Work and Energy Notes by Arun Umrao

How energy relates to work? to know more, read this notes. Suitable for CBSE board students. Sections contains work energy, work energy units, work energy conversion

1. 1
ENERGY&WORK
A SHORTNOTE
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY- GPL LICENSING

2. 2 Energy & Work
Contents
1 Energy & Work 3
1.1 Measurement of Energy . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Measurement of Energy . . . . . . . . . . . . . . . . . 3
1.1.2 Kilo Watt Hour . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.1 Energy Exchange . . . . . . . . . . . . . . . . . . . . . 4
1.2.2 Methods of Energy Exchange . . . . . . . . . . . . . . 5
1.2.3 Type of Energy Source . . . . . . . . . . . . . . . . . . 6
1.2.4 Energy Equilibrium . . . . . . . . . . . . . . . . . . . . 6
1.2.5 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.6 Specific Heat Capacity . . . . . . . . . . . . . . . . . . 8
1.2.7 Heat Exchange . . . . . . . . . . . . . . . . . . . . . . 9
1.2.8 Phases of Matter . . . . . . . . . . . . . . . . . . . . . 11
1.2.9 Latent Energy . . . . . . . . . . . . . . . . . . . . . . . 11
1.3 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.3.1 Work In Linear Motion . . . . . . . . . . . . . . . . . . 14
1.3.2 Work Energy Relation . . . . . . . . . . . . . . . . . . 14
1.3.3 Work In Vertical Motion . . . . . . . . . . . . . . . . . 15
1.4 Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.4.1 Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.4.2 Efficiency in Parallel Network . . . . . . . . . . . . . . 22
1.4.3 Efficiency in Series Network . . . . . . . . . . . . . . . 23
1.4.4 Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3. 1.1. MEASUREMENT OF ENERGY 3
1
Energy & Work
1.1 Measurement of Energy
Energy is a physical quantity that has ability to do equal amount of work.
For efficient use of energy and its conversion into work equivalent, itrequires
to be measured in physical parameters.
1.1.1 Measurement of Energy
Energy is measured in two units.
Calorieis CGS unit of energy measurement. One Calorie of energy is
equal to the energy required to increase the temperature of onegram of water
from 14.5
◦
C to 15.5
◦
C at one atmospheric pressure.
Jouleis MKS unit of energy measurement. One Calorie of heat equals
to the 4.2 Joule.
Solved Problem 1.1Find the energy equivalent in Joules of two Calories.
SolutionOne calorie is equal to 4.2 Joules. Therefore, two calories would
be equal to 8.4 Joules.
Solved Problem 1.2Convert sixty joules into calories.
Solution4.2 Joules are equal to one calorie. Therefore, sixty Joules
would be equal to 14.29 calories.
1.1.2 Kilo Watt Hour
It is unit of measurement of electrical energy. The kilowatt-hour (kWh)
energy is equivalent to one kilowatt (1 kW) of power consumed in one hour
(1 h) of time.
1kWh= 1kW×1 hour (1.1)

4. 4 Energy & Work
In terms of second unit
1kWh= 1kW×
nseconds
3600
(1.2)
Solved Problem 1.3A LED bulb of 5 watt is used in a park street pole. It
lights up for 8 hours daily. Find the electric energy consumed by it in KWh
unit for the month of Jun.
SolutionIn KWh unit of electric energy measurement, the power should
be in kilo watt unit and time should be in hours. Now, in Jun month there
are 30 days. LED bulb lights up for 8 hours daily. So, total time is
t= 30×8 = 240hr
Total LED bulb wattage is 5 watt or 5×10
−3
KW. Now, total electric energy
is consumed by LED bulb in month of Jun is
E= 5×10
−3
KW×240hr= 1.2KWh
This is total unit of electric consumption by LED bulb.
1.2 Energy
1.2.1 Energy Exchange
It is scientifically proven that each body/object/material, whose temperature
is greater than 0K, radiates energy as well as absorbs energy simultaneously.
The quantity of radiated energy may be greater than or equal to or less than
the absorbed energy. It depends on the environment where object is placed.
If object is placed in an environment whose temperature is less thantemper-
ature of object then object would radiate more energy to environment that
absorbing of the energy from the environment. This radiation and absorp-
tion of the energy by a body is called energy exchange with environment.
For simplicity of the concept, here we generally assume that objectwith rel-
atively higher temperature (hot object) always emits energy while object at
relatively lower temperature (cold object) always absorbs energy.

5. 1.2. ENERGY 5
T1
A
T2
B
From above figure, assume that two platesAandBare placed from each
other at finite distance. Temperature of plateAis higher than temperature
of plateB, i.e.T1> T2. In the simplest form, plateAwould emit energy
while plateBwould absorb energy radiated from plateA. The exchange of
heat takes place in form of radiation.
1.2.2 Methods of Energy Exchange
There are three methods by which energy exchange takes place.
ConductionIn conduction, only heat/energy transfers from one point
to other point in object under heating or cooling. The material particles
do not move from their equilibrium position but they oscillate and collide
with the adjacent material particles. This phenomenon is visible mostly in
solids. The vibrational amplitude of material particles at hot region ismore
than vibrational amplitude of material particles at cold region. For example,
heating of iron rod, stones etc.
ConvectionIn convection, energy transmission takes place by movement
of material particles. In convection, particles move from higher temperature
region to lower temperature region. Convection is visible mostly in liquid
and gas. The kinetic energy of material particles at hot region is more than
kinetic energy of material particles at cold region. For example, heating of
water and air etc.
RadiationIn radiation, energy is transported from hot material to cold
material in form of quanta of electro-magnetic waves. Colder objects emit
waves with lower frequencies while hot objects emit waves with higherfre-
quencies. For example, energy transmission by light wave, radio waves, mi-
crowaves etc. In contrast to the conduction and convection, radiation can
be occurred even if two objects are placed in vacuum or they are not in
physically contact.

6. 6 Energy & Work
Solved Problem 1.4A hot and a cold plates are placed in air at finite dis-
tance so that they can exchange heat. Write the names of methodof energy
exchange take place between them.
SolutionTwo plates are placed in air medium, hence convection method
would take place for exchange of heat. As radiation occurred bothin medium
and in vacuum, therefore, radiation method would also take place for ex-
change of heat. Two plates are not in contact physically, hence conduction
method would not take place for exchange of heat.
1.2.3 Type of Energy Source
In this chapter we have assumed that energy sources or energy sinks are
mostlynon-constant temperature sources, it means their temperature
either fall or rise with time respectively to reach equilibrium when theyare
place in thermal contact. For example, a hot bead is cooled down by placing
it in open air, or a bead is placed in sun light to heat up it.
Another type of energy source areconstant temperature sources
whose temperature do not change with time. For example, sun, a blast
furness, heating coil etc.
1.2.4 Energy Equilibrium
Suppose two or more bodies at different temperatures are kept in thermal
contact
1
and they start exchanging energy among themselves. During the
process body at higher temperature losses energy and body at lower temper-
ature gains the energy. This exchange of energy take place until temperature
of all bodies become equal. When any of all bodies neither gain or loss energy,
then it is said that the system is inenergy equilibrium.
Always remember that in heat and thermodynamics, temperature
means only temperature in Kelvin temperature scale even if there is no
variation in answers on using Celsius scale.
1
Type of arrangement of bodies so that exchange of heat/energytakes place among
them.

7. 1.2. ENERGY 7
1.2.5 Heat Capacity
Let ‘Q’ quantity of energy is given to a body of mass ‘m’ and the temperature
of the body is increased by ∆T. Now energy stored by the body is
dQ=m c∆T (1.3)
Wherecis specific heat capacity of a substance. ∆Tis always computed
as subtracting final temperature from initial temperature.If ∆T
is negative then substance is releasing heat and if ∆Tis positive then sub-
stance is absorbing heat. Sometimes, specific heat capacity of a substance is
represented bys.
SubstancecinJ Kg
−1
K
−1
cinCal gm
−1
K
−1
Aluminum 900 0.215
Copper 386 0.092
Gold 126 0.030
Lead 128 0.030
Silver 233 0.055
Tungsten 134 0.032
Zinc 387 0.092
Mercury 140 0.033
Water 4186 1.000
Steam 2010 0.480
Ice 2050 0.490
Glass 840 0.200
Table 1.1: Specific heats of various substances.
Unit of specific heat capacity is Joule per Kg per degree Centigrade or
Calorie per gram per degree Centigrade in Centigrade scale or Joule per Kg
per Kelvin or Calorie per gram Kelvin in Kelvin scale. These two different
units are equal as one degree difference is equal for both Celsius and Kelvin

8. 8 Energy & Work
scale.(Alarm!)Note that specific heat is dependent of Kelvin temperature
scale. Therefore, it is good habit to convert Celsius temperature scale into
Kelvin temperature scale even if there is no difference between final results
when we use either Celsius temperature scale or Kelvin temperature scale.
From the relation 1.3, if specific heat is not constant but it depends on
temperature, i.e. it is function of temperature, then relation 1.3 would be
written as
dQ=m c(T)dT (1.4)
Where temperature is in Kelvin scale. Now, if temperature changes from
T1K toT2K, then
Q=
Z
T2
T1
m c(T)dT (1.5)
1.2.6 Specific Heat Capacity
From the relationdQ=m c∆T, specific heat capacity can be written as
c=
dQ
m∆T
IfdQheat is given to raise ∆T= 1K temperature ofm= 1 kilogram
substance ordQheat is released to drop ∆T= 1K temperature bym= 1
kilogram substance, then value ofdQin Joule is value of specific heat capacity
of that substance.
Solved Problem 1.5A virtual substance releases 25J energy by dropping its
temperature from 300K to 290K. If mass of this virtual substance is 0.025
kilogram then find its specific heat capacity.
SolutionThe drop in temperature of virtual substance is
∆T= 290−300 =−10K
Negative sign shows that there is release of heat from the virtual substance.
Mass of virtual substance is 0.025 kilogram. Now from the definition of
specific heat capacity
c=
25J
0.025Kg×10K
= 100J Kg
−1
K
−1
This is specific heat capacity of the virtual substance.

9. 1.2. ENERGY 9
Solved Problem 1.6An unknown substance releases 25J energy by dropping
its temperature from 300K to 292.85K. If mass of this unknown substance is
0.025 kilogram then find its specific heat capacity. Can you find the name of
substance?
SolutionThe drop in temperature of unknown substance is
∆T= 292.85−300 =−7.14K
Negative sign shows that there is release of heat from the unknownsubstance.
Mass of unknown substance is 0.025 kilogram. Now from the definition of
specific heat capacity
c=
25J
0.025Kg×7.14K
= 140.05J Kg
−1
K
−1
This is specific heat capacity of the unknown substance. From the table 1.1,
it is approximately equal to the specific heat capacity of Mercury, therefore
we can say that this unknown substance is Mercury.
1.2.7 Heat Exchange
As we know that different substances have different specific heat capacities.
It means if equal mass of substances are given equal amount of heat energy
then due to their different capacities to hold this energy, rise in temper-
ature would be different. For example, if 10 joules of energy is given to
10 gram aluminum and copper both at 20
◦
C, then temperature of copper
would rise more than temperature of aluminum as copper has lower specific
heat capacity than aluminum. This is why, during heat exchange between
two different substances, fall of temperature in heat transmitter substance is
different than the rise in temperature of heat receiving substance. Mathe-
matically,|Q1|=|Q2|but|∆T1| 6=|∆T2|for same mass of heat exchanging
substances (m1=m2).
Solved Problem 1.7In terms of price gold is more precious than copper as 10
gram gold costs about fifty thousands while copper costs just twenty rupees.
But in terms of specific heat capacity, copper is more precious thangold.
Copper can store more amount of heat energy than gold. This is why, most
of heat absorbers in electronic devices are made of copper. Now, tell that
what is heat capacity of copper and gold?

10. 10 Energy & Work
SolutionThe specific heat capacity of copper and gold are 386J Kg
−1
K
−1
and 126J Kg
−1
K
−1
respectively.
Solved Problem 1.8Convert the specific heat capacity of copper
386J Kg
−1
K
−1
intoJ gm
−1
K
−1
.
SolutionThe specific heat capacity of copper is 386J Kg
−1
K
−1
. In
J gm
−1
K
−1
, it is
c= 386J Kg
−1
K
−1
= 386J
−
10
3
gm
·
−1
K
−1
= 0.386J gm
−1
K
−1
This is required result.
Solved Problem 1.910 Joule energy is given 5 gram copper and gold each
when they are at temperature of 20
◦
C. Find their final temperatures.
SolutionThe specific heats of the copper and gold are 0.386 Joule per
Gram per Kelvin and 0.126 Joule per Gram per Kelvin respectively. The
energy supplied to these two metal would increase their temperature by ∆T
Kelvin. Now, for copper
10 = 5×10
−3
×386×∆T1
On solving it, we have
∆T1= 5.18K= 5.18
◦
C
One degree difference of Kelvin temperature scale and Celsius temperature
scale is equal. Now, the final temperature of the copper would be
T1f= 20 + 5.18 = 25.18
◦
C
Now, for gold
10 = 5×10
−3
×126×∆T2
On solving it, we have
∆T2= 15.87K= 15.87
◦
C
One degree difference of Kelvin temperature scale and Celsius temperature
scale is equal. Now, the final temperature of the gold would be
T2f= 20 + 15.87 = 35.87
◦
C
T1fandT2fare the final temperatures of copper and gold respectively. These
results also indicate that temperature of materials with low specific heat
capacity rise more than the temperature of materials with high specific heat
capacity when they are supplied equal amount of energy.

11. 1.2. ENERGY 11
Solved Problem 1.10Temperature of two kilogram water is raised from 25
◦
C
to 50
◦
C. Find the amount of additional energy accumulated by water.
SolutionWhen water is heated, it absorbs energy and raises its tem-
perature. The amount of energy required to raise the temperature of two
kilogram of water from 25
◦
C to 50
◦
C is given by
∆Q=mc∆T= 2×4186×(50−25)
Note that the temperature difference in Celsius temperature scaleor Kelvin
temperature scale is equal, our concept of using ∆Tin Celsius unit is not
mathematically wrong but it is conceptually wrong. So
∆Q= 2×4186×(323−298) = 209.3kJ
This is amount of energy require to raise the temperature of two kilogram
water from 25
◦
C to 50
◦
C.
1.2.8 Phases of Matter
There are three phases of a material. (i) Solid phase, (ii) Liquid phaseand
(iii) Gaseous phase. When state of a matter changes from solid to liquid
or from liquid to gas then it is to be said that matter is changing its phase.
During the change of phase, temperature of the substance does not
change. Energy supplied to solid matter to change its phase into liquid, It
is consumed to broke the bonds responsible for solid phase of the matter.
Similarly, during the change of liquid to gaseous phase, energy is consumed
in breaking of inter-molecular forces among liquid molecules.
1.2.9 Latent Energy
Energy required to change the phase of substance from solid to liquid or
liquid to gas is known as latent energy of the substance. Let ‘Q’ heatis
required to melt ‘m’ kilogram mass from solid to liquid then
Q=mL (1.6)
WhereLis latent heat of the substance. If phase of substance changes from
solid to liquid then latent heat is known aslatent heat of melting. If phase
of substance changes from liquid to gas then latent heat is known aslatent

12. 12 Energy & Work
heat of evaporation. The unit if latent heat is Joule per Kg or Calorie per
gram. 334 Joule of energy are required to melt one gram of ice at 0
◦
C. The
latent heat of water at 100
◦
C for vaporization is approximately 2230 joules
per gram.
Suppose a ice cube is at temperature of−T
◦
C and head is supplied
to melt it. A point is to be remembered that the temperature of ice cube
increases to 0
◦
C ice as whole and then it start melting to liquid of 0
◦
C.
During the processit is assumed that heat absorbed by ice cube is
homogeneously distributed through out the ice cube. In numerical
problems avoid taking fraction of ice that is heated and melt to liquid form
though the rest of ice is in solid form.
Solved Problem 1.1120gwater at temperature 10
◦
C is mixed with 50gwater
at temperature 50
◦
C. Find the equilibrium temperature of the mixture.
Solution
100
◦
C
0
◦
C
Solid Liquid Gas
50
◦
C
10
◦
C
bc
b
T
Here, hot water is mixed up with cold water and both substances are
in same phase. During mixing up, no state change takes place. Thereis
only exchange of heat from hot water and cold water. Thus the temperature
of cold water would increase and temperature of hot water would decrease.
Let the common temperature of the water after complete mixed upand
equilibrium isTin Celsius. Now, heat exchange between these two different
temperature level water should be same. So,
20×10
−3
×c×(T−10) = 50×10
−3
×c×(50−T)
Or
2×(T−10) = 5×(50−T)

13. 1.2. ENERGY 13
On solving it, we getT= 38.5
◦
C (approx).
Solved Problem 1.1220gice at temperature−25
◦
C is mixed with 50gwater
at temperature 50
◦
C. Find the equilibrium temperature of the mixture.
Solution
100
◦
C
0
◦
C
Solid Liquid Gas
50
◦
C
−25
◦
C
b
bcbc
bc
T
Here, hot water is mixed up with chilled ice. During mixing up, first ice
temperature would change from−25
◦
C to 0
◦
C and then ice would melt into
0
◦
C water. Finally the temperature of cold water would increase from 0
◦
C to
an equilibrium temperature (sayT). A phase change is take up from ice into
water, so first we would check whether hot water has sufficient heat energy to
first increase the temperature of ice from−25
◦
C (ice) to 0
◦
C (ice) and then
melt it completely. For this purpose we take reference temperature 0
◦
C. So
heat transferred from hot water to cold water in MKS system is
Qh= 50×4.18×(50−0) = 10.45×10
3
J
The heat required by ice to raise its temperature from
Qi= 20×2.05×(0−(−25)) = 1.03×10
3
J
Qh> Qi, hence there is remaining heat that may melt ice at 0
◦
C into water
at 0
◦
C. Now, we would compute the amount of heat required to melt the 20g
ice completely.
Qim= 20×334 = 6.68×10
3
J
Now,Qim<(Qh−Qi) hence ice would completely melted. And there is
further remaining heat that would raise the temperature of 0
◦
C water. Let
the final equilibrium temperature isT. Finally
20×2.05×25 + 20×334 + 20×4.18×T= 50×4.18×(50−T) (1.7)
On solving it, we haveT= 9.40
◦
.

14. 14 Energy & Work
1.3 Work
1.3.1 Work In Linear Motion
It is said to be work done by external or internal agents if there is change in
position of object or object get deformed. If position of object ischanging
then work done is
W=F×d
It is scalar quantity, hence vector form of work relation is
W=
~
F·
~
d (1.8)
1.3.2 Work Energy Relation
Sometime a sudden force changes the velocity of a moving object without
loss of any energy. In this case, work done is equal to the change inthe
momentum of the object. In real life, when external or internal forces are
applied on the system, there is loss of some energy which does not contribute
to the work. So, work is also equivalent to the change in the kinetic energy
when an object is suffered sudden impact with loss of non-working energy
(work done by non conservative forces).
W=
1
2
mv
2
−
1
2
mu
2
(1.9)
For the work and kinetic energy relation 1.9, the direction of initial and final
velocities do not matter. For example, if a force by a system is appliedon the
object in the same direction to which object was moving then definitely its
velocity will increase. If applied force by a system is in opposite direction to
the motion of object (say speedu) then object would work on the system till
it comes in rest (say speedv= 0). Consequently, system gains some amount
of the work/energy (saydW). If system bounce back the object with same
speedv=ufromu= 0 in opposite direction, then system needed only same
mount of work/energy (dW) that it was received from the object previously.
So, net work done by the system is zero (gain of energy is equal to work
done). Therefore, directions of velocities before and after workdone on the
moving object do not matter. Only magnitudes of the velocities matter for
the relation 1.9.

15. 1.3. WORK 15
Solved Problem 1.13A ball of 2 kilogram is moving with speed of 20m/s.
On applying an external force, its speed increases to 30m/s. Find the work
done by external force.
SolutionThe work done by external force on the ball is given by
W=
1
2
mv
2
−
1
2
mu
2
=
1
2
×2×
−
30
2
−20
2
·
It givesW= 500J.
1.3.3 Work In Vertical Motion
In vertical plane, work is always equals to the change in total energy of the
object when object changes its vertical position. Total energy at a point in
a vertical plane is equal to the sum of kinetic energy and potential energy of
the object at that point. In terms of position of the object
W=mg(h2−h1) (1.10)
Solved Problem 1.14A rain drop of radiusrmm falls from a height ofhm
above the ground. It falls with decreasing acceleration (due to viscous re-
sistance of the air) until at half its original height, it attains its maximum
(terminal) speed, and moves with uniform speed thereafter. Whatis the
work done by the gravitational force on the drop in the first 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 isvm/s.
SolutionIn the first case drop falls due to gravity of the earth. Again
viscous resistance of air increases as the velocity of falling drop increases.
This means downward net force decreases and acceleration of thedrop also
decreases. When drops falls by height ofh/2, its velocity becomes constant
and net force become zero. Acceleration after vanishing of net force becomes
zero and drop falls with constant speed. Now assuming velocity of drop at
heighthis zero and initial acceleration of falling drop is maximum becomes
0 after falling byh/2. Let retarding acceleration isa. Now Net force on the
drop at any instant is
R=mg−F (1.11)

16. 16 Energy & Work
Net forceRwould let the fall drop with accelerationaand the equation
(1.11) becomes
ma=mg−F (1.12)
Now
F=mg−ma
During the fall of drop from a height ofhtoh/2 time for work done to both
drop and and viscous resistance is same. Assume drop is at depth ofxfrom
maximum height after timetand atx+dxafter timet+dt. Now work done
by viscous resistance when distance cover by drop bydxis
dWf1P=F·dx
Or
dWf1P= (mg−ma)·dx
Integrating above relation for height fromhtoh/2 and
Z
Wf1P=
Z
h/2
h
(mg−ma)·dx
Wf1P= [mgx−max]
h/2
h
=
≤
1
2
mgh−mgh−
1
2
mah+mah

=
≤
−
1
2
mgh+
1
2
mah

Work done by viscous force due to change in kinetic energy is
Wf1K=
1
2
mv
2
−
1
2
mu
2
(1.13)
Initial velocity is 0 and
Wf1K=
1
2
mv
2
(1.14)
Substituting the value ofvfrom the third equation of linear motion for height
h/2 under the effect of retarding accelerationa. So
v
2
=u
2
+ 2a
h
2
v
2
=ah (1.15)

17. 1.3. WORK 17
Substituting the value ofv
2
in equation (1.14)
Wf1K=−
1
2
mah
Total work done by viscous force is
Wf1=Wf1K+Wf1P
Wf1=−
1
2
mgh (1.16)
This is the work done by viscous resistance force for half motion. Negative
sigh shows that the direction of viscous force is upward and work done by it
is negative. If drop is moving with constant retardation accelerationathen
net force on the ball is (g−a)mand it is equal to the viscous resistance
force and work down by drop weight is equal to theWf1but opposite in sign
(drop’s motion is opposite to the direction of viscous resistance force) hence
Wd=
1
2
mv
2
(1.17)
During the second half, velocity of the drop is constant hence net force on
the drop is zero and
0 =mg−F (1.18)
F=mg
During the fall of drop from a height ofh/2 to 0 time for work done to both
drop and and viscous resistance is same. Assume drop is at depth ofxfrom
h/2 height after timetand atx+dxafter timet+dt. Now work done by
viscous resistance when distance cover by drop bydxis
dWf2=F·dx
Or
dWf2P=mg·dx

18. 18 Energy & Work
On solving above equation by integrating it for the height limitsh/2 to 0
Z
dWf2P=
Z
0
h/2
mg·dx
Wf2P= [mgx]
0
h/2
=
≤
0−
1
2
mgh

=
≤
−
1
2
mgh

Wf2P=−mg
h
2
(1.19)
Work done by the viscous force is equal to the change in potential energy of
the drop in second half motion. In term of kinetic energy, substituting the
value ofgfrom the relation
v
2
=v
2
+ 2g
h
2
for second half motion.
g= 0
Though the gravity is acting on the body but gravitational acceleration that
increases velocity of the drop is zero. Hence there is no change in velocity of
the drop. Now substituting the value ofgin equation (1.19)
Wf2K= 0
Hence work done by viscous force sum of work done in kinetic energyform
and in potential energy form so
Wf1=Wf2K+Wf2P
Wf1=−mg
h
2
(1.20)

19. 1.4. EFFICIENCY 19
1.4 Efficiency
Mechanical work is amount of energy expend or found when there isvisible
change in the position of the object. In other words, work is the product of
force and displacement of the ideal rigid object in the direction of force.
W=Fd
F
d
Unit of work in MKS system is Joule and it is scalar quantity. This is
standard definition of the work. Sometime object does not displacefrom
its equilibrium position but there is still work/energy loss. To solve this
confusion, a modified form of work definition can be stated as “ workis
equal to the dot product of force vector and vector of displacement of force
point”. Force point is the point on the object where force is acted upon. In
case of non ideal rigid objects, when a force is applied on them, theydo not
displace, but they got compressed. In this case, work is defined asproduct
of force vector and displacement of force along displacement vector.
F F F F
d
If force vector and displacement vector are
~
Fand
~
drespectively, then
work done is given by
W=
~
F·
~
d (1.21)
~
F
~
d
On solving equation 1.21, we have
W=|
~
F| |
~
d|cosθ
Here,θis angle between force and displacement vectors. When both force
and displacement vectors are parallel, then work isW=F·d. When force

20. 20 Energy & Work
and displacement vectors are normal, then work isW= 0. According to the
second case, where force and displacement vectors are normal to each other,
like a walking man, man carrying weight do zero work. Is it true? Physics
says Yes, bio-physics says No.
Walking ManA walking man pushes earth by his one leg and put his
second leg in forward location continuously to move forward as shown in the
first part of below figure.
dy
d d
When we see the walking man, we observe two arrangement of his legs.
One is triangular and other is straight line. In triangular arrangement, legs
form two side of triangle and earth makes base. In linear arrangement, both
legs are placed close together along the vertical axis, as shown in second part
of above figure. In cycle of each step, man lifts his weight bydyvertically
upward. Thus work done by him in each cycle is
dW=mg×dy
This is work done by walking man in each cycle of step.
Weight Carrying ManA man is carrying a load massmover his head as
shown in the first part of below figure. To keep the position of mass intact,
he applied upward forceF=mg. He is moving in horizontal plane by a
distancedeither left or right.
~F
~
F
′
dy
d d
At a certain time, another massmis added to the load. Due to this extra
mass, his hands drop down by a depth ofdy. To keep the original level of
masses, he must do workdW= 2m×dyinstantaneously and thereafter he

21. 1.4. EFFICIENCY 21
must applied forceF
′
= 2mgcontinuously to keep the level of load intact.
Bio-physics said that, human muscles and bones are not perfectly rigid as
machines are. Hence a man carrying weight continuously holds mass up and
down by a vertical displacementdy. Works done by him in each vertical
cycle is
dW= 2mg×dy
This continuous work tires the man.
1.4.1 Efficiency
Q
W
R
F
S
Efficiency of a system is its ability to convert energy into mechanical work or
vice-versa. A system is not free from friction and perfectly insulated, hence
the amount of heat taken by a system is not completely converted into work
or vice-versa. A fraction of energy is lost due to radiation and friction. From
above figure,
Q=W+F+R
The efficiency of the system is
η=
W
Q
(1.22)
HereW≤Q, so the efficiency of a system can not be more than 1. It is
unit-less quantity. The range of efficiency is from zero to one. i.e.
0≤η≤1
Efficiency percentage is given by
η=
W
Q
×100% (1.23)

22. 22 Energy & Work
1.4.2 Efficiency in Parallel Network
Assume there are three systems connected parallaly as shown in the figure
below. The energy supplied to each of the system isQ1,Q2andQ3. The
efficiencies of these three systems areη1,η2andη3.
η1Q1
η2Q2
η3Q3
W
The outputs of these three systems individually are
W1=Q1×η1
W2=Q2×η2
and
W3=Q3×η3
η1
η2
η3
η
v Net output isW=W1+W2+W3. So
W=Q1×η1+Q2×η2+Q3×η3
Net input isQ=Q1+Q2+Q3. Ifηis equivalent efficiency of the whole
system, then
η=
W
Q

23. 1.4. EFFICIENCY 23
Now,
η=
Q1×η1+Q2×η2+Q3×η3
Q1+Q2+Q3
(1.24)
In short form, fornsystems
η=
n
X
i=1
Qiηi
Qi
(1.25)
1.4.3 Efficiency in Series Network
Assume a series network, each has difference coefficient of efficiency as shown
in the figure below. EnergyQis supplied to the first system of network. The
output of this system is
Q1=Q×η1
This outputQ1is passed to next cascaded network of efficiencyη2. Now, the
output by this system is
Q2= [Q1]×η2
Or
Q2=Q×η1×η2
η1 η2 η3 η4
Q1 Q2 Q3Q W
Similarly the final outputWwill be
W=Q×η1×η2×. . .×η4
It gives
η1×η2×. . .×η4=
W
Q
η1 η2 η3 η4
η

24. 24 Energy & Work
If equivalent efficiency of the system isηthenη=W/Qand
η=η1×η2×. . .×η4 (1.26)
In short form, fornsystems
η=
n
Y
i=1
ηi (1.27)
1.4.4 Gain
Gain is a proportional value that shows the relationship between themag-
nitude of the input to the magnitude of the output signals. Gain differs
from the efficiency as efficiency is relation for input to output for mechanical
systems while gain is relation for input to output for the electrical system-
s/devices. In mechanical systems, source is defined for input only, while in
electrical systems, sources are separated for input and output.