Work, Energy & Power Questions with Answers Class 11 Notes | EduRev

Class 11 : Work, Energy & Power Questions with Answers Class 11 Notes | EduRev

 Page 1


PWEP – 1
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
WORK, ENERGY AND POWER
6.1 Introduction :
The Scalar Product :
Q. Find the angle between force ) k
ˆ
5 j
ˆ
4 i
ˆ
3 ( F ? ? ?
?
 unit and displacement ) k
ˆ
3 j
ˆ
4 i
ˆ
5 ( d ? ? ?
?
 unit.
Also find the projection of d on F
? ?
. [NCERT Solved Example 6.1]
Solution : ? = cos
–1
 0.32, 16/ ?50
6.2 Notions of Work and Kinetic Energy : The Work Energy Theorem :
Q. How are the work and kietic energy related to each other ? Explain.
OR
Q. Write work energy theorem and prove this theorem when a constant force is applied on a body.
Solution : Relation between work and kinetic energy is given by work energy theorem for which the
statement is following : The change in kinetic energy of a particle is equal to the work done on it by the net
force.
Proof of this theorem :
The relation for rectilinear motion under constant acceleration a is given by v
2
 – u
2
 = 2 as, where u and v are
the initial and final speeds and s the distance transversed. Multiplying both sides by m/2, we have
W Fs s ) ma ( mu
2
1
mv
2
1
2 2
? ? ? ?
A ‘W’ is the work done by force F on the body over the displacement s. This theorem can be written as :
K
f
 – K
i
 = W, where K
f
 and K
i
 are respectively the initial and final kinetic energies of the object.
Q. It is well known that a raindrop or a small pebble falls under the influence of the downward
gravitational force and the opposing resistive force. The later is known to be proportional to the
speed of the drop but is otherwise undetermined. Consider a drop or small pebble 1.00 g falling from
a cliff of height 1.00 km. It hits the ground with a speed of 50.0 m s
–1
. What is the work done by the
unknown resistive force ? [NCERT Solved Example 6.2]
Solution : –8.75 J
6.3 Work :
Q. Define work by the constant force.
Solution : 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.
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 = d . F
? ?
Q. Discuss different cases when work is not done.
Solution : No work is done if : (i) the displacement is zero. (ii) the force is zero. (iii) The force and
displacement are mutually perpendicular. This is so since, for ? = ?/2 rad (=90
0
), cos ( ?/2) = 0.
Q. A block is moving on a horizontal table, what is the work done by gravitational force Mg ?
Solution : For the block moving on a smooth horizontal table, the gravitational force mg does not work
since it acts at right angles to the displacement.
Page 2


PWEP – 1
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
WORK, ENERGY AND POWER
6.1 Introduction :
The Scalar Product :
Q. Find the angle between force ) k
ˆ
5 j
ˆ
4 i
ˆ
3 ( F ? ? ?
?
 unit and displacement ) k
ˆ
3 j
ˆ
4 i
ˆ
5 ( d ? ? ?
?
 unit.
Also find the projection of d on F
? ?
. [NCERT Solved Example 6.1]
Solution : ? = cos
–1
 0.32, 16/ ?50
6.2 Notions of Work and Kinetic Energy : The Work Energy Theorem :
Q. How are the work and kietic energy related to each other ? Explain.
OR
Q. Write work energy theorem and prove this theorem when a constant force is applied on a body.
Solution : Relation between work and kinetic energy is given by work energy theorem for which the
statement is following : The change in kinetic energy of a particle is equal to the work done on it by the net
force.
Proof of this theorem :
The relation for rectilinear motion under constant acceleration a is given by v
2
 – u
2
 = 2 as, where u and v are
the initial and final speeds and s the distance transversed. Multiplying both sides by m/2, we have
W Fs s ) ma ( mu
2
1
mv
2
1
2 2
? ? ? ?
A ‘W’ is the work done by force F on the body over the displacement s. This theorem can be written as :
K
f
 – K
i
 = W, where K
f
 and K
i
 are respectively the initial and final kinetic energies of the object.
Q. It is well known that a raindrop or a small pebble falls under the influence of the downward
gravitational force and the opposing resistive force. The later is known to be proportional to the
speed of the drop but is otherwise undetermined. Consider a drop or small pebble 1.00 g falling from
a cliff of height 1.00 km. It hits the ground with a speed of 50.0 m s
–1
. What is the work done by the
unknown resistive force ? [NCERT Solved Example 6.2]
Solution : –8.75 J
6.3 Work :
Q. Define work by the constant force.
Solution : 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.
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 = d . F
? ?
Q. Discuss different cases when work is not done.
Solution : No work is done if : (i) the displacement is zero. (ii) the force is zero. (iii) The force and
displacement are mutually perpendicular. This is so since, for ? = ?/2 rad (=90
0
), cos ( ?/2) = 0.
Q. A block is moving on a horizontal table, what is the work done by gravitational force Mg ?
Solution : For the block moving on a smooth horizontal table, the gravitational force mg does not work
since it acts at right angles to the displacement.
PWEP – 2
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. What is the work by Earth’s gravitational force acting on moon ?
Solution : If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravi-
tational force does not work. The moon’s instantaneous displacement is tangential while the earth’s force is
radially inwards and ? = ?/2.
Q. Can work be positive or negative ? Explain.
Solution : Work can be both positive and negative. If ? is between 0
0
 and 90
0
, cos ? in Eq. is positive. If ?
is between 90
0
 and 180
0
, cos ? is negative. In many examples the frictional force opposes displacement and
? = 180
0
. Then the work done by friction is negative (cos 180
0
 = –1).
Q. Does the work by frictional force always negative ? Explain.
Solution : The work done by frictional force can be negative or positive. Walking is possible only due to
frictional force and frictional force will do positive work.
Q. Write down the units and dimensional formula of work and energy.
Solution : The Work and energy have the same dimensions, [ML
2
T
–2
]. The SI unit of these is joule (J).
Q. 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 ? [NCERT Solved Example 6.3]
Solution : (a) –2000 J (b) 0
6.4 Kinetic Energy :
Q. Define kinetic energy. Is it a scalar or vector quantity ?
Solution : The energy due to the motion of the particle is known as kinetic energy. If an object of mass m
has velocity 
v
?
, its kinetic eneregy K is 
2
mv
2
1
v . v m
2
1
K ? ?
? ?
. Kinetic energy is a scalar quantity. .
Q. A particle of mass m is projected with speed u at an angle of ? with the vertical plane. What is the
kinetic energy of the particle at the highest point of the trajectory ?
Solution : At the highest point of the trajectory, the vertical component of the velocity will become zero but
horizontal component of velocity will remain. The horizontal component of velocity equals to u sin ? just
after the projection which remains constant during the motion. Hence at the highest point of trajectory the
kinetic energy of the particle equals to 
2
1
mu
2
sin
2
?.
Q. In a ballistics demostration a police officer fires a bullet of mass 50.0 g with speed 200 m s
–1
 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 ? [NCERT Solved Example 6.4]
Solution : 63.2 m/s
6.5 Work Done by a Variable Force :
Q. Define work done by the variable force. How can you find work done on force displacement
graph ?
Solution : The work done by variable force F(x) is given by 
?
?
f
i
x
x
dx ) x ( F W . Thus, for a varying force the
work done can be expressed as a definite integral of force over displacement. The shaded part represents
the work done by the variable force in the following graph :
Page 3


PWEP – 1
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
WORK, ENERGY AND POWER
6.1 Introduction :
The Scalar Product :
Q. Find the angle between force ) k
ˆ
5 j
ˆ
4 i
ˆ
3 ( F ? ? ?
?
 unit and displacement ) k
ˆ
3 j
ˆ
4 i
ˆ
5 ( d ? ? ?
?
 unit.
Also find the projection of d on F
? ?
. [NCERT Solved Example 6.1]
Solution : ? = cos
–1
 0.32, 16/ ?50
6.2 Notions of Work and Kinetic Energy : The Work Energy Theorem :
Q. How are the work and kietic energy related to each other ? Explain.
OR
Q. Write work energy theorem and prove this theorem when a constant force is applied on a body.
Solution : Relation between work and kinetic energy is given by work energy theorem for which the
statement is following : The change in kinetic energy of a particle is equal to the work done on it by the net
force.
Proof of this theorem :
The relation for rectilinear motion under constant acceleration a is given by v
2
 – u
2
 = 2 as, where u and v are
the initial and final speeds and s the distance transversed. Multiplying both sides by m/2, we have
W Fs s ) ma ( mu
2
1
mv
2
1
2 2
? ? ? ?
A ‘W’ is the work done by force F on the body over the displacement s. This theorem can be written as :
K
f
 – K
i
 = W, where K
f
 and K
i
 are respectively the initial and final kinetic energies of the object.
Q. It is well known that a raindrop or a small pebble falls under the influence of the downward
gravitational force and the opposing resistive force. The later is known to be proportional to the
speed of the drop but is otherwise undetermined. Consider a drop or small pebble 1.00 g falling from
a cliff of height 1.00 km. It hits the ground with a speed of 50.0 m s
–1
. What is the work done by the
unknown resistive force ? [NCERT Solved Example 6.2]
Solution : –8.75 J
6.3 Work :
Q. Define work by the constant force.
Solution : 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.
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 = d . F
? ?
Q. Discuss different cases when work is not done.
Solution : No work is done if : (i) the displacement is zero. (ii) the force is zero. (iii) The force and
displacement are mutually perpendicular. This is so since, for ? = ?/2 rad (=90
0
), cos ( ?/2) = 0.
Q. A block is moving on a horizontal table, what is the work done by gravitational force Mg ?
Solution : For the block moving on a smooth horizontal table, the gravitational force mg does not work
since it acts at right angles to the displacement.
PWEP – 2
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. What is the work by Earth’s gravitational force acting on moon ?
Solution : If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravi-
tational force does not work. The moon’s instantaneous displacement is tangential while the earth’s force is
radially inwards and ? = ?/2.
Q. Can work be positive or negative ? Explain.
Solution : Work can be both positive and negative. If ? is between 0
0
 and 90
0
, cos ? in Eq. is positive. If ?
is between 90
0
 and 180
0
, cos ? is negative. In many examples the frictional force opposes displacement and
? = 180
0
. Then the work done by friction is negative (cos 180
0
 = –1).
Q. Does the work by frictional force always negative ? Explain.
Solution : The work done by frictional force can be negative or positive. Walking is possible only due to
frictional force and frictional force will do positive work.
Q. Write down the units and dimensional formula of work and energy.
Solution : The Work and energy have the same dimensions, [ML
2
T
–2
]. The SI unit of these is joule (J).
Q. 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 ? [NCERT Solved Example 6.3]
Solution : (a) –2000 J (b) 0
6.4 Kinetic Energy :
Q. Define kinetic energy. Is it a scalar or vector quantity ?
Solution : The energy due to the motion of the particle is known as kinetic energy. If an object of mass m
has velocity 
v
?
, its kinetic eneregy K is 
2
mv
2
1
v . v m
2
1
K ? ?
? ?
. Kinetic energy is a scalar quantity. .
Q. A particle of mass m is projected with speed u at an angle of ? with the vertical plane. What is the
kinetic energy of the particle at the highest point of the trajectory ?
Solution : At the highest point of the trajectory, the vertical component of the velocity will become zero but
horizontal component of velocity will remain. The horizontal component of velocity equals to u sin ? just
after the projection which remains constant during the motion. Hence at the highest point of trajectory the
kinetic energy of the particle equals to 
2
1
mu
2
sin
2
?.
Q. In a ballistics demostration a police officer fires a bullet of mass 50.0 g with speed 200 m s
–1
 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 ? [NCERT Solved Example 6.4]
Solution : 63.2 m/s
6.5 Work Done by a Variable Force :
Q. Define work done by the variable force. How can you find work done on force displacement
graph ?
Solution : The work done by variable force F(x) is given by 
?
?
f
i
x
x
dx ) x ( F W . Thus, for a varying force the
work done can be expressed as a definite integral of force over displacement. The shaded part represents
the work done by the variable force in the following graph :
PWEP – 3
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of
100 N over a distance of 10 m. Thereafter she gets progressively tired and her applied force reduces
linearly with distance to 50 N. The total distance by which the trunk has been moved is 20 m. Plot the
force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate
the work done by the two forces over 20 m. Also find the final kinetic energy.
[NCERT Solved Example 6.5]
Solution :
1750 J, –1000 J
6.6 The Work-Energy Theorem for a Variable Force :
Q. Prove the work-energy theorem for variable force.
Solution : From newton’s second law f = ma
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
dt
dx
dx
dv
m
dt
dv
m F ? v
dx
dv
m F ?
?
?
?
?
?
?
?
? ? ?
? ?
f
i
f
i
f
i
x
x
v
v
v
v
vdv m mvdv dx F ?
f
i
v
v
2
2
v
m W
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
v
2
v
m W
2
i
2
f
?
i f
2
i
2
f
K K mv
2
1
mv
2
1
W ? ? ? ?
? W = ?K
Thus, the work energy theorem is proved for a variable force.
Q. A block of mass m = 1 kg, moving on a horizontal surface with speed v
t
 = 2 ms
–1
 enters a rough
patch ranging from x = 0.10 m to x = 2.01 m. The ratarding force F
r
 on the block in this range is
inversely proportional to x over this range, 
x
k
F
r
?
? for 0.1 < x < 2.01 m
= 0 for x < 0.1 m and x > 2.01 m, where k = 0.5 J. What is the final kinetic energy and speed v
f
 of the
block as it crosses this patch  ? (ln 20.1 = 3)  [NCERT Solved Example 6.6]
Solution : 0.5 J, 1 m/s
6.7 The Concept of Potential Energy :
Q. What is potential energy ? Write down the dimensions and unit of potential energy.
Solution : Potential energy is the ‘stored energy’ by virtue of the position of configuration of a body. The
dimensions of potential energy are [ML
2
T
–2
] and the unit is joule (J).
Q. What is conservative force OR Write down the properties of conservative force.
Solution : (i) The work done (change in kinetic energy) by a conservative force depends on the initial and
final positions only. (ii) Work done by conservative force for closed path equals to zero. (iii) The
mechanical energy of a system will be constant in the presence of conservative forces.
Q. Give some examples of conservative force.
Solution : Gravitational force, Spring force and Electrostatic force are the examples of conservative force.
Page 4


PWEP – 1
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
WORK, ENERGY AND POWER
6.1 Introduction :
The Scalar Product :
Q. Find the angle between force ) k
ˆ
5 j
ˆ
4 i
ˆ
3 ( F ? ? ?
?
 unit and displacement ) k
ˆ
3 j
ˆ
4 i
ˆ
5 ( d ? ? ?
?
 unit.
Also find the projection of d on F
? ?
. [NCERT Solved Example 6.1]
Solution : ? = cos
–1
 0.32, 16/ ?50
6.2 Notions of Work and Kinetic Energy : The Work Energy Theorem :
Q. How are the work and kietic energy related to each other ? Explain.
OR
Q. Write work energy theorem and prove this theorem when a constant force is applied on a body.
Solution : Relation between work and kinetic energy is given by work energy theorem for which the
statement is following : The change in kinetic energy of a particle is equal to the work done on it by the net
force.
Proof of this theorem :
The relation for rectilinear motion under constant acceleration a is given by v
2
 – u
2
 = 2 as, where u and v are
the initial and final speeds and s the distance transversed. Multiplying both sides by m/2, we have
W Fs s ) ma ( mu
2
1
mv
2
1
2 2
? ? ? ?
A ‘W’ is the work done by force F on the body over the displacement s. This theorem can be written as :
K
f
 – K
i
 = W, where K
f
 and K
i
 are respectively the initial and final kinetic energies of the object.
Q. It is well known that a raindrop or a small pebble falls under the influence of the downward
gravitational force and the opposing resistive force. The later is known to be proportional to the
speed of the drop but is otherwise undetermined. Consider a drop or small pebble 1.00 g falling from
a cliff of height 1.00 km. It hits the ground with a speed of 50.0 m s
–1
. What is the work done by the
unknown resistive force ? [NCERT Solved Example 6.2]
Solution : –8.75 J
6.3 Work :
Q. Define work by the constant force.
Solution : 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.
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 = d . F
? ?
Q. Discuss different cases when work is not done.
Solution : No work is done if : (i) the displacement is zero. (ii) the force is zero. (iii) The force and
displacement are mutually perpendicular. This is so since, for ? = ?/2 rad (=90
0
), cos ( ?/2) = 0.
Q. A block is moving on a horizontal table, what is the work done by gravitational force Mg ?
Solution : For the block moving on a smooth horizontal table, the gravitational force mg does not work
since it acts at right angles to the displacement.
PWEP – 2
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. What is the work by Earth’s gravitational force acting on moon ?
Solution : If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravi-
tational force does not work. The moon’s instantaneous displacement is tangential while the earth’s force is
radially inwards and ? = ?/2.
Q. Can work be positive or negative ? Explain.
Solution : Work can be both positive and negative. If ? is between 0
0
 and 90
0
, cos ? in Eq. is positive. If ?
is between 90
0
 and 180
0
, cos ? is negative. In many examples the frictional force opposes displacement and
? = 180
0
. Then the work done by friction is negative (cos 180
0
 = –1).
Q. Does the work by frictional force always negative ? Explain.
Solution : The work done by frictional force can be negative or positive. Walking is possible only due to
frictional force and frictional force will do positive work.
Q. Write down the units and dimensional formula of work and energy.
Solution : The Work and energy have the same dimensions, [ML
2
T
–2
]. The SI unit of these is joule (J).
Q. 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 ? [NCERT Solved Example 6.3]
Solution : (a) –2000 J (b) 0
6.4 Kinetic Energy :
Q. Define kinetic energy. Is it a scalar or vector quantity ?
Solution : The energy due to the motion of the particle is known as kinetic energy. If an object of mass m
has velocity 
v
?
, its kinetic eneregy K is 
2
mv
2
1
v . v m
2
1
K ? ?
? ?
. Kinetic energy is a scalar quantity. .
Q. A particle of mass m is projected with speed u at an angle of ? with the vertical plane. What is the
kinetic energy of the particle at the highest point of the trajectory ?
Solution : At the highest point of the trajectory, the vertical component of the velocity will become zero but
horizontal component of velocity will remain. The horizontal component of velocity equals to u sin ? just
after the projection which remains constant during the motion. Hence at the highest point of trajectory the
kinetic energy of the particle equals to 
2
1
mu
2
sin
2
?.
Q. In a ballistics demostration a police officer fires a bullet of mass 50.0 g with speed 200 m s
–1
 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 ? [NCERT Solved Example 6.4]
Solution : 63.2 m/s
6.5 Work Done by a Variable Force :
Q. Define work done by the variable force. How can you find work done on force displacement
graph ?
Solution : The work done by variable force F(x) is given by 
?
?
f
i
x
x
dx ) x ( F W . Thus, for a varying force the
work done can be expressed as a definite integral of force over displacement. The shaded part represents
the work done by the variable force in the following graph :
PWEP – 3
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of
100 N over a distance of 10 m. Thereafter she gets progressively tired and her applied force reduces
linearly with distance to 50 N. The total distance by which the trunk has been moved is 20 m. Plot the
force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate
the work done by the two forces over 20 m. Also find the final kinetic energy.
[NCERT Solved Example 6.5]
Solution :
1750 J, –1000 J
6.6 The Work-Energy Theorem for a Variable Force :
Q. Prove the work-energy theorem for variable force.
Solution : From newton’s second law f = ma
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
dt
dx
dx
dv
m
dt
dv
m F ? v
dx
dv
m F ?
?
?
?
?
?
?
?
? ? ?
? ?
f
i
f
i
f
i
x
x
v
v
v
v
vdv m mvdv dx F ?
f
i
v
v
2
2
v
m W
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
v
2
v
m W
2
i
2
f
?
i f
2
i
2
f
K K mv
2
1
mv
2
1
W ? ? ? ?
? W = ?K
Thus, the work energy theorem is proved for a variable force.
Q. A block of mass m = 1 kg, moving on a horizontal surface with speed v
t
 = 2 ms
–1
 enters a rough
patch ranging from x = 0.10 m to x = 2.01 m. The ratarding force F
r
 on the block in this range is
inversely proportional to x over this range, 
x
k
F
r
?
? for 0.1 < x < 2.01 m
= 0 for x < 0.1 m and x > 2.01 m, where k = 0.5 J. What is the final kinetic energy and speed v
f
 of the
block as it crosses this patch  ? (ln 20.1 = 3)  [NCERT Solved Example 6.6]
Solution : 0.5 J, 1 m/s
6.7 The Concept of Potential Energy :
Q. What is potential energy ? Write down the dimensions and unit of potential energy.
Solution : Potential energy is the ‘stored energy’ by virtue of the position of configuration of a body. The
dimensions of potential energy are [ML
2
T
–2
] and the unit is joule (J).
Q. What is conservative force OR Write down the properties of conservative force.
Solution : (i) The work done (change in kinetic energy) by a conservative force depends on the initial and
final positions only. (ii) Work done by conservative force for closed path equals to zero. (iii) The
mechanical energy of a system will be constant in the presence of conservative forces.
Q. Give some examples of conservative force.
Solution : Gravitational force, Spring force and Electrostatic force are the examples of conservative force.
PWEP – 4
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. Define conservative force in terms of the potential energy.
Solution : A force F(x) is conservative if it can be derived from a scalar quantity V(x), known as potential
energy, by the relation given by 
dx
dV
) x ( F ? ? . This implies that 
? ?
? ? ? ?
f
i
f
i
x
x
V
V
f i
V V dV dx ) x ( F .
The change in potential energy, for a conservative force, ?V is equal to the negative of the work done by the
force ?V = –F(x) ?x
Q. Define non-conservative force and give some examples.
Solution : If the work done or the kinetic energy did depend on other factors such as the velocity or the
particular path taken by the object, the force would be called non-conservative. The examples of
non-conservative forces are : Frictional force, air friction, viscous force etc.
Q. What is gravitational potential energy ?
Solution : The work done by the external agent to move (very slowly) a particle of mass m against the
gravitational force is mgh, where h is the vertical displacement. This work gets stored as potential energy.
Hence, the gravitational potential energy equals to mgh.
6.8 The Conservation of Mechanical Energy :
Q. What is mechanical energy ?
Solution : The sum of kinetic and potential energy of the body is known as mechanical energy.
Q. Write down the statement of conservation of mechanical energy. Prove this theorem.
Solution : Statement of conservation of mechanical energy : The total mechanical energy of a system is
conserved if the forces, doing work on it, are conservative.
Proof of theorem : Suppose that a body undergoes displacement ?x under the action of a conservative
force F. Then from the WE theorem we have, ?K = F(x) ?x
If the force is conservative, the potential energy function V(x) can be defined such that – ?V = F(x) ?x
the above equations imply that ?K + ?V = 0 ? ?(K + V) = 0, which means that K + V , the sum of the
kinetic and potential energies of the body is a constant. Over the whole path, x
i
 to x
f
, this means that
K
i
 + V(x
i
) = K
f
 + V(xf)
The quantity K + V(x), is called the total mechanical energy of the system.
Q. Illustrate the conservation of mechanical energy for a ball of mass ‘m’ is droped from a height H.
Solution : A ball of mass m is dropped from a height H. The total mechanical energies E
0
, E
h
 and E
H
 of the
ball at the indicated heights zero (ground level), h and H, respectively are :
E
H
 = mgH, E
h
 = mgh + 
2
h
mv
2
1
, E
0
 = (1/2) mv
f
2
As the mechanical energy is conserved, thus E
H
 = E
0
 or, mgH = 
2
h
mv
2
1
, gH 2 v
f
?
This is the result of speed for a particle in free fall after covering the height H.
Further, E
H
 = E
h
 which implies, v
h
2
 = 2g(H – h) and is a familier result from kinematics for the speed of free
fall particle.
At the height H, the energy is purely potential. It is partially converted to kinetic at height h and is fully
kinetic at ground level. This illustrates the conservation of mechanical energy.
Page 5


PWEP – 1
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
WORK, ENERGY AND POWER
6.1 Introduction :
The Scalar Product :
Q. Find the angle between force ) k
ˆ
5 j
ˆ
4 i
ˆ
3 ( F ? ? ?
?
 unit and displacement ) k
ˆ
3 j
ˆ
4 i
ˆ
5 ( d ? ? ?
?
 unit.
Also find the projection of d on F
? ?
. [NCERT Solved Example 6.1]
Solution : ? = cos
–1
 0.32, 16/ ?50
6.2 Notions of Work and Kinetic Energy : The Work Energy Theorem :
Q. How are the work and kietic energy related to each other ? Explain.
OR
Q. Write work energy theorem and prove this theorem when a constant force is applied on a body.
Solution : Relation between work and kinetic energy is given by work energy theorem for which the
statement is following : The change in kinetic energy of a particle is equal to the work done on it by the net
force.
Proof of this theorem :
The relation for rectilinear motion under constant acceleration a is given by v
2
 – u
2
 = 2 as, where u and v are
the initial and final speeds and s the distance transversed. Multiplying both sides by m/2, we have
W Fs s ) ma ( mu
2
1
mv
2
1
2 2
? ? ? ?
A ‘W’ is the work done by force F on the body over the displacement s. This theorem can be written as :
K
f
 – K
i
 = W, where K
f
 and K
i
 are respectively the initial and final kinetic energies of the object.
Q. It is well known that a raindrop or a small pebble falls under the influence of the downward
gravitational force and the opposing resistive force. The later is known to be proportional to the
speed of the drop but is otherwise undetermined. Consider a drop or small pebble 1.00 g falling from
a cliff of height 1.00 km. It hits the ground with a speed of 50.0 m s
–1
. What is the work done by the
unknown resistive force ? [NCERT Solved Example 6.2]
Solution : –8.75 J
6.3 Work :
Q. Define work by the constant force.
Solution : 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.
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 = d . F
? ?
Q. Discuss different cases when work is not done.
Solution : No work is done if : (i) the displacement is zero. (ii) the force is zero. (iii) The force and
displacement are mutually perpendicular. This is so since, for ? = ?/2 rad (=90
0
), cos ( ?/2) = 0.
Q. A block is moving on a horizontal table, what is the work done by gravitational force Mg ?
Solution : For the block moving on a smooth horizontal table, the gravitational force mg does not work
since it acts at right angles to the displacement.
PWEP – 2
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. What is the work by Earth’s gravitational force acting on moon ?
Solution : If we assume that the moon’s orbits around the earth is perfectly circular then the earth’s gravi-
tational force does not work. The moon’s instantaneous displacement is tangential while the earth’s force is
radially inwards and ? = ?/2.
Q. Can work be positive or negative ? Explain.
Solution : Work can be both positive and negative. If ? is between 0
0
 and 90
0
, cos ? in Eq. is positive. If ?
is between 90
0
 and 180
0
, cos ? is negative. In many examples the frictional force opposes displacement and
? = 180
0
. Then the work done by friction is negative (cos 180
0
 = –1).
Q. Does the work by frictional force always negative ? Explain.
Solution : The work done by frictional force can be negative or positive. Walking is possible only due to
frictional force and frictional force will do positive work.
Q. Write down the units and dimensional formula of work and energy.
Solution : The Work and energy have the same dimensions, [ML
2
T
–2
]. The SI unit of these is joule (J).
Q. 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 ? [NCERT Solved Example 6.3]
Solution : (a) –2000 J (b) 0
6.4 Kinetic Energy :
Q. Define kinetic energy. Is it a scalar or vector quantity ?
Solution : The energy due to the motion of the particle is known as kinetic energy. If an object of mass m
has velocity 
v
?
, its kinetic eneregy K is 
2
mv
2
1
v . v m
2
1
K ? ?
? ?
. Kinetic energy is a scalar quantity. .
Q. A particle of mass m is projected with speed u at an angle of ? with the vertical plane. What is the
kinetic energy of the particle at the highest point of the trajectory ?
Solution : At the highest point of the trajectory, the vertical component of the velocity will become zero but
horizontal component of velocity will remain. The horizontal component of velocity equals to u sin ? just
after the projection which remains constant during the motion. Hence at the highest point of trajectory the
kinetic energy of the particle equals to 
2
1
mu
2
sin
2
?.
Q. In a ballistics demostration a police officer fires a bullet of mass 50.0 g with speed 200 m s
–1
 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 ? [NCERT Solved Example 6.4]
Solution : 63.2 m/s
6.5 Work Done by a Variable Force :
Q. Define work done by the variable force. How can you find work done on force displacement
graph ?
Solution : The work done by variable force F(x) is given by 
?
?
f
i
x
x
dx ) x ( F W . Thus, for a varying force the
work done can be expressed as a definite integral of force over displacement. The shaded part represents
the work done by the variable force in the following graph :
PWEP – 3
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of
100 N over a distance of 10 m. Thereafter she gets progressively tired and her applied force reduces
linearly with distance to 50 N. The total distance by which the trunk has been moved is 20 m. Plot the
force applied by the woman and the frictional force, which is 50 N versus displacement. Calculate
the work done by the two forces over 20 m. Also find the final kinetic energy.
[NCERT Solved Example 6.5]
Solution :
1750 J, –1000 J
6.6 The Work-Energy Theorem for a Variable Force :
Q. Prove the work-energy theorem for variable force.
Solution : From newton’s second law f = ma
? ?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
dt
dx
dx
dv
m
dt
dv
m F ? v
dx
dv
m F ?
?
?
?
?
?
?
?
? ? ?
? ?
f
i
f
i
f
i
x
x
v
v
v
v
vdv m mvdv dx F ?
f
i
v
v
2
2
v
m W
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
v
2
v
m W
2
i
2
f
?
i f
2
i
2
f
K K mv
2
1
mv
2
1
W ? ? ? ?
? W = ?K
Thus, the work energy theorem is proved for a variable force.
Q. A block of mass m = 1 kg, moving on a horizontal surface with speed v
t
 = 2 ms
–1
 enters a rough
patch ranging from x = 0.10 m to x = 2.01 m. The ratarding force F
r
 on the block in this range is
inversely proportional to x over this range, 
x
k
F
r
?
? for 0.1 < x < 2.01 m
= 0 for x < 0.1 m and x > 2.01 m, where k = 0.5 J. What is the final kinetic energy and speed v
f
 of the
block as it crosses this patch  ? (ln 20.1 = 3)  [NCERT Solved Example 6.6]
Solution : 0.5 J, 1 m/s
6.7 The Concept of Potential Energy :
Q. What is potential energy ? Write down the dimensions and unit of potential energy.
Solution : Potential energy is the ‘stored energy’ by virtue of the position of configuration of a body. The
dimensions of potential energy are [ML
2
T
–2
] and the unit is joule (J).
Q. What is conservative force OR Write down the properties of conservative force.
Solution : (i) The work done (change in kinetic energy) by a conservative force depends on the initial and
final positions only. (ii) Work done by conservative force for closed path equals to zero. (iii) The
mechanical energy of a system will be constant in the presence of conservative forces.
Q. Give some examples of conservative force.
Solution : Gravitational force, Spring force and Electrostatic force are the examples of conservative force.
PWEP – 4
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. Define conservative force in terms of the potential energy.
Solution : A force F(x) is conservative if it can be derived from a scalar quantity V(x), known as potential
energy, by the relation given by 
dx
dV
) x ( F ? ? . This implies that 
? ?
? ? ? ?
f
i
f
i
x
x
V
V
f i
V V dV dx ) x ( F .
The change in potential energy, for a conservative force, ?V is equal to the negative of the work done by the
force ?V = –F(x) ?x
Q. Define non-conservative force and give some examples.
Solution : If the work done or the kinetic energy did depend on other factors such as the velocity or the
particular path taken by the object, the force would be called non-conservative. The examples of
non-conservative forces are : Frictional force, air friction, viscous force etc.
Q. What is gravitational potential energy ?
Solution : The work done by the external agent to move (very slowly) a particle of mass m against the
gravitational force is mgh, where h is the vertical displacement. This work gets stored as potential energy.
Hence, the gravitational potential energy equals to mgh.
6.8 The Conservation of Mechanical Energy :
Q. What is mechanical energy ?
Solution : The sum of kinetic and potential energy of the body is known as mechanical energy.
Q. Write down the statement of conservation of mechanical energy. Prove this theorem.
Solution : Statement of conservation of mechanical energy : The total mechanical energy of a system is
conserved if the forces, doing work on it, are conservative.
Proof of theorem : Suppose that a body undergoes displacement ?x under the action of a conservative
force F. Then from the WE theorem we have, ?K = F(x) ?x
If the force is conservative, the potential energy function V(x) can be defined such that – ?V = F(x) ?x
the above equations imply that ?K + ?V = 0 ? ?(K + V) = 0, which means that K + V , the sum of the
kinetic and potential energies of the body is a constant. Over the whole path, x
i
 to x
f
, this means that
K
i
 + V(x
i
) = K
f
 + V(xf)
The quantity K + V(x), is called the total mechanical energy of the system.
Q. Illustrate the conservation of mechanical energy for a ball of mass ‘m’ is droped from a height H.
Solution : A ball of mass m is dropped from a height H. The total mechanical energies E
0
, E
h
 and E
H
 of the
ball at the indicated heights zero (ground level), h and H, respectively are :
E
H
 = mgH, E
h
 = mgh + 
2
h
mv
2
1
, E
0
 = (1/2) mv
f
2
As the mechanical energy is conserved, thus E
H
 = E
0
 or, mgH = 
2
h
mv
2
1
, gH 2 v
f
?
This is the result of speed for a particle in free fall after covering the height H.
Further, E
H
 = E
h
 which implies, v
h
2
 = 2g(H – h) and is a familier result from kinematics for the speed of free
fall particle.
At the height H, the energy is purely potential. It is partially converted to kinetic at height h and is fully
kinetic at ground level. This illustrates the conservation of mechanical energy.
PWEP – 5
Einstein Classes,  Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
Q. A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity v
0
at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the
string becoming slack only on reaching the topmost point C. This is shown in figure.
Obtain an expression for (i) v
0
; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies
(K
B
/K
C
) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C.
[NCERT Solved Example 6.7]
Solution : (i) gL 5 (ii) gL 3 (iii) 
1
3
6.9 The Potential Energy of a Spring :
Q. What is the nature of spring force ?
Solution : The spring force is a variable force which is conservative.
Q. What is ideal spring ?
Solution : An ideal spring is light and may be treated as massless.
Q. What is Hook’s law of the spring ?
Solution : In an ideal spring, the spring force F
s
 is proportional to x and x is the extension or compression
in the spring from the equilibrium position. This force law for the spring is called Hooke’s law and is
mathematically stated as F
s
 = –kx, here k is known as spring constant. The spring is said to be stiff if k is
large and soft if k is small.
Q. Derive the expression for work done by spring force and hence derive the expression for the
potential energy of a spring.
Solution :
Figure shows a block attached to a spring and resting on a smooth horizontal surface. The other end of the
spring is attached to a rigid wall.
Suppose that we pull the block outwards as in figure. If the extension is x
m
, the work done by the spring
force is 
? ?
? ? ?
m m
x
0
x
0
s s
kxdx dx F W = 
2
kx
2
m
? .
Note that the work done by the external pulling force F is positive since it overcomes the spring force. This
work done by external agent will be stored in the form of potential energy in the spring, known as spring
potential energy. Hence the spring potential equals to 
2
m
kx
2
1
.
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