NCERT Textbook - Work, Energy and Power Class 11 Notes | EduRev

Physics Class 11

Created by: Infinity Academy

Class 11 : NCERT Textbook - Work, Energy and Power Class 11 Notes | EduRev

 Page 1


CHAPTER SIX
WORK, ENERGY AND POWER
6.1  INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
a competitive examination, an artist painting a beautiful
landscape, all are said to be working. In physics, however,
the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
day is said to have a large stamina or energy. We admire a
long distance runner for her stamina or energy. Energy is
thus our capacity to do work. In Physics too, the term ‘energy’
is related to work in this sense, but as said above the term
‘work’ itself is defined much more precisely. The word ‘power’
is used in everyday life with different shades of meaning. In
karate or boxing we talk of ‘powerful’ punches. These are
delivered at a great speed. This shade of meaning is close to
the meaning of the word ‘power’ used in physics. We shall
find that there is at best a loose correlation between the
physical definitions and the physiological pictures these
terms generate in our minds. The aim of this chapter is to
develop an understanding of these three physical quantities.
Before we proceed to this task, we need to develop a
mathematical prerequisite, namely the scalar product of two
vectors.
6.1.1 The Scalar Product
We have learnt about vectors and their use in Chapter 4.
Physical quantities like displacement, velocity, acceleration,
force etc. are vectors. We have also learnt how vectors are
added or subtracted. W e now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A
.
B (read
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The concept of potential
energy
6.8 The conservation of
mechanical energy
6.9 The potential energy of a
spring
6.10 Various forms of energy : the
law of conservation of energy
6.11 Power
6.12 Collisions
Summary
Points to ponder
Exercises
Additional exercises
Appendix 6.1
2015-16(20/01/2015)
Page 2


CHAPTER SIX
WORK, ENERGY AND POWER
6.1  INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
a competitive examination, an artist painting a beautiful
landscape, all are said to be working. In physics, however,
the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
day is said to have a large stamina or energy. We admire a
long distance runner for her stamina or energy. Energy is
thus our capacity to do work. In Physics too, the term ‘energy’
is related to work in this sense, but as said above the term
‘work’ itself is defined much more precisely. The word ‘power’
is used in everyday life with different shades of meaning. In
karate or boxing we talk of ‘powerful’ punches. These are
delivered at a great speed. This shade of meaning is close to
the meaning of the word ‘power’ used in physics. We shall
find that there is at best a loose correlation between the
physical definitions and the physiological pictures these
terms generate in our minds. The aim of this chapter is to
develop an understanding of these three physical quantities.
Before we proceed to this task, we need to develop a
mathematical prerequisite, namely the scalar product of two
vectors.
6.1.1 The Scalar Product
We have learnt about vectors and their use in Chapter 4.
Physical quantities like displacement, velocity, acceleration,
force etc. are vectors. We have also learnt how vectors are
added or subtracted. W e now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A
.
B (read
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The concept of potential
energy
6.8 The conservation of
mechanical energy
6.9 The potential energy of a
spring
6.10 Various forms of energy : the
law of conservation of energy
6.11 Power
6.12 Collisions
Summary
Points to ponder
Exercises
Additional exercises
Appendix 6.1
2015-16(20/01/2015)
A dot B) is defined as
A
.
B = A B cos ? (6.1a)
where ?  is the angle between the two vectors as
shown in Fig. 6.1(a). Since A, B and cos ? are
scalars, the dot product of A and B is a scalar
quantity. Each vector, A and B, has a direction
but their scalar product does not have a
direction.
From Eq. (6.1a), we have
A
.
B  = A (B cos ? )
       = B (A cos ? )
Geometrically, B cos ? is the projection of B onto
A in Fig.6.1 (b) and A cos ?  is the projection of A
onto B in Fig. 6.1 (c). So, A
.
B is the product of
the magnitude of A and the component of B along
A. Alternatively, it is the product of the
magnitude of B and the component of A along B.
Equation (6.1a) shows that the scalar product
follows the commutative law :
A
.
B = B
.
A
Scalar product obeys the distributive
law:
A
.
 (B + C) = A
.
B + A
.
C
Further, A
.
 ( ? B) = ? (A
.
B)
where ? is a real number.
The proofs  of the above equations are left to
you as an exercise.
For unit vectors 
$ $ $
i,j,k we have
$ $ $ $ $ $
i i j j k k · = · = · =1
$ $ $ $ $ $
i j j k k i · = · = · = 0
Given two vectors
A i j k = + + A A A
x y z
$ $ $
B i j k = + + B B B
x y z
$ $ $
their scalar product is
    
ˆ ˆ ˆ ˆ ˆ ˆ . .
x y z x y z
A A A B B B     A B i j k i j k
 = + + A B A B A B
x x y y z z
(6.1b)
From the definition of scalar product and
(Eq. 6.1b) we have :
( i )
x x y y z z
A A A A A A   A A 
  
 
Or, A A A A
2
x
2
y
2
z
2
= + + (6.1c)
sinceA
.
A = |A ||A| cos 0 = A
2
.
(ii) A
.
B = 0, if A and B are perpendicular.
Example 6.1  Find the angle between force
F = 
ˆ
(3 i + 4 j – 5 k) 
$ $
unit and displacement
d = 
ˆ
(5 i + 4 j + 3 k) 
$ $
unit. Also find the
projection of F on d.
Answer F
.
d =
x x y y z z
F d F d F d  
= 3 (5) + 4 (4) + (– 5) (3)
= 16 unit
Hence F
.
d = cos F d  = 16 unit
Now F
.
F = 
2 2 2 2
  
x y z
F F F F   
= 9 + 16 + 25
= 50 unit
and d
.
d = d
2 
= 
2 2 2
  
x y z
d d d  
= 25 + 16 + 9
= 50 unit
  cos  = 
16 16
= =0.32
50 50 50
,
 = cos
–1
  0.32
Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A
.
B = A B cos ?.  (b) B cos ? is the projection
of B onto A. (c) A cos ? is the projection of A onto B.
WORK, ENERGY AND POWER 115
u
2015-16(20/01/2015)
Page 3


CHAPTER SIX
WORK, ENERGY AND POWER
6.1  INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
a competitive examination, an artist painting a beautiful
landscape, all are said to be working. In physics, however,
the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
day is said to have a large stamina or energy. We admire a
long distance runner for her stamina or energy. Energy is
thus our capacity to do work. In Physics too, the term ‘energy’
is related to work in this sense, but as said above the term
‘work’ itself is defined much more precisely. The word ‘power’
is used in everyday life with different shades of meaning. In
karate or boxing we talk of ‘powerful’ punches. These are
delivered at a great speed. This shade of meaning is close to
the meaning of the word ‘power’ used in physics. We shall
find that there is at best a loose correlation between the
physical definitions and the physiological pictures these
terms generate in our minds. The aim of this chapter is to
develop an understanding of these three physical quantities.
Before we proceed to this task, we need to develop a
mathematical prerequisite, namely the scalar product of two
vectors.
6.1.1 The Scalar Product
We have learnt about vectors and their use in Chapter 4.
Physical quantities like displacement, velocity, acceleration,
force etc. are vectors. We have also learnt how vectors are
added or subtracted. W e now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A
.
B (read
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The concept of potential
energy
6.8 The conservation of
mechanical energy
6.9 The potential energy of a
spring
6.10 Various forms of energy : the
law of conservation of energy
6.11 Power
6.12 Collisions
Summary
Points to ponder
Exercises
Additional exercises
Appendix 6.1
2015-16(20/01/2015)
A dot B) is defined as
A
.
B = A B cos ? (6.1a)
where ?  is the angle between the two vectors as
shown in Fig. 6.1(a). Since A, B and cos ? are
scalars, the dot product of A and B is a scalar
quantity. Each vector, A and B, has a direction
but their scalar product does not have a
direction.
From Eq. (6.1a), we have
A
.
B  = A (B cos ? )
       = B (A cos ? )
Geometrically, B cos ? is the projection of B onto
A in Fig.6.1 (b) and A cos ?  is the projection of A
onto B in Fig. 6.1 (c). So, A
.
B is the product of
the magnitude of A and the component of B along
A. Alternatively, it is the product of the
magnitude of B and the component of A along B.
Equation (6.1a) shows that the scalar product
follows the commutative law :
A
.
B = B
.
A
Scalar product obeys the distributive
law:
A
.
 (B + C) = A
.
B + A
.
C
Further, A
.
 ( ? B) = ? (A
.
B)
where ? is a real number.
The proofs  of the above equations are left to
you as an exercise.
For unit vectors 
$ $ $
i,j,k we have
$ $ $ $ $ $
i i j j k k · = · = · =1
$ $ $ $ $ $
i j j k k i · = · = · = 0
Given two vectors
A i j k = + + A A A
x y z
$ $ $
B i j k = + + B B B
x y z
$ $ $
their scalar product is
    
ˆ ˆ ˆ ˆ ˆ ˆ . .
x y z x y z
A A A B B B     A B i j k i j k
 = + + A B A B A B
x x y y z z
(6.1b)
From the definition of scalar product and
(Eq. 6.1b) we have :
( i )
x x y y z z
A A A A A A   A A 
  
 
Or, A A A A
2
x
2
y
2
z
2
= + + (6.1c)
sinceA
.
A = |A ||A| cos 0 = A
2
.
(ii) A
.
B = 0, if A and B are perpendicular.
Example 6.1  Find the angle between force
F = 
ˆ
(3 i + 4 j – 5 k) 
$ $
unit and displacement
d = 
ˆ
(5 i + 4 j + 3 k) 
$ $
unit. Also find the
projection of F on d.
Answer F
.
d =
x x y y z z
F d F d F d  
= 3 (5) + 4 (4) + (– 5) (3)
= 16 unit
Hence F
.
d = cos F d  = 16 unit
Now F
.
F = 
2 2 2 2
  
x y z
F F F F   
= 9 + 16 + 25
= 50 unit
and d
.
d = d
2 
= 
2 2 2
  
x y z
d d d  
= 25 + 16 + 9
= 50 unit
  cos  = 
16 16
= =0.32
50 50 50
,
 = cos
–1
  0.32
Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A
.
B = A B cos ?.  (b) B cos ? is the projection
of B onto A. (c) A cos ? is the projection of A onto B.
WORK, ENERGY AND POWER 115
u
2015-16(20/01/2015)
PHYSICS 116
?
known to be proportional to the speed of
the drop but is otherwise undetermined.
Consider a drop of mass 1.00 g falling from
a height 1.00 km.  It hits the ground with
a speed of 50.0 m s
-1
.  (a) What is the work
done by the gravitational force ? What is
the work done by the unknown resistive
force?
Answer (a) The change in kinetic energy of the
drop is
2
1
0
2
K m v   
=
1
2
10 50 50
-3
× × ×
= 1.25 J
where we have assumed that the drop is initially
at rest.
Assuming that g is a constant with a value
10 m/s
2
, the work done by the gravitational force
is,
W
g 
= mgh
      = 10
-3
 ×10 ×10
3
      = 10.0 J
(b) From the work-energy theorem
g r
K  W W     
where W
r
  is the work done by the resistive force
on the raindrop.  Thus
W
r 
 = ?K - W
g
     = 1.25 -10
      = - 8.75 J
is negative. ?
6.3  WORK
As seen earlier, work is related to force and the
displacement over which it acts.  Consider a
constant force F acting on an object of mass m.
The object undergoes a displacement d in the
positive x-direction as shown in Fig. 6.2.
Fig. 6.2 An object undergoes a displacement d
under the influence of the for ce F.
6.2 NOTIONS OF WORK AND KINETIC
ENERGY: THE WORK-ENERGY THEOREM
The following relation for rectilinear motion under
constant acceleration a has been encountered
in Chapter 3,
v
2
 - u
2
 = 2 as
where u and v  are the initial and final speeds
and s the distance traversed.  Multiplying both
sides by m/2, we have
2 2
1 1
2 2
mv mu mas Fs   (6.2a)
where the last step follows from Newton’s
Second Law. We can generalise Eq. (6.1)
to three dimensions by employing
vectors
v
2 
- u
2 
= 2 a.d
Once again multiplying both sides by m/2 , we
obtain
2 2
1 1
2 2
mv mu m    a.d F.d (6.2b)
The above equation provides a motivation for
the definitions of work and kinetic energy. The
left side of the equation is the difference in the
quantity ‘half the mass times the square of the
speed’ from its initial value to its final value. We
call each of these quantities the ‘kinetic energy’,
denoted by K. The right side is a product of the
displacement and the component of the force
along the displacement.  This quantity is called
‘work’ and is denoted by W.  Eq. (6.2b) is then
K
f 
-  K
i 
= W (6.3)
where K
i
  and K
f
  are respectively the initial and
final kinetic energies of the object. Work refers
to the force and the displacement over which it
acts. Work is done by a force on the body over
a certain displacement.
Equation (6.2) is also a special case of the
work-energy (WE) theorem : The change in
kinetic energy of a particle is equal to the
work done on it by the net force. We shall
generalise the above derivation to a varying force
in a later section.
Example 6.2  It is well known that a
raindrop falls under the influence of the
downward gravitational force and the
opposing resistive force.  The latter is
2015-16(20/01/2015)
Page 4


CHAPTER SIX
WORK, ENERGY AND POWER
6.1  INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
a competitive examination, an artist painting a beautiful
landscape, all are said to be working. In physics, however,
the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
day is said to have a large stamina or energy. We admire a
long distance runner for her stamina or energy. Energy is
thus our capacity to do work. In Physics too, the term ‘energy’
is related to work in this sense, but as said above the term
‘work’ itself is defined much more precisely. The word ‘power’
is used in everyday life with different shades of meaning. In
karate or boxing we talk of ‘powerful’ punches. These are
delivered at a great speed. This shade of meaning is close to
the meaning of the word ‘power’ used in physics. We shall
find that there is at best a loose correlation between the
physical definitions and the physiological pictures these
terms generate in our minds. The aim of this chapter is to
develop an understanding of these three physical quantities.
Before we proceed to this task, we need to develop a
mathematical prerequisite, namely the scalar product of two
vectors.
6.1.1 The Scalar Product
We have learnt about vectors and their use in Chapter 4.
Physical quantities like displacement, velocity, acceleration,
force etc. are vectors. We have also learnt how vectors are
added or subtracted. W e now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A
.
B (read
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The concept of potential
energy
6.8 The conservation of
mechanical energy
6.9 The potential energy of a
spring
6.10 Various forms of energy : the
law of conservation of energy
6.11 Power
6.12 Collisions
Summary
Points to ponder
Exercises
Additional exercises
Appendix 6.1
2015-16(20/01/2015)
A dot B) is defined as
A
.
B = A B cos ? (6.1a)
where ?  is the angle between the two vectors as
shown in Fig. 6.1(a). Since A, B and cos ? are
scalars, the dot product of A and B is a scalar
quantity. Each vector, A and B, has a direction
but their scalar product does not have a
direction.
From Eq. (6.1a), we have
A
.
B  = A (B cos ? )
       = B (A cos ? )
Geometrically, B cos ? is the projection of B onto
A in Fig.6.1 (b) and A cos ?  is the projection of A
onto B in Fig. 6.1 (c). So, A
.
B is the product of
the magnitude of A and the component of B along
A. Alternatively, it is the product of the
magnitude of B and the component of A along B.
Equation (6.1a) shows that the scalar product
follows the commutative law :
A
.
B = B
.
A
Scalar product obeys the distributive
law:
A
.
 (B + C) = A
.
B + A
.
C
Further, A
.
 ( ? B) = ? (A
.
B)
where ? is a real number.
The proofs  of the above equations are left to
you as an exercise.
For unit vectors 
$ $ $
i,j,k we have
$ $ $ $ $ $
i i j j k k · = · = · =1
$ $ $ $ $ $
i j j k k i · = · = · = 0
Given two vectors
A i j k = + + A A A
x y z
$ $ $
B i j k = + + B B B
x y z
$ $ $
their scalar product is
    
ˆ ˆ ˆ ˆ ˆ ˆ . .
x y z x y z
A A A B B B     A B i j k i j k
 = + + A B A B A B
x x y y z z
(6.1b)
From the definition of scalar product and
(Eq. 6.1b) we have :
( i )
x x y y z z
A A A A A A   A A 
  
 
Or, A A A A
2
x
2
y
2
z
2
= + + (6.1c)
sinceA
.
A = |A ||A| cos 0 = A
2
.
(ii) A
.
B = 0, if A and B are perpendicular.
Example 6.1  Find the angle between force
F = 
ˆ
(3 i + 4 j – 5 k) 
$ $
unit and displacement
d = 
ˆ
(5 i + 4 j + 3 k) 
$ $
unit. Also find the
projection of F on d.
Answer F
.
d =
x x y y z z
F d F d F d  
= 3 (5) + 4 (4) + (– 5) (3)
= 16 unit
Hence F
.
d = cos F d  = 16 unit
Now F
.
F = 
2 2 2 2
  
x y z
F F F F   
= 9 + 16 + 25
= 50 unit
and d
.
d = d
2 
= 
2 2 2
  
x y z
d d d  
= 25 + 16 + 9
= 50 unit
  cos  = 
16 16
= =0.32
50 50 50
,
 = cos
–1
  0.32
Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A
.
B = A B cos ?.  (b) B cos ? is the projection
of B onto A. (c) A cos ? is the projection of A onto B.
WORK, ENERGY AND POWER 115
u
2015-16(20/01/2015)
PHYSICS 116
?
known to be proportional to the speed of
the drop but is otherwise undetermined.
Consider a drop of mass 1.00 g falling from
a height 1.00 km.  It hits the ground with
a speed of 50.0 m s
-1
.  (a) What is the work
done by the gravitational force ? What is
the work done by the unknown resistive
force?
Answer (a) The change in kinetic energy of the
drop is
2
1
0
2
K m v   
=
1
2
10 50 50
-3
× × ×
= 1.25 J
where we have assumed that the drop is initially
at rest.
Assuming that g is a constant with a value
10 m/s
2
, the work done by the gravitational force
is,
W
g 
= mgh
      = 10
-3
 ×10 ×10
3
      = 10.0 J
(b) From the work-energy theorem
g r
K  W W     
where W
r
  is the work done by the resistive force
on the raindrop.  Thus
W
r 
 = ?K - W
g
     = 1.25 -10
      = - 8.75 J
is negative. ?
6.3  WORK
As seen earlier, work is related to force and the
displacement over which it acts.  Consider a
constant force F acting on an object of mass m.
The object undergoes a displacement d in the
positive x-direction as shown in Fig. 6.2.
Fig. 6.2 An object undergoes a displacement d
under the influence of the for ce F.
6.2 NOTIONS OF WORK AND KINETIC
ENERGY: THE WORK-ENERGY THEOREM
The following relation for rectilinear motion under
constant acceleration a has been encountered
in Chapter 3,
v
2
 - u
2
 = 2 as
where u and v  are the initial and final speeds
and s the distance traversed.  Multiplying both
sides by m/2, we have
2 2
1 1
2 2
mv mu mas Fs   (6.2a)
where the last step follows from Newton’s
Second Law. We can generalise Eq. (6.1)
to three dimensions by employing
vectors
v
2 
- u
2 
= 2 a.d
Once again multiplying both sides by m/2 , we
obtain
2 2
1 1
2 2
mv mu m    a.d F.d (6.2b)
The above equation provides a motivation for
the definitions of work and kinetic energy. The
left side of the equation is the difference in the
quantity ‘half the mass times the square of the
speed’ from its initial value to its final value. We
call each of these quantities the ‘kinetic energy’,
denoted by K. The right side is a product of the
displacement and the component of the force
along the displacement.  This quantity is called
‘work’ and is denoted by W.  Eq. (6.2b) is then
K
f 
-  K
i 
= W (6.3)
where K
i
  and K
f
  are respectively the initial and
final kinetic energies of the object. Work refers
to the force and the displacement over which it
acts. Work is done by a force on the body over
a certain displacement.
Equation (6.2) is also a special case of the
work-energy (WE) theorem : The change in
kinetic energy of a particle is equal to the
work done on it by the net force. We shall
generalise the above derivation to a varying force
in a later section.
Example 6.2  It is well known that a
raindrop falls under the influence of the
downward gravitational force and the
opposing resistive force.  The latter is
2015-16(20/01/2015)
WORK, ENERGY AND POWER 117
?
Table 6.1   Alternative Units of Work/Energy in J
Example 6.3  A cyclist comes to a skidding
stop in 10 m.  During this process, the force
on the cycle due to the road is 200 N and
is directly opposed to the motion.  (a) How
much work does the road do on the cycle ?
(b) How much work does the cycle do on
the road ?
Answer  Work done on the cycle by the road is
the work done by the stopping (frictional) force
on the cycle due to the road.
(a) The stopping force and the displacement make
an angle of 180
o
  (p rad) with each other.
Thus, work done by the road,
W
r
 =  Fd cos?
     =  200 × 10 × cos p
      = – 2000 J
It is this negative work that brings the cycle
to a halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and
opposite force acts on the road due to the
cycle. Its magnitude is 200 N. However, the
road undergoes no displacement.  Thus,
work done by cycle on the road is zero.            ?
The lesson of Example 6.3 is that though the
force on a body A exerted by the body B is always
equal and opposite to that on B by A (Newton’s
Third Law); the work done on A by B is not
necessarily equal and opposite to the work done
on B by A.
6.4  KINETIC ENERGY
As noted earlier, if an object of mass m has
velocity v, its kinetic energy K  is
2
K m mv
  
  
  
v v .                       (6.5)
Kinetic energy is a scalar quantity. The kinetic
energy of an object is a measure of the work an
The work done by the force is defined to be
the product of component of the force in the
direction of the displacement and the
magnitude of this displacement.  Thus
W = (F cos ? )d = F.d (6.4)
We see that if there is no displacement, there
is no work done even if the force is large.  Thus,
when you push hard against a rigid brick wall,
the force you exert on the wall does no work.  Yet
your muscles are alternatively contracting and
relaxing and internal energy is being used up
and you do get tired.  Thus, the meaning of work
in physics is different from its usage in everyday
language.
No work is done if :
(i) the displacement is zero as seen in the
example above. A weightlifter holding a 150
kg mass steadily on his shoulder for 30 s
does no work on the load during this time.
(ii) the force is zero.  A block moving on a smooth
horizontal table is not acted upon by a
horizontal force (since there is no friction), but
may undergo a large displacement.
(iii) the force and displacement are mutually
perpendicular. This is so since, for ? = p/2 rad
(= 90
o
), cos (p/2) = 0.  For the block moving on
a smooth horizontal table, the gravitational
force mg  does no work since it acts at right
angles to the displacement. If we assume that
the moon’s orbits around the earth is
perfectly circular then the earth’s
gravitational force does no work.  The moon’s
instantaneous displacement is tangential
while the earth’s force is radially inwards and
?  = p/2.
Work can be both positive and negative.  If ?  is
between 0
o
 and 90
o
, cos ?  in Eq. (6.4) is positive.
If  ?  is  between 90
o
 and 180
o
,   cos ?  is negative.
In many examples the frictional force opposes
displacement and ?  = 180
o
. Then the work done
by friction is negative (cos 180
o
 = –1).
From Eq. (6.4) it is clear that  work and energy
have the same dimensions,  [ML
2
T
–2
]. The SI unit
of these is joule (J), named after the famous British
physicist James Prescott Joule  (1811-1869). Since
work and energy are so widely used as physical
concepts, alternative units abound and some of
these are listed in Table 6.1.
2015-16(20/01/2015)
Page 5


CHAPTER SIX
WORK, ENERGY AND POWER
6.1  INTRODUCTION
The terms ‘work’, ‘energy’ and ‘power’ are frequently used
in everyday language. A farmer ploughing the field, a
construction worker carrying bricks, a student studying for
a competitive examination, an artist painting a beautiful
landscape, all are said to be working. In physics, however,
the word ‘Work’ covers a definite and precise meaning.
Somebody who has the capacity to work for 14-16 hours a
day is said to have a large stamina or energy. We admire a
long distance runner for her stamina or energy. Energy is
thus our capacity to do work. In Physics too, the term ‘energy’
is related to work in this sense, but as said above the term
‘work’ itself is defined much more precisely. The word ‘power’
is used in everyday life with different shades of meaning. In
karate or boxing we talk of ‘powerful’ punches. These are
delivered at a great speed. This shade of meaning is close to
the meaning of the word ‘power’ used in physics. We shall
find that there is at best a loose correlation between the
physical definitions and the physiological pictures these
terms generate in our minds. The aim of this chapter is to
develop an understanding of these three physical quantities.
Before we proceed to this task, we need to develop a
mathematical prerequisite, namely the scalar product of two
vectors.
6.1.1 The Scalar Product
We have learnt about vectors and their use in Chapter 4.
Physical quantities like displacement, velocity, acceleration,
force etc. are vectors. We have also learnt how vectors are
added or subtracted. W e now need to know how vectors are
multiplied. There are two ways of multiplying vectors which
we shall come across : one way known as the scalar product
gives a scalar from two vectors and the other known as the
vector product produces a new vector from two vectors. We
shall look at the vector product in Chapter 7. Here we take
up the scalar product of two vectors. The scalar product or
dot product of any two vectors A and B, denoted as A
.
B (read
6.1 Introduction
6.2 Notions of work and kinetic
energy : The work-energy
theorem
6.3 Work
6.4 Kinetic energy
6.5 Work done by a variable
force
6.6 The work-energy theorem for
a variable force
6.7 The concept of potential
energy
6.8 The conservation of
mechanical energy
6.9 The potential energy of a
spring
6.10 Various forms of energy : the
law of conservation of energy
6.11 Power
6.12 Collisions
Summary
Points to ponder
Exercises
Additional exercises
Appendix 6.1
2015-16(20/01/2015)
A dot B) is defined as
A
.
B = A B cos ? (6.1a)
where ?  is the angle between the two vectors as
shown in Fig. 6.1(a). Since A, B and cos ? are
scalars, the dot product of A and B is a scalar
quantity. Each vector, A and B, has a direction
but their scalar product does not have a
direction.
From Eq. (6.1a), we have
A
.
B  = A (B cos ? )
       = B (A cos ? )
Geometrically, B cos ? is the projection of B onto
A in Fig.6.1 (b) and A cos ?  is the projection of A
onto B in Fig. 6.1 (c). So, A
.
B is the product of
the magnitude of A and the component of B along
A. Alternatively, it is the product of the
magnitude of B and the component of A along B.
Equation (6.1a) shows that the scalar product
follows the commutative law :
A
.
B = B
.
A
Scalar product obeys the distributive
law:
A
.
 (B + C) = A
.
B + A
.
C
Further, A
.
 ( ? B) = ? (A
.
B)
where ? is a real number.
The proofs  of the above equations are left to
you as an exercise.
For unit vectors 
$ $ $
i,j,k we have
$ $ $ $ $ $
i i j j k k · = · = · =1
$ $ $ $ $ $
i j j k k i · = · = · = 0
Given two vectors
A i j k = + + A A A
x y z
$ $ $
B i j k = + + B B B
x y z
$ $ $
their scalar product is
    
ˆ ˆ ˆ ˆ ˆ ˆ . .
x y z x y z
A A A B B B     A B i j k i j k
 = + + A B A B A B
x x y y z z
(6.1b)
From the definition of scalar product and
(Eq. 6.1b) we have :
( i )
x x y y z z
A A A A A A   A A 
  
 
Or, A A A A
2
x
2
y
2
z
2
= + + (6.1c)
sinceA
.
A = |A ||A| cos 0 = A
2
.
(ii) A
.
B = 0, if A and B are perpendicular.
Example 6.1  Find the angle between force
F = 
ˆ
(3 i + 4 j – 5 k) 
$ $
unit and displacement
d = 
ˆ
(5 i + 4 j + 3 k) 
$ $
unit. Also find the
projection of F on d.
Answer F
.
d =
x x y y z z
F d F d F d  
= 3 (5) + 4 (4) + (– 5) (3)
= 16 unit
Hence F
.
d = cos F d  = 16 unit
Now F
.
F = 
2 2 2 2
  
x y z
F F F F   
= 9 + 16 + 25
= 50 unit
and d
.
d = d
2 
= 
2 2 2
  
x y z
d d d  
= 25 + 16 + 9
= 50 unit
  cos  = 
16 16
= =0.32
50 50 50
,
 = cos
–1
  0.32
Fig. 6.1 (a) The scalar product of two vectors A and B is a scalar : A
.
B = A B cos ?.  (b) B cos ? is the projection
of B onto A. (c) A cos ? is the projection of A onto B.
WORK, ENERGY AND POWER 115
u
2015-16(20/01/2015)
PHYSICS 116
?
known to be proportional to the speed of
the drop but is otherwise undetermined.
Consider a drop of mass 1.00 g falling from
a height 1.00 km.  It hits the ground with
a speed of 50.0 m s
-1
.  (a) What is the work
done by the gravitational force ? What is
the work done by the unknown resistive
force?
Answer (a) The change in kinetic energy of the
drop is
2
1
0
2
K m v   
=
1
2
10 50 50
-3
× × ×
= 1.25 J
where we have assumed that the drop is initially
at rest.
Assuming that g is a constant with a value
10 m/s
2
, the work done by the gravitational force
is,
W
g 
= mgh
      = 10
-3
 ×10 ×10
3
      = 10.0 J
(b) From the work-energy theorem
g r
K  W W     
where W
r
  is the work done by the resistive force
on the raindrop.  Thus
W
r 
 = ?K - W
g
     = 1.25 -10
      = - 8.75 J
is negative. ?
6.3  WORK
As seen earlier, work is related to force and the
displacement over which it acts.  Consider a
constant force F acting on an object of mass m.
The object undergoes a displacement d in the
positive x-direction as shown in Fig. 6.2.
Fig. 6.2 An object undergoes a displacement d
under the influence of the for ce F.
6.2 NOTIONS OF WORK AND KINETIC
ENERGY: THE WORK-ENERGY THEOREM
The following relation for rectilinear motion under
constant acceleration a has been encountered
in Chapter 3,
v
2
 - u
2
 = 2 as
where u and v  are the initial and final speeds
and s the distance traversed.  Multiplying both
sides by m/2, we have
2 2
1 1
2 2
mv mu mas Fs   (6.2a)
where the last step follows from Newton’s
Second Law. We can generalise Eq. (6.1)
to three dimensions by employing
vectors
v
2 
- u
2 
= 2 a.d
Once again multiplying both sides by m/2 , we
obtain
2 2
1 1
2 2
mv mu m    a.d F.d (6.2b)
The above equation provides a motivation for
the definitions of work and kinetic energy. The
left side of the equation is the difference in the
quantity ‘half the mass times the square of the
speed’ from its initial value to its final value. We
call each of these quantities the ‘kinetic energy’,
denoted by K. The right side is a product of the
displacement and the component of the force
along the displacement.  This quantity is called
‘work’ and is denoted by W.  Eq. (6.2b) is then
K
f 
-  K
i 
= W (6.3)
where K
i
  and K
f
  are respectively the initial and
final kinetic energies of the object. Work refers
to the force and the displacement over which it
acts. Work is done by a force on the body over
a certain displacement.
Equation (6.2) is also a special case of the
work-energy (WE) theorem : The change in
kinetic energy of a particle is equal to the
work done on it by the net force. We shall
generalise the above derivation to a varying force
in a later section.
Example 6.2  It is well known that a
raindrop falls under the influence of the
downward gravitational force and the
opposing resistive force.  The latter is
2015-16(20/01/2015)
WORK, ENERGY AND POWER 117
?
Table 6.1   Alternative Units of Work/Energy in J
Example 6.3  A cyclist comes to a skidding
stop in 10 m.  During this process, the force
on the cycle due to the road is 200 N and
is directly opposed to the motion.  (a) How
much work does the road do on the cycle ?
(b) How much work does the cycle do on
the road ?
Answer  Work done on the cycle by the road is
the work done by the stopping (frictional) force
on the cycle due to the road.
(a) The stopping force and the displacement make
an angle of 180
o
  (p rad) with each other.
Thus, work done by the road,
W
r
 =  Fd cos?
     =  200 × 10 × cos p
      = – 2000 J
It is this negative work that brings the cycle
to a halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and
opposite force acts on the road due to the
cycle. Its magnitude is 200 N. However, the
road undergoes no displacement.  Thus,
work done by cycle on the road is zero.            ?
The lesson of Example 6.3 is that though the
force on a body A exerted by the body B is always
equal and opposite to that on B by A (Newton’s
Third Law); the work done on A by B is not
necessarily equal and opposite to the work done
on B by A.
6.4  KINETIC ENERGY
As noted earlier, if an object of mass m has
velocity v, its kinetic energy K  is
2
K m mv
  
  
  
v v .                       (6.5)
Kinetic energy is a scalar quantity. The kinetic
energy of an object is a measure of the work an
The work done by the force is defined to be
the product of component of the force in the
direction of the displacement and the
magnitude of this displacement.  Thus
W = (F cos ? )d = F.d (6.4)
We see that if there is no displacement, there
is no work done even if the force is large.  Thus,
when you push hard against a rigid brick wall,
the force you exert on the wall does no work.  Yet
your muscles are alternatively contracting and
relaxing and internal energy is being used up
and you do get tired.  Thus, the meaning of work
in physics is different from its usage in everyday
language.
No work is done if :
(i) the displacement is zero as seen in the
example above. A weightlifter holding a 150
kg mass steadily on his shoulder for 30 s
does no work on the load during this time.
(ii) the force is zero.  A block moving on a smooth
horizontal table is not acted upon by a
horizontal force (since there is no friction), but
may undergo a large displacement.
(iii) the force and displacement are mutually
perpendicular. This is so since, for ? = p/2 rad
(= 90
o
), cos (p/2) = 0.  For the block moving on
a smooth horizontal table, the gravitational
force mg  does no work since it acts at right
angles to the displacement. If we assume that
the moon’s orbits around the earth is
perfectly circular then the earth’s
gravitational force does no work.  The moon’s
instantaneous displacement is tangential
while the earth’s force is radially inwards and
?  = p/2.
Work can be both positive and negative.  If ?  is
between 0
o
 and 90
o
, cos ?  in Eq. (6.4) is positive.
If  ?  is  between 90
o
 and 180
o
,   cos ?  is negative.
In many examples the frictional force opposes
displacement and ?  = 180
o
. Then the work done
by friction is negative (cos 180
o
 = –1).
From Eq. (6.4) it is clear that  work and energy
have the same dimensions,  [ML
2
T
–2
]. The SI unit
of these is joule (J), named after the famous British
physicist James Prescott Joule  (1811-1869). Since
work and energy are so widely used as physical
concepts, alternative units abound and some of
these are listed in Table 6.1.
2015-16(20/01/2015)
PHYSICS 118
?
object can do by the virtue of its motion. This
notion has been intuitively known for a long time.
The kinetic energy of a fast flowing stream
has been used to grind corn. Sailing
ships employ the kinetic energy of the wind. Table
6.2 lists the kinetic energies for various
objects.
Example 6.4  In a ballistics demonstration
a police officer fires a bullet of mass 50.0 g
with speed 200 m s
-1
 (see Table 6.2) on soft
plywood of thickness 2.00 cm.  The bullet
emerges with only 10% of its initial kinetic
energy.  What is the emergent speed of the
bullet ?
Answer  The initial kinetic energy of the bullet
is mv
2
/2 = 1000 J.  It has a final kinetic energy
of 0.1×1000 = 100 J.  If v
f
  is the emergent speed
of the bullet,
1
2
= mv
f
2
100 J
kg 05 . 0
 J 100 2 ×
=
f
v
                =  63.2 m s
–1
The speed is reduced by approximately 68%
(not 90%).                                                                      ?
6.5  WORK DONE BY A VARIABLE FORCE
A constant force is rare.  It is the variable force,
which is more commonly encountered.  Fig. 6.2
is a plot of a varying force in one dimension.
If the displacement ?x is small, we can take
the force F(x)  as approximately constant and
the work done is then
?W =F (x) ?x
Table 6.2  Typical kinetic energies (K)
This is illustrated in Fig. 6.3(a).  Adding
successive rectangular areas in Fig. 6.3(a) we
get the total work done as
( )
?
? ?
f
i
x
x
x x F W
(6.6)
where the summation is from the initial position
x
i
 
 
to the final position x
f
.
If the displacements are allowed to approach
zero, then the number of terms in the sum
increases without limit, but the sum approaches
a definite value equal to the area under the curve
in Fig. 6.3(b). Then the work done is
          
  d
f
i
x
x
F x x 
 (6.7)
where ‘lim’ stands for the limit of the sum when
?x  tends to zero.  Thus, for a varying force
the work done can be expressed as a definite
integral of force over displacement (see also
Appendix 3.1).
lim W =
lim
x ? ?
( )
?
?
f
i
x
x
x x F
0
Fig. 6.3(a)
2015-16(20/01/2015)
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Dynamic Test

Content Category

Related Searches

Energy and Power Class 11 Notes | EduRev

,

shortcuts and tricks

,

NCERT Textbook - Work

,

MCQs

,

Previous Year Questions with Solutions

,

ppt

,

pdf

,

Objective type Questions

,

NCERT Textbook - Work

,

mock tests for examination

,

Important questions

,

study material

,

Energy and Power Class 11 Notes | EduRev

,

Extra Questions

,

Free

,

Viva Questions

,

Semester Notes

,

video lectures

,

Sample Paper

,

practice quizzes

,

Exam

,

past year papers

,

Energy and Power Class 11 Notes | EduRev

,

NCERT Textbook - Work

,

Summary

;