6.

WORK, ENERGY AND POWER

Work:
It is defined as the product of displacement and component of applied force in the direction of
displacement.
Work done by the constant force:
In the figure,
F is the constant force acting on the body, x is the displacement
of the body,
q is the angle between the directions of F and x ,
F cos q is the component of F along displacement ,
F sin q is the component of F perpendicular to displacement.
Let m be the mass of the body.
By the definition,
Work done = component of force in the direction of displacement ´ displacement
W = F cos q ´ x
Þ W = Fx cos q
! !
In vector form W = F • x
! !
i.e. the dot product of F and x is work done.
Special cases:
1. If q = 0o Þ cos 0o = 1 \ W = F.x---- maximum work done.
2. If q = 90 Þ cos 90 = 0 \ W = 0 ---- minimum work done.
o o

Work done is zero when the force and displacement are perpendicular to each other.
Examples:
1. The work done by the coolie in carrying a suitcase on horizontal smooth surface is zero. Since the
applied force and displacement are perpendicular to each other (Ideal case).
2. The work done by the centripetal force on the rotating body is zero. Since at each instant the centripetal
force and displacement are perpendicular to each other.
3. The tension (force) in the string of a simple pendulum is always perpendicular to displacement. So,
work done by the tension is zero.
4. If the displacement of the body is zero, work done is zero. For example, when we push a rigid wall,
there is no displacement of the wall . Therefore, work done is zero.
Positive work:
Consider the relation, W = Fx cos q .
When 0o £ q £ 90o , cos q is positive. Hence W = Fx cos q is positive.
Work done by the force is said to be positive, if the applied force and the displacement are in the same direction.
Examples:
1. When a horse pulls a cart, the applied force and the displacement are in the same direction. So work
done by the horse is positive.
2. When a body is lifted, the lifting force and displacements are in the same direction. So work done by
the lifting force is positive.
3. When a spring is stretched, the stretching force and displacement (extension) are in the same direction.
So work done by the stretching force is positive
4. When a body falls freely, the force of gravity and displacement are in the same direction. So work done
by the gravity is positive.
Negative work:
Consider the relation, W = Fx cos q .
When 90o £ q £ 180o , cos q is negative.
Hence W = Fx cos q is negative
Work done by the force is said to be negative, if the force and the displacement are in the opposite
directions.

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Examples:
1. When the body is dragged on a rough surface, the work done by the frictional force is negative. This
is because the frictional force and the displacements are in opposite directions.
2. When a body is lifted, the work done by the gravity is negative. This is because the gravitational force
acts vertically downwards and the displacement is in the vertically upwards.
3. When breaks are applied to the moving vehicle, the work done by the breaking force is negative. This
is because the breaking force and the displacements are in opposite directions.
Units of Work done:
SI unit is joule . Symbol is J
Define joule or define SI unit of work:
Consider W = Fx cos q ,
If F: lN , x =lm, q = 0o , then W = 1Nm = lJ
Work done is said to be l joule, when a force of lN displaces a body through lm in the direction of force.
Note:
Work done is a scalar quantity. Hence, if number of forces is acting on the body, the total work done
is the algebraic sum of the work done by the individual forces.
Dimensional formula for work done is [M1L2T-2]
Other units of work and energy:
CGS unit ----- erg, FPS unit---- foot
1 erg = 10 -7 J
1 electron volt (eV) = 1.6 ´ 10 -16 J
1 Calorie (cal) = 4,186 J
1 kilowatt hour (1 kWh) = 3.6 ´ 106 J
Graphical representation of work done by a constant force.( Force -
displacement curve):
The graph shows the force F is constant with increasing the
displacement d
The work done by a force = area under force-displacement curve.
W = OA ´ OC Þ W = F ´ x
Work done by a variable force:
The figure shows the variation of force F acting on the body with its
position x. In this case the force is not constant from P to Q. Therefore, the
region below the curve PQ is divided in to small rectangles like ABCD each
of width dx which is extremely small. Through the displacement dx, F is
constant.
Work done during displacement dx is given by,
dW = Area under the curve AB.
dW = AC ´ CD Þ dW = F dx
The total work done from initial position x1 to the final position x2 is equal to the sum of the areas of all the
rectangles. It is given by,
x2
W = ò dW ! F and x are continuous variables.
x1
x2
W = ò Fdx ! F is the function of x.
x1

Energy:
It is the capacity to do work.
Mechanical energy:
It is the energy possessed by the body by virtue of its motion (or) configuration
Note:
Energy is a scalar quantity. The unit and dimensions are same as that of work done.

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Types of mechanical energy:
1) Kinetic energy and 2) Potential Energy.
Kinetic energy:
It is the energy possessed by the body by virtue of its motion.
Ex:
1. Flowing water possess kinetic energy. (This kinetic energy is used to run water mills).
2. Moving air (wind) possess kinetic energy. (This kinetic energy is used to run wind mills).
3. A moving hammer possesses kinetic energy. (This kinetic energy is used to drive the nails into wood).
4. A bullet fired from a gun possesses kinetic energy. (Due to this kinetic energy, the bullet penetrates
into a target)
In general, all moving bodies possess kinetic energy.
Kinetic energy of a body:
It is defined as the half of the product of mass and square of the velocity of the body.
Note:
1
1. The expression for kinetic energy is KE = mv 2 is applicable for variable force also. Thus, the
2
expression is valid irrespective of how the body acquires the velocity v.
2. We observe that KE µ m and also KE µ v 2 . The heavier body and fast moving body possess greater
kinetic energy.
3. Kinetic energy is always positive. It can never be negative.
4. Kinetic energy of a body depends upon the frame of reference. For example, kinetic energy of a person
1
of mass m sitting in a train with a velocity v is mv 2 in the frame of earth and kinetic energy of the
2
same person = 0, in the same frame of the train.
5. Relation between momentum and KE (Ek ) , P = 2mEk
6. The absolute unit of kinetic energy in SI is joule and in cgs is erg.
Work-energy theorem:
The work done by the net force in displacing a body is equal to the change in kinetic energy of the
body.
Consider a body of mass m moving with an initial velocity u
along a straight line. Let F be the constant force acts on the body in the
same direction. Let v be the final velocity of the body after covering a
distance x.
We have equation of motion,
v 2 = u 2 + 2ax Þ v 2 - u 2 = 2ax
m
Multiplying both sides of above equation by , we get
2
m
( ) m
2
1
2
1
Þ v 2 - u 2 = 2ax Þ mv 2 - mu 2 = m ax
2 2
Final KE - Initial KE = F x (! ma = F )
Change in kinetic energy = work done (! Fx = workdone)
This is work-energy theorem.
Potential Energy:
It is the energy possessed by the body by virtue of its position or configuration.
Types of potential energy
1) Gravitational potential energy.
2) Elastic potential energy.
3) Electrostatic potential energy.

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Gravitational potential energy:
It is the energy possessed by the body by virtue of its position above the earth's surface.
PE = V= mgh where m is the mass of the body, g is the acceleration due to gravity and h is the height of the
body from the surface of the earth.
Ex:
1. A stone placed on the roof of the house possess potential energy.
2. Water stored in the overhead tank possesses potential energy.
Conservative force:
A force is said to be conservative if the work done by the force depends on the initial and final positions
of the body.
Ex: (a) Gravitational force, (b) electrostatic force and (c) spring force.
Properties of conservative force:
1. Work done by this force depends on the initial and final positions of the body.
2. Work done in a round trip is zero.
3. Work done is completely recoverable.
4. It is central in nature
Non-conservative force:
A force is said to be non-conservative if the work done by the force depends on the path followed
between initial and final positions of the body.
Ex: (a) Frictional force, (b) air resistance and (c) viscous force.
Properties of non-conservative force:
1. Work done by this force depends on the path followed by body.
2. Work done in a round trip is not zero.
3. Work done is not completely recoverable
4. It is retarding in nature
Distinguish between conservative and non-conservative forces:
S.No Conservative force Non-conservative force
1 Work done by this force depends Work done by this force depends
on the initial and final positions of on the path followed by the body.
the body.
2 Work done in a round trip is Work done in a round trip is not
Zero. zero.
3 Work done is completely Work done is not recoverable.
recoverable.
4 It is central in nature. It is retarding in nature

Law of conservation of energy:

Energy can neither be created nor destroyed in any process, but it can be converted into one form to
another form. (or)
The total mechanical energy of an isolated system remains constant in any process.
(or)
The sum of all the energies in the universe remains constant.
(or)
Energy is always conserved when it transformed from one form to another.
Law of conservation of mechanical energy:
The total mechanical energy of an isolated system remains constant.
Verification of law of conservation of mechanical energy in the case of freely falling body:
Consider a body of mass m dropped from point A at a height h above the ground
At the point A:
From the law of conservation of energy
Total energy of the body, TE = PE + KE

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 Þ TE = mgh + mv A2 Þ TE = mgh + 0 (! v A = 0)
1
2
Þ TE = mgh ------------ (1)
At the point B:
As the body falls, its PE decreases but KE increases.
Let vB be the velocity of the body at the point B and x be the
distance traveled from A to B
TE = PE + KE
1
Þ TE = mg (h - x ) + mvB2 ------ (2)
2
Using the equation,
v 2 = u 2 + 2ax for the motion from A to B, we get
vB2 = 0 + 2 gx = 2 gx
1
(2) Þ TE = mg (h - x ) + m 2 gx Þ TE = mgh - mgx + m gx Þ TE = mgh ---------- (3)
2
At the point C:
Let vC be the velocity of the body at the point C and h be the distance traveled from A to C.
TE = PE + KE
1
Þ TE = 0 + mvC2 ------ (4) (PE = 0,! h = 0)
2
Using the equation,
v 2 = u 2 + 2ax for the motion from A to C, we get vC2 = 0 + 2 gh = 2 gh
1
(4) Þ TE = m ´ 2 gh
2
Þ TE = mgh ---------- (5)
From equations (1), (3) and (5), it is found that the total energy of the body remains constant at all points in
the path of the body. Thus the law of conservation of energy is verified
Elastic potential energy:
It is the energy possessed by the body due to its configuration.
Ex:
1) A wound spring in the clock can possess elastic potential energy.
2) A stretched bow can possess elastic potential energy.
Potential Energy of a spring:
It is the energy stored in the spring, when it is stretched or compressed.
In the figure,
F is the applied force
Fs is the spring force
x is the displacement of the block
Consider a spring OA of
negligible mass placed on a frictionless
surface. The end O of the spring is fixed
to rigid support and a block is attached
to the free end A. Let the spring is stretched from A to B through a distance x.
The spring is stretched through extremely small distance dx so that the applied force F remain constant.
Work done in stretching the spring through a distance dx is given by
dW = F ´ dx where F is the applied force
Work done in stretching the spring from x = 0 to x = x is given by

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 x x
W = ò dW Þ W = ò F ´ dx
0 0

The magnitude of applied force is equal to the magnitude of spring force, ie F = kx . where k is the spring
constant.
x
x
é x2 ù é x2 0 ù
Þ W = ò kx dx = k ê ú = k ê - ú
0 ë 2 û0 ë 2 2û
1
Þ W = kx 2
2
This work done in the spring stored as the potential energy of the spring and is given by
1
PE = kx 2
2
PE of a spring by graphical method:

The work done in stretching a spring and hence the PE can be calculated by F-x graph.
Work done in stretching a spring = Area of the shaded triangle
1 1 1
W = (OC ´ CA) Þ W = (x ´ kx ) Þ W = kx 2
2 2 2
This work done is stored in the spring as potential energy
1
PE = U = kx 2
2
Variation of PE, KE and TE of body attached to the vibrating spring (for a particle having SHM):

In the graph, the curves AOB and CED represents PE and KE respectively. PE is maximum at extreme
at positions and it is zero at mean position KE is maximum at mean position and it is zero at extreme positions.
The line AEB represents TE It is constant at all points between A and B.
Motion of a body in a vertical plane:
A bob of mass m is suspended by a light string of length L. It is imparted
a horizontal velocity v A at the lowest point A such that it completes a semicircular
trajectory in the vertical line with the string becoming slack only on reaching the
top most point C.
Given figure
At the lowest point A:
From the law of conservation of energy,
Altered figure

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 TE = PE + KE
1
Þ TE = 0 + mv A2
2
1 2
Þ TE = mv A ------------ (1)
2
At the highest point C:
We have, TE = PE + KE
1
Þ TE = mg 2 L + mvC2 ------------ (2)
2
The string slackens and the tension in the string becomes zero
at C.
Centripetal force is at the point C is given by
mv 2
Þ C = mg Þ vC2 = Lg Þ vC = Lg ------------ (3)
L
1
(3) in (2) Þ TE = mg 2 L + m Lg
2
æ 1ö æ5ö
Þ TE = mgLç 2 + ÷ Þ TE = mgLç ÷ ------- (4)
è 2ø è2ø
Compare (1) and (4), we get
1 2 æ5ö
mv A = mgLç ÷ Þ v A2 = gL5
2 è2ø
Þ v A = 5Lg -------- (5)
At the point B:
We have, TE = PE + KE
1
Þ TE = mgL + mvB2 ------------ (6)
2
æ5ö 1
Compare (4) and (6) Þ mgLç ÷ = mgL + mvB2
è2ø 2
æ5ö 1 æ5 ö 1 æ3ö 1
Þ mgLç ÷ - mgL = mvB2 Þ mgLç -1÷ = mvB2 Þ mgLç ÷ = mvB2
è2ø 2 è2 ø 2 è2ø 2
Þ vB2 = 3Lg Þ vB = 3Lg ------------ (7)
From (3), (5) and (7), v A : vB : vC = 5 : 3 : 1 and KA : KB : KC = 5 : 3 : 1
Power:
It is the rate of doing work
(or)
It is defined as the ratio of the work done and the time taken.
work done W
Power = ÞP=
Time t
SI unit of power is watt. Symbol is W
watt:
W
Consider the relation P = , If W =1J, t=1 second, then P = 1 watt.
t
The power is said to be 1 watt, when 1J of work is done in 1 second.
Horse power (H.P):
The power is said to be 1 HP, when 550ft poundal work is done in 1 second.
It can be shown that, 1 HP » 746 W

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 1. CGS unit of power is erg/s
2. FPS unit of power is ft poundal/s
3. Power is the scalar quantity.
4. Dimensional formula for power is [M1L2T-3]
5. The electrical energy is measured in kilowatt-hour (kWh). It is the electrical energy consumed in 1 hr
at the rate of 1000 J per second.
i.e. 1 kilowatt-hour = 1 kilowatt ´ 1 hour.
= 103 J/s ´ 3600 s
Þ 1kilowatt-hour = 36 ´ 105 J
Power is the dot product of force and velocity:
work done F x cos q
We have, Power = ÞP= (!W = F x )
Time t
æ x ö
Þ P = F v cos q ç! = v ÷
è t ø
! !
Þ P = F •v
Collision:
Collision is the physical striking between the bodies.
Ex:
1. Collision between the atoms, collision between the bodies. Etc.
2. Atoms and molecules of a gas enclosed in a container collide each other and collide with the walls of
the container.
3. When or particles (positively charged) are made to pass near the nucleus (positively charged), they
deflect from their normal path due to the electrostatic force of repulsion. Here physical collision is not
taking place, but a particle changes its path and hence an example for collision.
Elastic collision:
It is the collision in which both the momentum and kinetic energy of the system are conserved.
Ex:
1. Collision between the two glass balls or ivory balls is nearly elastic collision.
2. Collision between the atoms, molecules and sub atomic particles is nearly elastic.
Characteristics of elastic collision:
1. The total energy of the system is conserved.
2. The momentum is conserved.
3. Kinetic energy is conserved.
Inelastic collision:
It is the collision in which momentum is conserved but kinetic energy of the system is not conserved.
Ex: collision between two vehicles
Completely inelastic collision:
It is the collision in which two colliding bodies stick and move together after collision.
Ex:
1. Collision between the two wet clay balls.
2. Collision between a proton and an electron.
Characteristics of in-elastic collision:
1. The total energy of the system is conserved.
2. The momentum is conserved.
3. Kinetic energy is not conserved.
Head on collision (or) Collision in one dimension:

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 It is collision in which the initial velocities and final velocities of the two colliding bodies are along
the same straight line.
Expressions for final velocities of two bodies in one
dimensional elastic:
Consider two bodies A and B of masses m1
and m2 respectively moving in a straight line with
initial velocities u1 and u2 respectively before
collision. Let v1 and v2 be their final velocities after the collision. Assume m1 > m2 and u1 > u2 .
From the law of conservation of momentum,
Momentum of the system before collision = Momentum of the system after collision
m1 u1 + m2 u2 = m1 v1 + m2 v2 ----- (1)
Þ m1 u1 - m1 v1 = m2 v2 - m2 u2 Þ m1 ( u1- v1) = m2( v2 - u2) ----------- (2)
From the law of conservation of energy,
KE of the system before collision = KE of the system after collision.
1 1 1 1
m1u12 + m2u22 = m1v12 + m2 v22 Þ m1u12 - m1v12 = m2 v22 - m2u22
2 2 2 2
( ) ( )
Þ m1 u1 - v1 = m2 v22 - u22 ----------- (3)
2 2

(3) m (u - v ) m (v - u )
2 2
(u + v )(u - v ) = (v + u )(v - u )
2 2
\ Þ 1 1
= 1
Þ 2 2 2 1 1 1 1 2 2 2 2
(2) m (u - v ) m (v - u )
1 1 1 (u - v )
2 2 2(v - u ) 1 1 2 2

Þ (u + v ) = (v + u ) Þ v = v + u - u ------------- (4)
1 1 2 2 1 2 2 1

(4) in (1) Þ m u + m u = m (v + u - u )+ m v
1 1 2 2 1 2 2 1 2 2
Þ m1 u1 + m2 u2 = m1 v2 + m1 u2 - m1 u1+ m2 v2 Þ 2 m1 u1+ u2( m2- m1) = v2( m1+ m2)
2m u + u (m - m1 )
Þ v2 = 1 1 2 2 ------- (5)
(m1 + m2 )
Similarly for v1 Interconvert suffixes 1 and 2 in equation (5), we get
2m u + u (m - m2 )
Þ v1 = 2 2 1 1 ------- (6)
(m1 + m2 )
Case 1:
When the masses of the two bodies are equal, i.e. m1 = m2
2m2u2
Then, v1 = Þ v1 = u2
2m2
2m1u1
And v2 = Þ v2 = u1
2m1
If two bodies of equal masses undergo elastic collision in one dimension, the bodies exchange their
velocities after collision.
Case 2:
When the target body (B) is initially at rest, i.e. u2 = 0
u (m - m2 ) 2m1u1
Then, v1 = 1 1 and v2 =
(m1 + m2 ) (m1 + m2 )
Case 2(a):
When m1 >> m2 i.e. when the mass of B negligible as compared to the mass of the body A.
Then v1 » u1 and v2 » 2u1
Means, the heavier body moves without change in its velocity and lighter body moves with a velocity
two times the initial velocity of heavier body.
Case 2(b):

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 When m1 << m2 i.e. when the mass of A negligible as compared to the mass of the body B.
Then v1 » -u1 and v2 » 0
Means, the lighter body rebounds with almost same period and the heavier body gets negligible velocity.
Case 2(c):
When m1 = m2 i.e. the colliding bodies are of the same mass. Then v1 » 0 and v2 » u1
Means, the velocity of the body A is zero after collision and the velocity of the body B after collision is equal
to the velocity of the body before collision. If lighter bodies collide heavier bodies collide with lighter bodies
at rest, the motion of heavier body does not alter much. The lighter bodies experiences larger change in motion.
Expressions for common velocity of the two bodies and loss of kinetic energy in the case of one
dimensional inelastic collision (head-on collision):
Consider two bodies A and B of masses m1 and m2
respectively. Let the body A moving in a straight line with
an initial velocity u1 collides with the body B at rest ( u2 =0).
After the collision, the two bodies move together with a
common velocity V in the same direction.
Common velocity of the two bodies:
In the case of inelastic collision, momentum is conserved.
From the law of conservation of momentum,
Momentum of the system before collision = Momentum of the system after collision
m1 u1 + m2 (0) = m1 v+ m2 v Þ m1 u1 = ( m1 + m2 )v
m1u1
Þv = ----------- (1)
(m1 + m2 )
The above equation indicates common velocity of the two bodies.
For loss of kinetic energy:
In the case of inelastic collision, kinetic energy is not conserved. Some KE is lost in other forms of
energies ie KE before collision is greater than KE after collision.
\ Loss in KE = KE of the system before collision - KE of the system after collision
1 1 1 1 1 1
DK = m1u12 + m2 (0) - ( m1v 2 + m2 v 2 ) Þ DK = m1u12 - (m1 + m2 )v 2
2 2 2 2 2 2
m1u1
But v =
(m1 + m2 )
1é æ m1u1 ö ù 1 é
2
æ m12u12 öù
\ DK = êm1u1 - (m1 + m2 )çç
2
÷÷ ú = êm1u1 - (m1 + m2 )ç
2
÷
2ê (m + m ) ú 2 ç (m + m )2 ÷ú
ë è 1 2 ø û ëê è 1 2 øûú
1é æ m12u12 öù 1 é m1 ù 1 é m2 ù
Þ DK = êm1u12 - çç ÷÷ú = m1u12 ê1 - ú = m1u12 ê ú
2ë è (m1 + m2 ) øû 2 ë (m1 + m2 ) û 2 ë (m1 + m2 ) û
é m m u2 ù
Þ DK = ê 1 2 1 ú
ë 2(m1 + m2 ) û
If the velocity of the B is u2 before collision, then
( )
é m m u 2 - u22 ù
DK = ê 1 2 1 ú
ë 2(m1 + m2 ) û

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Elastic collisions in two dimensions:
Consider two perfectly elastic balls A and B
of masses m1 and m2 moving in the same direction
(say X-axis). Let u1 and u2 ( u1 > u2 ) be the velocities
of A and B respectively before the collision. After
the collision, the bodies move in different directions
with velocities v1 and v2 . Let the bodies A and B be
deflected at angles q1 & q 2 respectively with the
initial direction ( ie X-axis). The velocities v1 and v2
are resolved as shown in the figure.
From conservation of momentum,
Algebraic sum of X-components before collision = Algebraic sum of X-components after collision
m1 u1 + m2 u2 = m1 v1 cosq1 + m2 v2 cos q 2
Algebraic sum of Y-components before collision = Algebraic sum of Y-components after collision.
0 = m1 v1 sin q1 + m2 v2 sin q 2
We have to take difference of Y components after collision, because they are in opposite directions
From conservation of KE,
KE of the system before collision = KE of the system after collision.
1 1 1 1
m1u12 + m2u22 = m1v12 + m2 v22
2 2 2 2
Motion after collision involves four unknown quantities, they are v1 , v2 , q1 and q 2 . To calculate the
values of these four quantities, one of these quantities should be known.
Impact parameter:
The perpendicular distance (b) between the lines passing through the centers of
bodies is called the impact parameter. It is equal to the distance between the centers of
bodies during collision. This is a measure of the closeness of the collision. Impact
parameter is zero for head-on collision.
Coefficient of restitution (e):
The collisions which are neither perfectly elastic nor perfectly inelastic are called inelastic collisions.
The degree of elasticity of a collision is determined by a quantity called coefficient of restitution. Its value can
be calculated using law of restitution.
Law of restitution:
Relative velocity of separation after collision µ Relative velocity of approach before collision
Relative velocity of separation after collision = e ´ Relative velocity of approach before collision
Relative velocity of separation after collision
e=
Relative velocity of approach before collision
where e is the proportionality constant and is called as coefficient of restitution.
Coefficient of restitution (e):
It is the ratio of relative velocity of separation after collision to relative velocity of approach before
collision.
Let u1 and u2 be the velocities of approach before collision and v1 and v2 be the velocities of
separation after collision.
v -v
Then, e = 2 1
u1 - u2
For perfectly elastic collisions, e = 1
For perfectly inelastic collisions, e = 0
For inelastic collisions, e < l

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