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CHP # 6
WORK AND ENERGY
Q.1
Define work and its unit?
A.
WORK
It can be define as “when we applied a force on a body and the body covers a
distance in the direction of force, then we say that work is done.” It is a scalar
quantity. It is denoted by “W”.
It can also be defined as “the product of force and displacement in the direction
of force is called work.” Mathematically
Work  Force  displacement
 
W  F S
When there is an angle “  ” between the force and displacement, then
W  FS cos
The SI unit of work is joule.
Joule
It is defined as the work done is said to be one joule “if a force of one Newton is
applied to displace an object through one meter in the direction of force.”
Mathematically
1J  1N 1m
J  N m
Q.2
What is energy? Write its different types.
A.
Energy
The ability or capacity of a body to do work is called energy. It is a scalar
quantity. It is denoted by “E”. its SI unit is joule.
Some common types of energy are;
Mechanical energy
The energy due to which the different parts of the machines are
running is called mechanical energy.
Heat energy
The energy which flows from a hotter body towards a colder body
due to temperature difference is called heat energy.
Electrical energy
The energy associated with the charges in a conductor is called
electrical energy.
Chemical energy
The energy present in molecules or compounds to hold their bond
formation is called chemical energy. When the bond breaks this energy is
released in the form of heat energy.
(1)
(2)
(3)
(4)
(5)
Solar energy
The energy obtained from the sun is called solar energy. It is a big
source of energy.
Q.3
What is kinetic energy? Derive its mathematical expression.
A.
Kinetic energy
The energy posses by a body due to virtue of its motion are called
kinetic energy. It is denoted by “ K  E ”. Mathematically
K E 
1
mv 2
2
Proof
Consider a body of mass “m” is moving with an initial velocity of “v”
on a horizontal surface. It comes to rest due to the friction force “f” after
Covering a distance of “s”.
The work done against “F” on the object is given as;
W  F  S ----------- (1)
As we know from Newton’s second law,
F  ma
Also from third equation of motion,
2
2
2aS  v f  vi
By putting values we get
2(a) S  0 2  v 2
v2
2a
Now putt the value of “F” and “s” in equation (1)
v2
W  ma 
2a
1
W  mv 2
2
This work done is stored in the body as kinetic energy, so
1
K  E  mv 2
2
S 
Q.4
What is potential energy? Derive its mathematical expression.
A.
Potential energy
The energy posses by a body due to virtue of its position or
Configuration is called potential energy. It is denoted by “ P  E ”.
Mathematically
P  E  mgh
Proof
Consider a body of mass “m” is lying at a distance “h” from the
ground. The force acting on the body from the earth is;
F  w  mg
Let the body is lifted from height “h”. The work done by the force in lifting
this body through the height “h” is
W  F h
F  mg
But
W  mgh
This work done is stored in the body as potential energy, so
P  E  mgh
Q.5
State the law of conservation of energy with examples?
A.
Law of conservation of energy
This law state that “Energy can neither be created nor destroyed
in any process, but only can be change from one form to another form.”
For examples
(1) The chemical energy stored in the food is converted into heat energy as a
result of digestion in the body. This heat energy keeps our body warm and
enables us to do work.
(2) In dams the stored potential energy of water due to height is converted into
kinetic energy to run the turbine for producing electrical energy.
Q.6
A.
How electrical energy can be produced?
In the electric power station the energy of a fuel is used to boil water. The
light pressure steam is used to turn a turbine, which turns a dynamo to
generate electricity.
Q.7
Define power and its unit?
A.
Power
The time rate of doing work is called power. It is a scalar quantity. It
is denoted by “P”. Mathematically
work
power 
time
W
P 
t
E
P 
t
Unit of power
The SI unit of power is watt and can be define as “the power of the
body which does a work of one joule in one second time.” Mathematically
1J
1watt 
1 sec
Horse power is also a unit of the power which is stated as;
1hp  746watts
Conceptual Questions
Q.1
The mathematical relation for work and power are given by
W  FS
W
P 
t
using the above two relations derive a relation between power and force.
Solution
Given that
W
P 
---------- (1)
t
W  FS
By putting the value of “W” in equation (1) we get
FS
P 
t
As we know that
S
V 
t
So
P  F V
This is the required relation between power and force.
Q.2
A.
What type of energy conversion can be observed in water dams?
In dams the stored potential energy of water due to height is converted into
kinetic energy to run the turbine for producing electrical energy.
Q.3
What happen to the kinetic energy of a bullet when it penetrates into the
sand bags?
When the bullet hit the sand bags, its kinetic energy is used in doing work to
penetrate certain distance in sand bag before coming to rest.
A.
Q.4
A.
A meteor enters into earth’s atmosphere and burns. What happen to its
kinetic energy?
The meteor enters into earth’s atmosphere with very high velocity. it burns due
to the friction force of the air on it. As a result the kinetic energy of the meteor is
converted into heat energy.
Q.5
What do you understand by the term energy transformation support your
answer by giving two examples?
A.
Energy can be transferred from one form to another. For examples
(1) The chemical energy stored in the food is converted into heat energy as a
result of digestion in the body. This heat energy keeps our body warm and
enables us to do work.
(2) In dams the stored potential energy of water due to height is converted
into kinetic energy to run the turbine for producing electrical energy.
Q.6
A.
Why the efficiency of an engine can not be 100%?
Efficiency
It can be as “the ratio of the work done by the machine to the work done on the
machine is called efficiency. Or
The ratio of the output to the input of a machine is called efficiency. It is always
taken in percentage. Mathematically
output
Efficiency 
 100%
input
The efficiency of a machine can not be 100% because due to friction force
some of the work done on the machine is losses.
Q.7
A.
State the energy changes taking place in each of the following cases.
a) The bulb of a torch glows, when it is switched on
b) The burning of the fuels.
The energy changes taking place as;
a) When the torch is switched on, the chemical energy of the cell is
converted into electrical energy, then the bulb of the torch changes this
electrical energy into light and heat energy.
b) The fuels are basically coal, oil, gas or wood. These materials are made
from the atoms bonded together. When it is burnt the stored chemical
energy is converted into heat and light energy.
Q.8
A.
(a)
(b)
(c)
(d)
What kind of energy is possessed in the following situation?
a) A stone lying on the top of a roof. b) A flying aero plane.
c) A wound up spring of a toy car. d) A speeding train.
The energy possessed can be stated as;
The stone lying on the top of the roof has gravitation potential energy.
A flying aero plane has kinetic energy due to motion and potential energy
due to its position.
A wound up spring of a toy has elastic potential energy.
A speeding train has kinetic energy.
Q.9
In each of the following situation the system consist of a ball and earth.
Describe the work done and changes in energy forms.
a) You through a ball horizontally
b) The horizontally thrown by a fielder
c) A ball is thrown vertical and it comes to rest at top of its flight.
d) The ball falls back to earth, where you catch.
A.
(a)
The work done and energy changes can be stated as;
When we throw a ball horizontally, it possesses kinetic energy due to its
motion. This kinetic energy of the ball is used in doing work.
The horizontally thrown ball by the fielder possessed kinetic and gravitation
potential energy. The kinetic energy is used to do work against air resistance
while the gravitation potential energy remains constant.
When a ball is thrown vertically, work is done against gravity. This work done is
stored in the ball as gravitation potential energy.
When the ball falls back to earth the gravitation potential energy change to
kinetic energy to do work.
(b)
(c)
(d)
Q.10 State weather the following objects possess KE, PE or both.
a) A man climbing a hill
b) A stationary stone lying at the top of a hill.
c) A flying bird
A.
a) A man climbing a hill possessed both kinetic and potential energy.
b) A stationary stone lying at the top of a hill possess potential energy.
c) A flying bird possessed both kinetic and potential energy.
Q.11 Which would have a greater effect on the K.E of an object, doubling the
mass or doubling the velocity?
A.
Kinetic energy can be defined as;
1
K E 
mv 2
2
This equation shows that kinetic energy can be effect more by doubling the
velocity.
Numerical Questions
(1)
(i)
ii)
iii)
A 2Kg object is released from rest from a height of 10m above the ground.
Calculate
the initial potential energy at the moment of release
the kinetic energy at the moment it reaches 4m above the ground
the speed of the object just before impact with the ground
After the impact the object finally comes to rest. Explain what happens to the
lost kinetic energy. [Take g=10m/s2]
Given data
m  2 Kg
i)
h  10m
g  10m / s 2
PE  ?
P  E  mgh
P  E  2  10  10
P  E  200 j
ii) m  2 Kg
g  10m / s 2
S  10  4  6m
K E ?
vi  0m / s
Vf
2
 V 2  2 gS  2  10  6  120m / s
1
mV 2
2
1
K  E   2  120
2
K  E  120 j
K E 
iii) m  2 Kg
g  10m / s 2
S  10m
V 2  2 gS
V 2  2  10  10
V 2  200
V  14  1m / s
The kinetic energy is used in doing work to penetrate the object in ground.
(2)
(i)
In the following situations, calculate the power involved.
A force of 50N moves through a distance of 10m in 5s.
Given data
F  50 N
S  10m
t  5s
P ?
W
P 
t
FS
P 
t
50  10
P 
5
P  100watt
ii)
An object of mass 1Kg is lifted up vertically through 5m in 10s.
Given data
m  1Kg
h  5m
t  10s
P?
W
P
t
mgh
P
t
1  10  5
10
P  5watt
P
(3) A rubber ball of mass 0.12Kg is held at a height of 2.5m above the ground,
and then released.
(i)
Calculate the kinetic energy of the ball just before it hits the ground.
Given data
m  0.12Kg
h  2 .5 m
g  10m / s 2
K E ?
By law of conservation of energy we can write that;
K E
K E
K E
K E
(ii)
 PE
 mgh
 0.12  10  2.5
 3j
Calculate the velocity of the ball just before it hits the ground.
Given data
m  0.12Kg
S  2.5m
g  10m / s 2
V ?
V 2  2 gS
V  2 gS
V  2  10  2.5
(4)
V  50
V  7.1m / s
Two bodies of equal masses move with uniform velocity V and 3V respectively.
Find the ratio of their kinetic energies.
Given data
Mass of first body = m
Velocity of first body = V
1
K-E of first body = mV 2
2
Mass of second body = m
Velocity of second body = 3V
1
K-E of second body = m(3V ) 2
2
Ratios of the kinetic energies are
1
1
mV 2 : m(3V ) 2
2
2
1
: 9
(5)
A man whose mass is 75Kg walks up a flight of 12 steps each 20cm high
in 5seconds. Find the power he develops in watts.
Given data
m  75Kg
No of steps = n = 12
Height of each step = 20cm = 0.20m
Total height =h = 12  0.20m =2.4m
g  10m / s 2
t  5s
P?
W
P
t
mgh
P
t
75  10  2.4
P
5
P  360watts
(6)
A mason of 600N weight is climbing a ladder 10m high. Find his potential
energy at the middle of the ladder.
Given data
W  mg  600N
h  10m
PE  ?
Potential energy at the middle of the ladder can be stated as;
mgh
PE 
2
600  10
PE 
2
P  E  3000 j
(7)
A ball of weight 100N is moving on a frictionless surface with the velocity
of 10m/s. Compute its kinetic energy.
Given data
W  mg  100N
W
g
100
m
 10Kg
10
V  10m / s
K E ?
1
K  E  mV 2
2
1
K  E   10  (10) 2
2
K  E  5  100  500 j
m
(8)
A car of mass 800Kg accelerates uniformly from rest to a speed of 25m/s
in 10 seconds. What is its power?
Given data
m  800Kg
V  25m / s
t  10s
P?
Energy
P
time
K E
P
t
1  mV 2
P
2t
800  (25) 2
P
2  10
800  625
P
 25000watts  25kwatts
20
(9)
A girl does a work of 800j, in lifting a 40Kg bucket from a well. If the
acceleration due to gravity is 10m/s2. Calculate the depth of the well.
Given data
W  800 j
m  40Kg
g  10m / s 2
h?
W  mgh
W
h
mg
800
h
40  10
800
h
400
h  2m