NCERT Solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 6
Chapter Name : Work, Energy And Power

Q6.1 The sign of work done by a force on a body is important to understand. State carefully if the
following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform
velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest

Answer. (a) Positive
In the given case, force and displacement are in the same direction. Hence, the sign of
work done is positive. In this case, the work is done on the bucket.
(b) Negative
In the given case, the direction of force (vertically downward) and displacement
(vertically upward) are opposite to each other. Hence, the sign of work done is negative.
(c) Negative
Since the direction of frictional force is opposite to the direction of motion, the work done
by frictional force is negative in this case.
(d) Positive
Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the
body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the
body. Since the applied force acts in the direction of motion of the body, the work done is
positive.
(e) Negative
The resistive force of air acts in the direction opposite to the direction of motion of the
pendulum. Hence, the work done is negative in this case.

Page : 134 , Block Name : Exercise

Q6.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of
7 N on a table with coef cient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s,

Mass of the body, m = 2kg

Applied force, F = 7N

Coefficient of kinetic friction, μ = 0.1

Initial velocity, u = 0

Time, t = 10s

The acceleration produced in the body by the applied force is given by Newton's seconc

law of motion as:

′ F 7 2
a = = = 3.5m/s
m 2

Frictional force is given as:

f = μmg

= 0.1 × 2 × 9.8 = −1.96N

The acceleration produced by the frictional force:
′ ′′
a = a + a

2
= 3.5 + (−0.98) = 2.52m/s

The distance travelled by the body is given by the equation of motion:

1 2
s = ut + at
2
1 2
= 0 + × 2.52 × (10) = 126m
2

(a) Work done by the applied force, Wa = F × s = 7 × 126 = 882J

(b) Work done by the frictional force, Wf = F × s = −1.96 × 126 = −247J

(c) Net force = 7 + (−1.96) = 5.04N

Work done by the net force, W net = 5.04 × 126 = 635 J

(d) From the first equation of motion, final velocity can be calculated as:

v = u + at

= 0 + 2.52 × 10 = 25.2m/s
1 2 1 2
Change in kinetic energy = mv − mu
2 2

1 2 2 2 2
= × 2 (v − u ) = (25.2) − 0 = 635J
2

Page : 134 , Block Name : Exercise

Q6.3 Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The
total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the
regions, if any, in which the particle cannot be found for the given energy. Also, indicate the
minimum total energy the particle must have in each case. Think of simple physical contexts for
which these potential energy shapes are relevant

(a) x > a; 0

Total energy of a system is given by the relation:

E = P . E. +K. E

∴ K ⋅ E. = E − P. E

Kinetic energy of a body is a positive quantity, It cannot be negative. Therefore, the
particle will not exist in a region where K.E. becomes negative.
In the given case, the potential energy of the particle becomes greater than total
energy (E) for x > a. Hence, kinetic energy becomes negative in this region. Therefore,
the particle will not exist is this region. The minimum total energy of the particle is zero.
(b) All regions
In the given case, the potential energy (Vo) is greater than total energy (E) in all
regions. Hence, the particle will not exist in this region.
(c)x > a and x < b; −V1

In the given case, the condition regarding the positivity of K.E. is satis ed only in the
region between x > a and x < b.
The minimal potential energy in this case is —VI. Therefore, K E
Therefore, for the positivity of the kinetic energy, the total energy of the particle must
be greater than —VI. So, the minimum total energy the particle must have is —VI.
(d)− < x < ;
b a a
< x <
b
; −V 1
2 2 2 2

In the given case, the potential energy (Vo) of the particle becomes greater than the
total energy (E) for − < x < and − < x <
b b a a

2 2 2 2

Therefore, the particle will not exist in these region.
The minimum potential energy in this case is —VI. Therefore, K.E. E — (—VI) E + VI.
Therefore, for the positivity of the kinetic energy, the total energy of the particle must
be greater than -VI. So, the minimum total energy the particle must have is -VI.

Page : 134 , Block Name : Exercise

Q6.4 The potential energy function for a particle executing linear simple harmonic motion is
given by V (x) = kx /2, where k is the force constant of the oscillator. For k = 0.5 N m , the
2 −1

graph of V(x) versus x is shown in Fig. 6.12. Show that a particle of total energy 1 J moving under
this potential must ‘turn back’ when it reaches x = ± 2 m

Total energy of the particle, E = 1J

−1
Force constant, k = 0.5Nm

1 2
Kinetic energy of the particle, K = mv
2

According to the conservation law:

E = V + K

1 2 1 2
1 = kx + mv
2 2

At the moment of 'turn back', velocity (and hence K ) becomes zero.
1 2
1 = kx
2

1 2
× 0.5x = 1
2

2
x = 4

x = ±2

Hence, the particle turns back when it reaches x ± 2 m

Page : 134 , Block Name : Exercise

Q6.5 Answer the following :
(a) The casing of a rocket in ight burns up due to friction. At whose expense is the heat energy
required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet
due to the sun is not normal to the comet’s velocity in general. Yet the work done by the
gravitational force over every complete orbit of the comet is zero. Why ?
(c) An arti cial satellite orbiting the earth in very thin atmosphere loses its energy gradually due
to dissipation against atmospheric resistance, however small. Why then does its speed increase
progressively as it comes closer and closer to the earth ?
(d) In Fig. 6.13
(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig. 6.13
(ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a
mass of 15 kg hangs at its other end. In which case is the work done greater ?

Answer. (a) Rocket
The burning of the casing of a rocket in ight (due to friction) results in the reduction of
the mass of the rocket.
According to the conservation of energy:
Total Energy (T .E.) = Potential energy Kinetic energy (ICE.)
1 2
= mgh + mv
2

The reduction in the rocket's mass causes a drop in the total energy. Therefore, the heat
energy required for the burning is obtained from the rocket.
(b) Gravitational force is a conservative force. Since the work done by a conservative
force over a closed path is zero, the work done by the gravitational force over every
complete orbit of a comet is zero.
(c) 'When an arti cial satellite, orbiting around earth, moves closer to earth, its potential
energy decreases because of the reduction in the height. Since the total energy of the
system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the
velocity of the satellite increases. However, due to atmospheric friction, the total energy
of the satellite decreases by a small amount.
(d) In the second case
case (i)

Mass,

Mass, m = 15kg

Displacement, s = 2m

Work done, W = F s cos θ

Where, θ = Angle between force and displacement
∘
= mgs cos θ = 15 × 2 × 9.8 cos 90

= 0

Case (ii)

Mass, m = 15kg

Displacement, s = 2m

Here, the direction of the force applied on the rope and the direction of the displacement

of the rope are same.

∘
Therefore, the angle between them, θ = 0
∘
since cos 0 = 1

Work done, W = F s cos θ = mgs

= 15 × 9.8 × 2 = 294J

Hence, more work is done in the second case.

Page : 134 , Block Name : Exercise

Q6.6 Underline the correct alternative :
(a) When a conservative force does positive work on a body, the potential energy of the body
increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the
external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision
are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Answer. (a) Decreases
(b) Kinetic energy
(c) External force
(d) Total linear momentum
(a) A conservative force does a positive work on a body when it displaces the body in
the direction of force. As a result, the body advances toward the centre of force. It
decreases the separation between the two, thereby decreasing the potential energy Of
the body.
(b) The work done against the direction Of friction reduces the velocity Of a body. Hence,
there is a loss of kinetic energy of the body.
(c) Internal forces, irrespective of their direction, cannot produce any change in the total
momentum Of a body. Hence, the total momentum Of a many- particle system is
proportional to the external forces acting on the system.
(d) The total linear momentum always remains conserved whether it is an elastic
collision or an inelastic collision.

Page : 134 , Block Name : Exercise

Q6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on
the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the nal kinetic energy is always less than the initial kinetic energy of
the system

Answer. (a) False
(b) False
(c) False

(d) True
(a) In an elastic collision, the total energy and momentum of both the bodies, and not of
each individual body, is conserved.
(b) Although internal forces are balanced, they cause no work to be done on a body. It
is the external forces that have the ability to do work. Hence, external forces are able to
change the energy Of a system.
(c) The work done in the motion of a body over a closed loop is zero for a conservation
force only.
(d) In an inelastic collision, the nal kinetic energy is always less than the initial kinetic
energy Of the cistern. This is because in such collisions, there is always a loss Of energy
in the form of heat, sound, etc.

Page : 134 , Block Name : Exercise

Q6.8 Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the
short time of collision of the balls (i.e. when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two
balls ?
(c) What are the answers to
(a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance between
their centres, is the collision elastic or inelastic ? (Note, we are talking here of potential energy
corresponding to the force during collision, not gravitational potential energy).

Answer. (a) No
In an elastic collision, the total initial kinetic energy of the balls will be equal to the total
nal kinetic energy of the balls. This kinetic energy is not conserved at the instant the
two balls are in contact with each other. In fact, at the time of collision, the kinetic
energy of the balls will get converted into potential energy.
(b) Yes
In an elastic collision, the total linear momentum of the system always remains
conserved.
(c) No; Yes
In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic
energy Of the billiard balls before collision will always be greater than that after collision.
The total linear momentum of the system of billiards balls will remain conserved even in
the case Of an inelastic collision,
(d) Elastic
In the given case, the forces involved are conservation. This is because they depend on
the separation between the centres Of the billiard balls. Hence, the collision is elastic.

Page : 134 , Block Name : Exercise

Q6.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration.
The power delivered to it at time t is proportional to

(i) t1/2

(ii) t
(iii) t 3/2

(iv) t2

(ii) t

Mass of the body = m

Acceleration of the body = a

Using Newton's second law of motion, the force experienced by the body is given by the

equation:

F = ma

Both m and a are constants. Hence, force F will also be a constant.

F = ma = Constant … (i)

For velocity v, acceleration is given as,
dv
a = = Constant
dt

dv = Constant xdt

v = αt

Where, a is another constant

vort

Power is given by the relation:

P = F. v

Using equations (i) and (iii), we have:

p ∝ t

Hence, power is directly proportional to time.

Page : 134 , Block Name : Exercise

Q6.10 A body is moving unidirectionally under the in uence of a source of constant power. Its
displacement in time t is proportional to
(i)t1/2

(ii) t
(iii) t 3/2

(iv) t2

3

(iii) t 2

Power is given by the relation:

P = Fv
dv
= mav = mv = Constant ( say, k)
dt

k
∴ vdv = dt
m

Integrating both sides:

2
v k
= t
2 m

2kt
v = √
m

For displacement x of the body, we have:

dx 2k 2
v = = √ t
dt m

1
′
dx = k t 2 dt

′ 2k
Where k = √ = New constant
3

On integrating both sides, we get:
3
2 ′
x = k t2
3

3

∴ x ∝ t2

Page : 134 , Block Name : Exercise

Q6.11 A body constrained to move along the z-axis of a coordinate system is subject to a constant
force F given by
^ ^ ^
F = − i + 2 j + 3kN

where i,j,k are unit vectors along the x-, y- and z-axis of the system respectively. What is the work
done by this force in moving the body a distance of 4 m along the z-axis ?

^ ^ ^
Force exerted on the body, F = − i + 2 j + 3kN

^
Displacement, s = 4km

Work done, W = F. s

^ ^ ^ ^
= (− i + 2 j + 3k) ⋅ (4k)

= 0 + 0 − 3 × 4

= 12J

Hence, 12 J of work is done by the force on the body.

Page : 134 , Block Name : Exercise

Q6.12 An electron and a proton are detected in a cosmic ray experiment, the rst with kinetic
energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain
the ratio of their speeds. (electron mass = 9.11×10-31 kg, proton mass = 1.67×10–27 kg, 1 eV =
1.60 ×10–19 J).

Electron is faster; Ratio of speeds is 13.54 : 1

−31
Mass of the electron, me = 9.11 × 10 kg

−27
Mass of the proton, mp = 1.67 × 10 kg

4
Kinetic energy of the electron, Eke = 10keV = 10 eV

4 −19
= 10 × 1.60 × 10

−15
= 1.60 × 10 J

5 −14
Kinetic energy of the proton, EKp = 100keV = 10 eV = 1.60 × 10 J

For the velocity of an electron ve , its kinetic energy is given by the relation:

1 2
EKc = mvc
2

2×EKe
∴ ve = √
m

−15
2×1.60×10 7
= √ −31
= 5.93 × 10 m/s
9.11×10

For the velocity of a proton vp , its kinetic energy is given by the relation:

1 2
EKp = mvp
2

2×Exp
vp = √
m

−14
2×1.6×10 6
∴ vp = √ −27
= 4.38 × 10 m/s
1.67×10

Hence, the electron is moving faster than the proton.

The ratio of their speeds:

7
vc 5.93×10
= 6
= 13.54 : 1
vp
4.38×10

Page : 134 , Block Name : Exercise

Q6.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with
decreasing acceleration (due to viscous resistance of the air) until at half its original height, it
attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the
work done by the gravitational force on the drop in the rst and second half of its journey ? What
is the work done by the resistive force in the entire journey if its speed on reaching the ground is
10 m s ? −1

−3
Radius of the raindrop, r = 2mm = 2 × 10 m

4 3
V = πr
3

3
4 −3 −3
= × 3.14 × (2 × 10 ) m
3

3 −3
Density of water, ρ = 10 kgm

Mass of the rain drop, m = ρV

3
4 −3 3
= × 3.14 × (2 × 10 ) × 10 kg
3

Gravitational force, F = mg
3
4 −3 3
= × 3.14 × (2 × 10 ) × 10 × 9.8N
3

The work done by the gravitational force on the drop in the first half of its journey:

W1 = FS

4 3
−3 3
= × 3.14 × (2 × 10 ) × 10 × 9.8
3

=0.082J

This amount of work is equal to the work done by the gravitational force on the drop in

the second half of its journey, i.e., WHI = 0.082J

As per the law of conservation of energy, if no resistive force is present, then the total

energy of the raindrop will remain the same.

∴ Total energy at the top:

Eτ = mgh + 0

3
4 −3 3
= × 3.14 × (2 × 10 ) × 10 × 9.8
3

= 0.1643

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10

m/s.

∴ Total energy at the ground:

1 2
EG = mv + 0
2
3
1 4 −3 3 2
= × × 3.14 × (2 × 10 ) × 10 × 9.8 × (10)
2 3

−3
= 1.675 × 10 J

∴ Resistive force = E6 − ET = −0.162J

Page : 134 , Block Name : Exercise

Q6.14 A molecule in a gas container hits a horizontal wall with speed 200 m s and angle 30°−1

with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is
the collision elastic or inelastic ?

Answer. Yes; Collision is elastic
The momentum of the gas molecule remains conserved whether the collision is elastic or
inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the
container, rebounding with the same speed.
It shows that the rebound velocity Of the wall remains zero. Hence, the total kinetic
energy of the molecule remains conserved during the collision. The given collision is an
example of an elastic collision.

Page : 134 , Block Name : Exercise

Q6.15 A pump on the ground oor of a building can pump up water to ll a tank of volume 30 m3
in 15 min. If the tank is 40 m above the ground, and the ef ciency of the pump is 30%, how much
electric power is consumed by the pump ?

3
Volume of the tank, V = 30m

Time of operation, t = 15min = 15 × 60 = 900s

Height of the tank, h = 40m

Efficiency of the purm, η = 30%

3 3
Density of water, ρ = 10 kg/m

3
Mass of water, m = ρV = 30 × 10 kg

3
Mass of water, m = ρV = 30 × 10 kg

Output power can be obtained as:

For input power P iu , ef ciency η,is given by the relation:
Work done mgh
P0 = =
Time t
3
30 × 10 × 9.8 × 40 3
= = 13.067 × 10 W
900

P0
η = = 30%
Pi
13.067
3
PI = × 100 × 10
30
5
= 0.436 × 10 W

= 43.6kW

Page : 134 , Block Name : Exercise

Q6.16 Two identical ball bearings in contact with each other and resting on a frictionless table
are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the
collision is elastic, which of the following (Fig. 6.14) is a possible result after collision ?

Answer. Case (ii)
It car. be observed that the total momentum before md after collision in each case is
constant.
For elastic collision, the total kinetic energy of a system remains conserved before
and after collision.
For mass Of each ball bearing m, we can write:
Total kinetic energy of the system before collision:

1 2 1
= mV + (2m)0
2 2

1 2
= mV
2

Case(i) Total kinetic energy of the system after collision:
2
1 1 V
= m × 0 + (2m)( )
2 2 2

1 2
= mV
4

Hence, the kinetic energy of the system is not conserved in case (i).
Case(ii) Total kinetic energy of the system after collision:
1 1 2
= (2m) × 0 + mV
2 2

1 2
= mV
2

Hence, the kinetic energy of the system is conserved in case (ii).

Case(iii) Total kinetic energy of the system after collision:
2
1 V
= (3m)( )
2 3

1 2
= mV
6

Hence, the kinetic energy of the system is not conserved in case (iii).

Page : 134 , Block Name : Exercise

Q6.17 The bob A of a pendulum released from 30o to the vertical hits another bob B of the same
mass at rest on a table as shown in Fig. 6.15. How high does the bob A rise after the collision ?
Neglect the size of the bobs and assume the collision to be elastic

Answer. Bob A will not rise at all In an elastic collision between two equal masses in which one is
stationary, while the other is moving with some velocity, the stationary mass acquires the same
velocity, while the moving mass immediately comes to rest after collision. In this case, a
complete transfer of momentum takes place from the moving mass to the stationary
Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest,
while bob B will move with the velocity of bob A at the instant of collision.

Page : 137 , Block Name : Exercise

Q6.18 The bob of a pendulum is released from a horizontal position. If the length of the
pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given

that it dissipated 5% of its initial energy against air resistance ?

Length of the pendulum, l = 1.5 m
Mass of the bob = m
Energy dissipated = 5%
According to the law of conservation of energy, the total energy of the system remains constant.
At the horizontal position:
Potential energy of the bob, EP = mgl
Kinetic energy of the bob, EK = 0
Total energy = mgl … (i)
At the lowermost point (mean position):
Potential energy of the bob, EP = 0
Kinetic energy of the bob, E = mv
1 2
K
2

Total energy E K =
1

2
mv
2
......(ii)
As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets
dissipated.
The total energy at the lowermost point is equal to 95% of the total energy at the horizontal
point, i.e.,
1 2 95
mv = × mgl
2 100

2×95×1.5×9.8
∴ v = √
100

= 5.28m/s

Page : 137 , Block Name : Exercise

Q6.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27
km/h on a frictionless track. After a while, sand starts leaking out of a hole on the oor of the
trolley at the rate of 0.05 kg s . What is the speed of the trolley after the entire sand bag is
−1

empty ?

Answer. The sandbag is placed on a trolley that is moving with a uniform speed Of 27 km/h. The
external forces acting on the system of the sandbag and the trolley is zero. When the sand starts
leaking from the bag, there will be no change in the velocity of the trolley. This is because the
leaking action does not produce any external force on the system. This is in accordance with
Newton's rst law of motion. Hence, the speed of the trolley will remain 27 km/h.

Page : 137 , Block Name : Exercise

3 1

Q6.20 A body of mass 0.5 kg travels in a straight line with velocity v = ax where a = 5m .
− −1
2 2 s

What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Mass of the body, m = 0.5kg
3 −1
−1
Velocity of the body is governed by the equation, v = ax 2 with a = 5m 2 s

Initial velocity, u( at x = 0) = 0

Final velocity v(atx = 2m) = 10√2m/s

Work done, W = change in kinetic energy

1 2 2
= m (v − u )
2

1 2 2
= × 0.5 [(10√2) − (0) ]
2

1
= × 0.5 × 10 × 10 × 2
2

= 50J

Page : 137 , Block Name : Exercise

Q6.21 The blades of a windmill sweep out a circle of area A.
(a) If the wind ows at a velocity v perpendicular to the circle, what is the mass of the air passing
through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A
= 30 m , v = 36 km/h and the density of air is 1.2 kg m . What is the electrical power produced ?
2 −3

Area of the circle swept by the windmill = A

Velocity of the wind = v

Density of air = ρ

(a) Volume of the wind flowing through the windmill per sec = AV

Mass of the wind flowing through the windmill per sec = ρAv

Mass m, of the wind flowing through the windmill in time t = ρAvt
1 2
(b) Kinetic energy of air = mv
2

1 2 1 3
= (ρAvt)v = ρAv t
2 2
2
(c) Area of the circle swept by the windmill = A = 30m

velocity of the wind = v = 36km/h

−3
Density of air, ρ = 1.2kgm

Electric energy produced = 25% of the wind energy
25
= × Kinetic energy of air
100

1 3
= ρAv t
8

Electrical energy
Electrical power =
Time
3
1 ρAv t 1 3
= = ρAv
8 t 8

1 3
= × 1.2 × 30 × (10)
8

3
= 4.5 × 10 W = 4.5kW

Page : 137 , Block Name : Exercise

Q6.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of
0.5 m each time. Assume that the potential energy lost each time she lowers the mass is
dissipated.
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 × 107J of energy per kilogram which is converted to mechanical energy with a
20% ef ciency rate. How much fat will the dieter use up?

(a) Mass of the weight, m = 10kg

Height to which the person lifts the weight, h = 0.5m

Number of times the weight is lifted, n = 1000

∴ Work done against gravitational force:

= n(mgh)

= 1000 × 10 × 9.8 × 0.5

3
= 49 × 10 J = 49kJ
7
(b) Energy equivalent of 1kg of fat = 3.8 × 10 J

Efficiency rate = 20%

Mechanical energy supplied by the person's body:

20 7
= × 3.8 × 10 J
100

1 7
= × 3.8 × 10 J
5

Equivalent mass of fat lost by the dieter:

1 3
= × 49 × 10
1 7
×3.8×10
5

245 −4
= × 10
3.8

−3
= 6.45 × 10 kg

Page : 137 , Block Name : Exercise

Q6.23 A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square
meter. If 20% of this energy can be converted to useful electrical energy, how large an area is
needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.

2
(a) 200m

3
(a) Power used by the family, P = 8kW = 8 × 10 W

Solar energy received per square metre = 200W

Efficiency of conversion from solar to electricity energy = 20%

Area required to generate the desired electricity = A

As per the information given in the question, we have:

3
8 × 10 = 20% × (A × 200)

20
= × A × 200
100

3
8×10 2
∴ A = = 200m
40

(b) The area of a solar plate required to generate 8kW of electricity is almost equivalent

to the area of the roof of a building having dimensions 14m × 14m.

Page : 137 , Block Name : Exercise

Q6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s strikes a block of wood of mass 0.4
−1

kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling
by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount
of heat produced in the block

Mass of the bullet, m = 0.012kg

Initial speed of the bullet, ub = 70m/s

Mass of the wooden block, M = 0.4kg

Initial speed of the wooden block, uB = 0

Final speed of the system of the bullet and the block = v

Applying the law of conservation of momentum:

mub + M uB = (m + M )v

0.012 × 70 + 0.4 × 0 = (0.012 + 0.4)v

0.84
∴ v = = 2.04m/s
0.412

For the system of the bullet and the wooden block:

′
Mass of the system, m = 0.412kg

Velocity of the system = 2.04m/s

Height up to which the system rises = h

Applying the law of conservation of energy to this system:

Potential energy at the highest point = Kinetic energy at the lowest point
′ 1 ′ 2
m gh = m v
2

2
1 v
∴ h = ( )
2 g

2
1 (2.04)
= ×
2 9.8

= 0.2123m

The wooden block will rise to a height of 0.2123m .

Heat produced = Kinetic energy of the bullet- Kinetic energy of the system

1 2 1 ′ 2
= mu − m v
2 2

1 2 1 2
= × 0.012 × (70) − × 0.412 × (2.04)
2 2

= 29.4 − 0.857 = 28.54J

Page : 137 , Block Name : Additional Exercises

Q6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two
stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the
bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 300 , θ2
= 600 , and h = 10 m, what are the speeds and times taken by the two stones ?

Answer. No; the stone moving down the steep plane will reach the bottom rst
Yes; the stones will reach the bottom with the same speed
vB = vC = 14 m/s
t = 2.86 s; t = 1.65 s
1 2

The given situation can be shown as in the following gure:

Here, the initial height (AD) for both the stones is the same (h). Hence, both will have the same
potential energy at point A.
As per the law of conservation of energy, the kinetic energy of the stones at points B and C will
also be the same, i. e.,
1 2 1 2
mv = mv
2 1 2 2

v1 = v2 = v, say

Where,
m = Mass of each stone
v = Speed of each stone at points B and C
Hence, both stones will reach the bottom with the same speed, v.
For stone I:
Net force acting on this stone is given by:
Frat = ma1 = mg sin θ1

a1 = g sin θ1

For stone II:
a2 = g sin θ2

∵ θ2 > θ1

∴ sin θ2 > sin θ1

∴ a2 > a1

Using the rst equation of motion, the time of slide can be obtained as:
v = u + at
v
∴ t = (∵ u = 0)
a

For stone I .
v
t1 =
a1

For stone II:
v
t2 =
a2

∵ a2 > a1

∴ a 2 < t1

Hence, the stone moving down the steep plane will reach the bottom rst.
The speed (v) of each stone at points B and C is given by the relation obtained from the law of
conservation of energy.
1
2
mgh = mv
2

∴ v = √2gh

= √2 × 9.8 × 10

= √196 = 14m/s

The times are given as:
v v 14 14
t1 = = = = = 2.86s
a1 g sin θ1 9.8×sin 30 1
9.8×
2

v v 14 14
t2 = = = = = 1.65s
a2 g sin θ2 9.8×sin 60 √3
9.8×
2

Page : 137 , Block Name : Additional Exercises

Q6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N
m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched
position. The block moves 10 cm down the incline before coming to rest. Find the coef cient of
friction between the block and the incline. Assume that the spring has a negligible mass and the
pulley is frictionless

Mass of the block, m = 1kg

−1
Spring constant, k = 100Nm

Displacement in the block, x = 10cm = 0.1m

The given situation can be shown as in the following figure.

At equilibrium:
∘
Normal reaction, R = mg cos 37

∘
Frictional force, f = μR = mg sin 37

Where, μ is the coefficient of friction
∘
Net force acting on the block = mg sin 37 − f

∘ ∘
= m gsin 37 − μm cos 37

∘ ∘
= mg (sin 37 − μ cos 37 )

At equilibrium, the work done by the block is equal to the potential energy of the spring,

i.e.,

∘ ∘ 1 2
mg (sin 37 − μ cos 37 ) x = kx
2

∘ ∘ 1
1 × 9.8 (sin 37 − μ cos 37 ) = × 100 × 0.1
2

0.602 − μ × 0.799 = 0.510

0.092
∴ μ = = 0.115
0.799

Page : 139 , Block Name : Additional Exercises

Q6.27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform
speed of 7 m s . It hits the oor of the elevator (length of the elevator = 3 m) and does not
−1

rebound. What is the heat produced by the impact ? Would your answer be different if the
elevator were stationary ?

Mass of the bolt, m = 0.3kg

Speed of the elevator = 7m/s

Height, h = 3m

since the relative velocity of the bolt with respect to the lift is zero, at the time of

impact, potential energy gets converted into heat energy.

Heat produced = Loss of potential energy

= mgh = 0.3 × 9.8 × 3

= 8.823

The heat produced will remain the same even if the lift is stationary. This is because of

the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Page : 139 , Block Name : Additional Exercises

Q6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A
child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m
s
−1
relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley.
What is the nal speed of the trolley ? How much has the trolley moved from the time the child
begins to run ?’

Mass of the trolley, M = 200kg

Speed of the trolley, v = 36km/h = 10m/s

Mass of the boy, m = 20kg

Initial momentum of the system of the boy and the trolley

= (M + m)v

= (200 + 20) × 10

= 2200kgm/s

′
Let v be the final velocity of the trolley with respect to the ground.

′
Final velocity of the boy with respect to the ground = v − 4
′ ′
Final momentum = M v + m (v − 4)

′ ′
= 200v + 20v − 80

′
= 220v − 80

As per the law of conservation of momentum:

Initial momentum = Final momentum
′
2200 = 220v − 80

′ 2280
∴ v = = 10.36m/s
220

Length of the trolley, l = 10m

′′
Speed of the boy, v = 4m/s
10
Time taken by the boy to run, t = = 2.5s
4

′′
:Distance moved by the trolley = v × t = 10.36 × 2.5 = 25.9m

Page : 139 , Block Name : Additional Exercises

Q6.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the
elastic collision of two billiard balls ? Here r is the distance between centres of the balls.

Answer. (i), (ii), (iii), (iv), and (vi)
The potential energy of a system of two masses is inversely proportional to the
separation between them. In the given case, the potential energy Of the system Of the
two balls will decrease as they come closer to each other, It will become zero (i.e., V(r)
= O) when the two balls touch each other, i.e., at r = 2R, where R is the radius of each
billiard ball, The potential energy curves given in gures (i), (ii), (iil), (iv), and (vi) do
not satisfy these two conditions. Hence, they do not describe the elastic collisions
between them.

Page : 139 , Block Name : Additional Exercises

Q6.30 Consider the decay of a free neutron at rest : n → p + e
Show that the two-body decay of this type must necessarily give an electron of xed energy and,
therefore, cannot account for the observed continuous energy distribution in the β-decay of a
neutron or a nucleus (Fig. 6.19).

[Note: The simple result of this exercise was one among the several arguments advanced by W.
Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is
known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e —, p or n), but is
neutral, and either massless or having an extremely small mass (compared to the mass of
electron) and which interacts very weakly with matter. The correct decay process of neutron is : n
gp+e–+ν]

The decay process of free neutron at rest is given as:

−
n → p + e

2
From Einstein's mass-energy relation, we have the energy of electron as Δmc

Where,

Δm = Mass defect = Mass of neutron - (Mass of proton + Mass of electron)

c = Speed of light

Δm and c are constants. Hence, the given two-body decay is unable to explain the

continuous energy distribution in the β -decay of a neutron or a nucleus. The presence of

neutrino von the LHS of the decay correctly explains the continuous energy distribution.

Page : 140 , Block Name : Additional Exercises