero reference, according to
o u r

Point : To define absolute PE or to express the value of PE of a body, we have to first assume
that the value of PE is zero at a particular level called zero reference. then h must be taken from
e
n
o
c
e.vnithat reference level. For positions above that level, h is given a positive sign and for
positions below that level, h is given a negative sign.

Consider the following example

Gravitational PE of a body can be positive, negative or zero.

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If ground is taken as zero reference

when body is at A when body is at B when body is at C
U = mgh1 U = mg(h1 + h2) U=0
If table top is taken as zero reference
When body is at A when body is at B when body is at c
u=0 u = mgh2 u = mg (-h1)
= -mgh1
ELASTIC POTENTIAL ENERGY
Elastic potential energy of a body is the potential energy gained by that body due to a deformation
happening on it.
For example, if we stretch or compress a spring, an external agent performs a work against the
spring force, which is conservative. So this work done is stored as PE in the spring.
POTENTIAL ENERGY OF A SPRING
Consider a spring of constant k fixed at one end. its free end is connected to a block of mass m. Let
the spring is at its natural length.

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Now the block is displaced towards right through a distance x by an external agent so that the spring
stretches through the same amount x. Then a spring force F= Kx is developed in it in a direction
opposite to the direction of stretching. This spring force is a conservative force and is a variable
force. Here external agent is performing a work against this conservative force without an acceleration.
So this work done will be stored as elastic PE in the spring. Now let us find that workdone. For that
consider a small element of length dx in the path. Then workdone by the external agent to displace
the mass through the element.
dw = Fext dx cos  = Kx dx cos 0
= kx dx
Total work done by external agent to displace Fext is the force given by external agent to overcome
the body from x = 0 (natural length) to x = x spring force kx Fext = kx, in the direction of dx
so  = 00

x x
 x2  1 1
w   kx dx  K    Kx 2 U  Kx 2
0  2 0 2 2

This work done is stored as PE in the spring where x is the elongation or compression in the spring.
Potential energy stored in a spring is always positive.
RELATION BETWEEN PE AND CONSERVATIVE FORCE
Potential energy is always associated with conservative forces only
(i) Gravitational PE
U = mgh; differentiating with respect to h and giving a (-) sign

dU  d
 (mgh)  mg 1   mg
dh dh

dU
  F,gravitational force F  mg
dh
= mg, downwards

dU
F 
dh

(ii) Elastic PE
PE stored in a spring;

1 2
U= Kx
2
differentiating with respect to x and giving a (-) sign.

dU  d  1 2  1
  K x    K 2x
dx dx  2  2

 k x

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= F, restoring force on spring


dv
 F
dx

So if we diffrentiate the equation for PE with respect to distance and give negative sign,
we get the corresponding conservative force.
conservative force - Differential of PE function with respect to distance

differential of a quantity with respect to distance is called gradient of that quantity

 conservative force   gradiant of PE

dv

F
dr

Note:

dU 

The equation F = F  is used when PE is given as a function of a single position co-ordinate x

dr
or y or z. But in many cases, U may be given as a function of more than one position co-ordinates. In
such a case we need to do an another type of mathematical calculation.
Negative sign always exist in the relations between potential energy and conservative force. this
negative sign indicates that, the direction of a conservative force is same as the direction in which PE
is reducing. So if we move in the direction of a conservative force, PE is found to be reducing.
Taking the reverse process ; change in PE is given by

 
r2

dU = -  F  dr (from position r1 to r2) and is known as the negative line integral of conservative force.
r1

PE OF A UNIFORM ROD PLACED VERTICALLY

Consider a uniform rod of mass m and length  placed vertically on the ground. Let PE is assumed
to be zero at ground level (zero reference)

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Here all the elements of the rod are not at the same height from the ground. So we have to use
integration method to find its PE.

m
Consider a small element of length dy at a height y from the surface mass of the element; dm =

m
dy PE of the element, du = (dm) g y = g dy


 
m m  y2 
 Total PE of the rod; U =  dU =  g  =  g  2 
ydy
0

m  2  mg 1
g   mg
 2 2 ;U= 2


That is ; U = mg  
2

So in these type of situations, where the size of the body is large, to calculate the gravitational PE
using the equation mgh, h must be taken as the height to the centre of mass of that body from the
zero reference level.
eg : Let a uniform disc of mass m and radius R is placed on the ground. Let ground be the zero reference
level.

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TRANSLATIONAL EQUILIBRIUM
If the net external force acting on a body is zero, it is in a state of translational equilibrium.
If Fnet = 0, using Newton’s law
ma = 0 or a - 0
So acceleration is zero for a body in translational equilibrium; This can happen in two situations
(i) When body is at a permanent rest
(ii) When body is in uniform motion
Equilibrium status are classified into three types.
1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equlibrium
This classification is made according to the potential energy values posessed by the bodies at these
equilibrium states. We know that

dU
Conservative force ; F =
dr
When body is in translational equilibrium
F=0

dU
 =0
dr
This can happen in three situations
(i) When U is a constant (Neutral equilibrium)
(ii) When U is maximum (unstable equlibrium)
(iii) When U is minimum (stable equlibrium)

dy dU
if we consider a graph connecting PE (U) and distance (r), then its slope ; 
dx dr
So at positions of equlibrium; slope = 0 so at those positions, graph will be parallel to x axis or r axis.
In this graph, slopes are found to do zero at positions 1, 2 and 3.

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The equlibrium status at which net force on the body is zero and PE is minimum is called stable
equilibrium. The equlibrium status at which net force is zero and PE is maximum is called unstable
equlibrium. The equilibrium status at which net force is zero and PE remains constant is called
neutral equilibrium.
STABLE EQULIBRIUM
Following properties are followed by a body in stable equilibrium.
(i) net force acting on the body is zero
(ii) PE of the body will be a minimum

du
  0 and slope of u - r graph = 0
dr

(iii) A ball is placed inside a smooth spherical shell as shown. This ball is in equilibrium at that position.
Now the ball is slightly displaced in the direction as shown and released. Then the ball will return to
the earlier equilibrium position. It means that, the earlier equilibrium position was stable and the ball
does not want to displace from that position. So when a body is displaced from its stable equilibrium
position, a net force starts acting on the body in a direction opposite to the direction of displacement
or towards the earlier equilibrium position. This force tends the body to bring back it to the earlier
equilibrium position. So not force developed here is said to be negative.
(iv) At stable equilibrium
U is minimum

du d2u
  0 and 2  ( )
dr dr
UNSTABLE EQUILIBRIUM
Following properties are followed by a body in unstable equilibrium
(i) Net force on the body is zero
(ii) PE of the body will be maximum

du
  0 and slope of U - r graph is zero
dr

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(iii) A ball is placed at the top of a smooth spherical shell as shown. This ball is in equilibrium at that
position . Now the ball is slrghtly displaced in the direction as shown and then released. Then the ball
will further move away. So the earlier position of equilibrium was unstable and the ball does not want
to stay there. So when a body is slightly displaced from unstable equilibrium position a net force is
developed in it in a direction same as the direction of displacement, away from equilibrium position.
That is why body further move away, when released. So net force developed here is said to be
positive.
(iv) At unstable equilibrium,

du d2u
u is maximum   0 and 2  ( )
dr dr
NEUTRAL EQUILIBRIUM
(i) Net force acting on the body is zero

du
(ii) PE of the body will be a constant   0 and slope of u-r graph = 0
dr
(iii) When a body is in neutral equilibrium, it will be in equilibrium over a region at a number of points.

Let a ball is placed on a

horizontal surface at point A. From there, it is displaced to point B as shown. At point A, ball was in
equilibrium. Now at point B also, the ball is in equlibrium. So it will not show any tendency to return or
move away, as no net force is developed on displacing.
(iv) At neutral equilibrium
U is a constant

du d2 u
  0 and 2  0
dr dr

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

So to check whether the given equilibrium is stable unstable or neutral, displace the body slightly
from the equilibrium position in the specified direction and then release it. Then a net force will be
developed on it and find the direction of that net force. If it is towards the direction of earlier equilibrium
position the body will show a tendency to return. Then the equlibrium was stable, in that direction.
But if the direction of net force is away from the equilibrium position, the body will further move away.
Then the equilibrium was unstable in that direction. But if no net force develops, body will continue in
its displaced position. Then the equilibrium was neutral.
PE- DISTANCE AND FORCE - DISTANCE GRAPHS
Consider a graph connecting PE and distance of a system as shown.

All the four points A1B1 C and D corresponds to equilibrium. Let us find the nature of equilibrium.

du du
Slope of U - r graph = at points A, B, C and D, graph is parallel to axis. So slope = 0. So at all
dr dr
these positions, body is in equilibrium.
Point A : Near point A, PE is a constant over a region. So equilibrium is neutral at A
Point B : At point B, PE is maximum. So equlibrium is unstable
Point C : At point C, PE is minimum. So equlibrium is stable.
point D : At point D, PE is maximum. So equlibrium is unstable.

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Consider the force -distance graph of a body as shown. In such a graph, the points at which graph
meets the x axis or distance axis, the force value will be zero. So those are the points of equilibrium.
So here points A,B,C are corresponding to equilibrium.
Point A : Take the next instant after point A. Then a force develops which is positive. so at A,
equilibrium was unstable
Point B : Take the next instant after point B. Then a force develops which is negative. So at A,
equilibrium was stable.
Point C : Take the next instant after point c. Then no force develops so equilibrium was neutral at C
WORK ENERGY THEOREM
Newton’s second law is a basic law that can be used to solve force related problems in mechanics.
Newton’s second law has a vector form. It does not consider internal forces during solving mechanical
systems. This demerit is overcome by suggesting work-energy theorem. It has a scalar form as it
suggest scalar quantities like work and energy.

Work - energy theorem is a scalar form of Newtons second law. This theorem consider
work done by internal forces also

Work - energy theorem states that; total work done by all the forces acting on a body or system is
equal to change in Kinetic energy of that body or system. So here we consider all the forces acting
on the system, which includes conservative nonconservative, external, internal, etc. forces. If we
apply this theorem from a non inertial frame, we consider the work done by pseudo forces also while
applying this theorem.

 work done by all force  Change in KE

This is the original form of work - energy theorem

SUPPLEMENTARY FORMS
Work - energy theorem has two supplementary forms.
(i) Let only conservative forces are performing work on the body.
For conservative force

dU
Fc =  dU   Fdr
dr C

This equation suggests that, when a body is displaced through a small distance dr under the influence
of a conservative force Fc, then change in its PE is du

Changein PE   Work doneby conservative forces


 PE   Wconservative force

(ii) Original form of work - energy theorem is W all =  KE

 W Conservative forces + W Forces other than conservative =  KE

but W conservative forces = -  PE

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

 -  PE + W forces other than conservative =  KE

 W forces other than conservative =  KE +  PE but  KE +  PE =  ME, change in mechanical

energy

 Work doneby allforces other than conservative  change inmechanical energy of system

So to change the total mechanical energy of a body, forces other than conservative
forces has to perform work

So consider the following three forms of energy change and work done.

1 KE  Work done by all forces

PE   work done by conservative forces only
ME  Work doneby all forces otherthan conservative

LAW OF CONSERVATION OF MECHANICAL ENERGY

The mechanical energy of a body is the sum of Kinetic energy and potential energy of the body
ME = KE + PE
We know that, to change the ME of a system, forces other than conservative force has to do work on
the system. So

If only conservative forces are performing work on a system, then the total mechanical
energy of that system remains a constant

This statement is known as law of conservation of mechanical energy.

Proof
Using work - energy theorem;
Work done by all forces = change in KE
Let us take all the forces either as conservative or forces other than conservative.

 work done by conservative forces 


    KE  (1)
work doneby forces other than conservative 

Let work done by forces other than conservative is zero. Also

Work done by conservative forces = - change in PE substituting in (1)

PE  O  KE
KE  PE  O
 ME  O  ME  cons tan t

It means that, individually, Kinetic energy and Potential energy may very from point to point, but their
sum is constant throughout.

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eg : Consider a small ball of mass m is falling under gravity from a height, from point A.

Consider two points B and C also as shown. Let only gravitational forces (conservative) is performing
work on the ball. So total ME of ball must be conserved.
Now let us calculate total ME of the body at points A, B and C

 MEA  MEB  MEc

This is because gravitational force is conservative from this, it is clear that, at point A, entire ME is
PE. But when it reaches point C, the entire ME appears as KE only. So from A to c, entire PE is
converting into KE. So we can conclude that, in cases where ME is conserved;

Loss in PE = gain in KE
also
Loss in KE = gain in PE

eg2: Oscillations of a simple pendulum

OA is the normal position of rest called mean position or equilibrium position. When the bob of the
pendulum is displaced to B, through a height, it is given PE = mgh, where m is the mass of bob. On
releasing the bob at B, it moves towards A. PE of the bob is being converted into KE. On reaching A,
the entire PE has been converted in to KE. The bob, therefore cannot stop at A. On account of
inertia, it overshoots the position A and reaches c at the same height h above A. The entire KE of the

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

bob at A is converted into PE at c. The whole process is repeated and the pendulum vibrates about
the equilibrium position A. At extreme position B and C, the bob is momentarily at rest. Hence KE =
0. So the entire energy at B and C is PE. At A entire energy is KE.

Point A
PE = 0

1
KE  m VA 2
2
MEA = kEA + PEA = mg  (1 - cos  )

VA 2  u2  2gh  2g  (1  cos )

1 2 1
= KE  2 m VA  2 m2g (1  cos )  mg  (1 cos )

point B
KE = 0

PE = mgh = mg AB1  mg(   cos )

=  mg  (   cos )

 MEB  mg  (1  cos )

ME A  mg  (1  cos  )

 ME A  MEB

When a pendulum oscillates, work is done only by component of gravitational force mg sin  . That is
why ME is conserved.

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eg3: Oscillations of a spring - block system

Consider a spring of constant K whose one end is fixed and other end is connected to a block of
mass m, as shown.

Let the block is displaced towards right through a distance x and released. Then system oscillates.
The only one force acting on the system is the spring force, which is conservative. So total ME of the
system will remain conserved. The variation of KE and PE with distance x given below.

Here also KE + PE = ME, a constant at every position of the oscillating system.

POWER
When a mechanical or any other work is performed effectively in a very short duration, it is called a
powerful performance. In science.

Power of an external agent or machine can be defined as the rate at which

work is done by that agent or machine.

Power = Rate of doing work

work done

time taken

usually this is taken as the average power. Thus power of a body measures how fast it can do the
work. When a body takes lesser time to do a particular amount of work, its power is said to be greater
and vice-versa.

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The power at a particular instant of time or instantaneous power is the ratio of small work done (dw)
to the small time interval (d t) around.

dw
P
dt
 
Let a force F is acting on a body to displace is through a small value ds in a small time interval dt.
t.
Then power delivered by the force.
 
dw  ds  ds  
P F  dt  v,ins tan tan eou s v elo city 
dt dt  

 
P F V
 
or P = FV cos  ,  is the angle b/w F and v .
 
If F is perpendicular to v , (  = 900), then power delivered will be zero. for example, power delivered
by a centripetal force is always zero. for doing work, energy has to be spend. So power can also be
defined as the rate at which energy is spend.
Unit of power

Joul
or Watt(W)
sec ond
Another popular unit of power is horse power (hp)

1h p  746W

Dimensions of power = ML2T 3

Efficiency of an engine is usually defined in terms of power, given as

output power

input power

eg: Let a motor is pumping water, through a pipe of cross sectional area A, with a velocity V.

Then force due to water jet F = AV2P

Where v is the relative velocity between water and the target on which it is hitting.
Then, power of water jet P = FV = AV2P V ; P = AV3P P V 3

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COLLISIONS
In common language, a collision is said to occur when objects crash into each other.
eg : collision between two automobiles, hammer striking a nail etc.
Often collisions are too brief in visible, though they involve significant distortion of the colliding bodies.
Generally, a collision can be defined as an isolated event happening between two or more bodies in
which relatively strong forces are exerted on each other for a relatively short time. Actual physical
contact between two bodies is not necessary for a collision. For example, an alpha particle speeding
towards a nucleus of a atom gets deflected by the electric force of repulsion. then scientifically, we
will say that, a collision is occured between them.
When a collision happen, there will be an exchange or transfer of amount of motion or momentum
between the bodies involved, but total momentum remain constant.

So scientifically, a collision is an isolated event happening between two or more bodies so that a
transfer or exchange of momentum happens between the bodies involved in that process such
that the total momentum of the systm remain conserved during that process.

During a collision process, forces are involved between the colliding bodies. For the entire system,
these forces are internal. But for individual bodies involved, these forces are external. So for the
entire system, total linear momentum will be surely conserved. But total kinetic energy of the system
may or may not be conserved. So
Two important physical quantities which determines the nature of a collision are momentum and
Kinetic energy.
Depending on the momentum and Kinetic energy, classification are made.
TYPE OF COLLISIONS
Collisions are classified into three types
(i) Elastic collision
(ii) Inelastic or partially elastic collision
(iii) Perfectly inelastic collision
(i) Elastic Collision
If both linear momentum and kinetic energy of the colliding system remain conserved during a collision,
it is called an elastic collision. Elastic collisions are also known as perfectly elastic collision.
The basic characterestics of an elastic collission are :
(i) Linear momentum of system is conserved
(ii) KE of system is conserved
(iii) Bodies regain their original size and shape after an elastic collision.
(iv) The forces involved during elastic collisions must be conservative.
(ii) Inelastic collisions
A collision in which, only linear momentum of the system remain conserved but kinetic energy does
not conserved is called an inelastic collision. So during such a collision, some loss of KE of the

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

system will occur as heat, sound etc. As there is always some loss of KE in most of the collisions,
therefore, collisions we come across in daily life are generally inelastic. the basic characterestics of
an inelastic collision are ;
(i) Total linear momentum of system is conserved
(ii) Total KE of the system is not conserved
(iii) Bodies do not regain their size and shape perfectly after this collision.
(iv) Some or all the forces involved in an inelastic collision may not be conservative.
(iii) Perfectly inelastic collisions
If two bodies stick together after colliding, that collision is said to be perfectly inelastics.
Its characterestics are
(i) Total linear momentum is conserved
(ii) Total KE of system is not conserved
(iii) The bodies are well deformed and they do not show any tendency to regain their original size
and shape.
(iv) All colliding forces are not conservative.
eg: Let a bullet hit a wooden block and stick to it. Collision is perfectly inelastic. If a meteorite
collides earth and buried in earth, collision is perfectly inelastic.
Reference Directions In a Collision
In a collision, ther are two basic reference directions. Consider two spherical bodies A and B under
going a collision, as shown.

There is a contact surface between the bodies

(i) Common tangent : It is a tangent drawn along the contact surface. Along common tangent,

the collision has no effect on the bodies. So the components of velocity of both the
bodies do not change along common tangent during collision.

(ii) Common normal or line of impact (LOI))

This direction is perpendicular to common tangent. The collision has maximum effect along LOI. So
the components of velocity of both the bodies suffer maximum change along LOI. So it is the most
important direction, we consider, to solve a collision.

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COEFFICIENT OF RESTITUTION (e)

This term gives a measure of amount of reformation happening to the bodies after collision. It is
defined as the ratio of impulse of reformation to impulse of deformation.

v 2  v1
e
u1  u2

Here v2 - v1 is the relative velocity of seperation between B and A, along LOI, after collision. Also u1
- u2 is the relative velocity of approach between B and A, along LOI, before collision.

relative velocity of separation after collision along LOI

e 
relative velocity of approach before collision along LOI

for elastic collision for inelastic collision

 e 1  0e  1

For perfectly inelastic collision [e depends on the nature of the surfaces coming in contact during
collision]

 e0
Head on collision
A collision is said to be head on or direct if the directions of the velocity of colliding objects are along
the line of action of impulses, acting at the instant of collision.

  
Here the directions of U1 and U2 are same as the direction in which impulsive force N is acting. if
just before collision, at least one of the colliding objects were moving in a direction different from the
line of action of the impulses, the collision is called oblique or indirect.
HEAD ON ELASTIC COLLISION IN ONE DIMENSION
Consider two bodies of masses m1 and m2, initially moving along the same straight line with velocities
u1 and u2 as shown.

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

They undergo a head on elastic collision. Let V1 and V2 are their respective velocities after collision.
Since collision is elastic, both linear momentum and KE are conserved.
Using conservation of linear momentum along LOI
m1 u1 + m2u2 = m1v1 + m2 v2
m2u2 + m2v2 = m1v1 - m1u1
m2 (u2 - V2) = m1 (v1 = u1)  (1)
Using conservation of KE

1 1 1 1
m1 u12  m2 u2 2  m1 v12  m2 v 2 2
2 2 2 2

m2u22  m2 v 2 2  m1v12  m1u12

m2 (u22  v 22 )  m1 (v12  u12 )  (2)

2 m (u 2  v 22 ) m1 (v12  u12 )
 2 2 
3 m2 (u2  v 2 ) m1 (v 1  u1 )

(u2  v 2 ) (u2  V2 ) (v1  u1 )  v1  u1 


(u2  v 2 ) (v1  u1 )

u2  v 2  u1  v1
v 2  v1  u1  u2  (3)

v 2  v1
1 v2-v1  relative velocity of separation after collision along LOI
u1  u2

v 2  v1
  e, the coefficient of restitution u1 - u2  relative velocity of approach before sion
u1  u2
before collision along LOI
Hence for an elastic collision e = 1
from (3); v2 = u1 - u2 + v1
sub in (1), m2 [u2 - (u1 - u2 + v1)] = m1 (v1 - u1)
m2u2 - m2u1 + m2u2 - m2v1 = m1v1 = m1u1
(m1 + m2) v1 = (m1 - m2) u1 + 2m2u2

(m1  m2 ) u1 2m2u2
V1  
m1  m2 m1  m2

(m2  m1 ) u2 2m1u1
Similarly we can set; V2  m1  m2

m1  m2

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Now let us consider some particular cases.

Case I : When masses of two bodies are equal,
ie, m1 = m2 = m, then

(m  m)u1 2mu2
v1  
mm mm

v1  u2  Velocity of A = velocity of B

after collision before collision

(m  m)u2 2mu1
Similarly ; v 2  
mm mm

v 2  u1  velocity of B Velocity of A

=
after collision before collision

Hence, when two bodies of equal masses undergo elastic head on collision in one dimension
their velocities are just inter changed.

Case 2 : Let target (m2) is initially at rest

 u2 = 0. Here we can consider different situations
(i) Let m1 = m2 = m

(m  m) 2m  0 (m  m)u2 2m ui
v1  u1  ; v2  
mm 2m mm m m

V1  01 V2  u1

So velocities are interchanged

(ii) let m1 > > m2 (target rest)

(m1  m2 )u1 2m2  0

v1   (m2 can be neglected as m << m )
m1  m2 m1  m2 2 1

 v1  u1

(m2  m1 ) 2m1 u1
v2   (m2 can be neglected)
m1  m2 m1  m2

v 2  2u1

Hence when a heavy body A undergoes an elastic head on collision with a light body B at rest, the

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

body A keeps on moving with the same velocity of its own and the body 13 starts moving with double
the initial velocity of A
(iii) Let m2 >> m1 (target rest)

(m1  m1 ) u1 2m2  0
v1   (m1 can be neglected as m < < m )
m1  m2 m1  m2 1 2

V1  u1

(m2  m1 )  0 2m1 u1
V2   (m1 can be neglected)
m1  m2 m1  m 2

V2  0

Hence when a light body A, undergoes an elastic head on collision with a heavy body B at rest, A
rebounds with its own velocity and B continues to be at rest.
HEAD ON INELASTIC COLLISION IN ONE DIMENSION

Here only momentum is conserved and not KE using conservation of linear momentum;

m1u1  m2 u2  m1 v1  m2 v 2  1

Let e be the coefficient of restitution, then

v 2  v1
e  v 2  v1  e(u1  u2 ) ; v  v  e (u  u )  (2)
u1  u2 2 1 1 2

Sub in (2) in (1) ; we get

(m1  em2 )u1 (m2  em2 )u2

v1  
m1  m2 m1  m2

(m2  em1 )u2 (m1  em1 ) u1

V2  
m1  m2 m1  m2

There will be a loss in KE given by

1 1 1 1 
KE of loss = m1 u12  m2 u22   m1 v 12  m2 v 2 2 
2 2 2 2 

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HEAD ON PERFECTLY INELASTIC COLLISION IN ONE DIMENSION

Since collision is perfectly inelastic, bodies stick together during collision and thereafter moves as a
single body with a common velocity v.
Here also only linear momentum is conserved

m1 u1  m2 u2   m1  m 2  v
m1u1  m2 u2
V
m1  m2

Loss in KE due to collision ;

2
1 1 1 1 1 1  m u  m2 u2 
E  m1 u12  m2u22  (m1  m2 )v 2 =  m1u12  m2 u2 2  (m1  m2 )  1 1 
2 2 2 2 2 2  m1  m2 

1  m1 m2  1 m m 
 u1  u2  2u1 u2   E   1 2  (u1  u2 )2
2 2
 

2  m1  m2  2  m1  m 2 

In all these cases of one dimensional head on collisions, u1 and u2 must be substituted with proper
signs. If both u1 and u2 are in the same directions, both are taken as positive. But if they are taken
in opposite directions, one is taken as positive and other negative.
COLLISIONS IN TWO DIMENSIONS
Generally, most of the practical collisions are two dimensional. In such a collision, the velocities
before and after collosion may lie in a plane. When we use law of conservation of linear momentum
to solve such collisions, equations are separately written in the two reference directions forming that
plane. This is because, linear momentum is a vector quantity. For example, if collision occurs in the
X Y plane, equations for momentum conservation can be written separately for x and y directions.
Consider an example for a two dimensional collision happening in the XY plane.

Let u1 is exactly in the x direction. Since collision is in the XY plane, we resolve the velocity of bodies
before and after collision as components in the X and Y directions. We will prepare equationd using

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

conservation of momentum in both X and Y directions seperately.

Using momentum conservation in x direction
(Px) initial = (Px) final

m1u1  m2 u2 cos   m1v1 cos 1  m2 v 2 cos 2  (1)

using momentum conservation in Y direction

(Py) initial = (Py) final

m2u2 sin   m1v 1 sin 1  m2 v 2 sin 2  (2)

These equations can be solved to obtain the unknown quantities. If collision is elastic, KE will also be
1 1 1 1
conserved as m1u12  m2u22  m1 v1 2  m2 v 22
2 2 2 2
CASES OF REBOUNDING
In most of the practical rebounding situations target ( like ground, wall etc) will be at rest.
NORMAL REBOUNDING
In normal rebounding, when a body hits a target like ground or wall, the direction of velocity of
colliding body will be perpendicular to the surface of the target. Re bounding can be elastic or inelastic.
If reubounding is elastic, after rebound the body will return to the same height from where it is
dropped. this is because no loss in KE happens during rebound. But if rebounding is inelastic, KE
loses and body cannot return to the initial height.
Let a ball is dropped from a height h to a horizontal floor, normally and the ball undergo continuous
rebounds from the floor inelastically. Let e be the coefficient of restitution between ball and ground.
Since ground is at rest e can be defined as

velocity of ball just after rebound

e
velocity of ball just before rebound

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Here V is the velocity with which ball hits the ground. Since collision is inelastic ball suffers a KE loss.
Let V1 be the velocity of ball just after the first rebound. Then

v1
e  v1  ev
v

Since v 1  v , it will reach a height h1 (< h) after first rebound. After that it will return and hits the ground
with same speed v1 . Then it again rebounds. Let v2 be the velocity after the second rebound. Then

v2
e  v 2  ev1  e (ev)
v1

 V2  e2 v
Since v2< v1, after the second rebound, the height reached by it h2 < h1. So maximum height reached
by body goes on decreasing after each rebound. Similarly, velocity after 3rd, 4th etc rebounds can be
written as :

v 3  e3 v, v 4  e4 v .....

So velocity after the nth rebound is given by;

Vn  en v Where v is the velocity with which the body first hits the target (ground)

n = 1, 2, 3 ...... here V  2gh

Maximum heights reached after each rebound can be calculated as ;

v12 (ev)2  v2 
h1    e2  
2g 2g  2g 

v  2gh
v2
h
2g

 h1  e2 h

v 22 (e2 v)2  v2 
h2    e 4    h2  e 4 h
2g 2g  2g 
Similarly we can write, h3 = e6h, h4 = e8h .....
So maximum height reached by the body after nth rebound

hn  e2nh h  initial height from which body is dropped

n = 1, 2, 3.......

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

Motion in vertical circle

This is an example of non-uniform circular motion. In this motion body is under the influence of gravity
of earth. When body moves from lowest point to highest point. Its speed decrease and becomes
minimum at highest point. Total mechanical energy of the body remains conserved and KE converts
into PE and vice versa.

(1) Velocity at any point on vertical loop:If u is the initial velocity imparted to body at lowest point
then. Velocity of body at height h is given by

w  u 2  2 gh  u 2  2 gl (1  cos  ) [ As h  1  l cos   1(1  cos  )]

Where l in the length of the string

(2) Tension at any point on vertical loop: Tension at general point P, According to Newton’s second
law of motion.

Net force towards center = centripetal force

mv 2 mv 2
T  mg cos   or T  mg cos  
l l
m 2
T [u  gl(2  3cos )] [As v  u 2  2gl(l  cos )]
l

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3) Velocity and tension in a vertical loop at different positions

It is clear form the table that : TA  TB  TC and TP  TP

TA  TB  3mg
TA  TC  6mg
and
TB  TC  3mg
5) Critical condition for vertical looping : If the tension at C is zero, then body will just complete revolution
in the vertical circle. This state of body is known as critical state. The speed of body in critical state is
called as critical speed.
mv 2
From the above table TC   5mg    u  5g
l
It means to complete the vertical circle the body must be projected with minimum velocity of 5gl at
the lowest point.
6) Various quantities for a critical condition in a vertical loop at different positions:

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 LECTURE NOTE 2025 - PHYSICS [FIRST YEAR]

Various conditions for vertical motion :

Work Done in Pulling the Chain against Gravity

A chain of length L and mass M is held on a frictionless table with (1/n) th of its length hanging over the
edge.
M
Let m  mass per unit length of the chain and y is the length of the chain hanging over the edge.
L
So the mass of the chain of length y will be ym and the force acting on it due to gravity will be mgy.
The work done in pulling the dy length of the chain on the table.

dW = F(– dy) [As y is decreasing]

i.e. dW = mgy (– dy)
So the work done in pulling the hanging portion on the table.
0 0
 y2  mg L2
W    mgy dy  mg   
L/n  2  L/n 2n 2

Mg L
W [As m = M/L]
2n 2

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Alternative method:
If point mass m is pulled through a height h then work done W = mgh
Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at
M
a height of L/(2n) from the lower end and mass of the hanging part of chain
n

So work done to raise the centre of mass of the chain on the table is given by
M L
W g [As W = mgh]
n 2n

MgL
or W
2n 2

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