COLLEGE OF TECHNOLOGICAL STUDIES
DEPARTMENT OF LABORATORY TECHNOLOGY
ASP 113 - GENERAL PHYSICS
(DIPLOMA)
LECTURE
PREPARED BY: PHYSICS STAFF
Date: June 2020
Student Name
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Table of Contents
CHAPTER 1 - UNITS AND DIMENSIONS ...................................................................... 1
1.1 Introduction ...................................................................................................................... 1
1.2 Derived Quantities ........................................................................................................... 1
1.3 Systems of Units .............................................................................................................. 1
1.4 Conversion of Units ......................................................................................................... 2
1.4.1 Conversion Method ................................................................................................... 3
1.5 Dimensions ...................................................................................................................... 4
CHAPTER 2 - KINEMATICS IN ONE DIMENSION ...................................................... 7
2.1 Average Velocity ............................................................................................................. 7
2.2 Instantaneous Velocity ..................................................................................................... 8
2.3 Acceleration ..................................................................................................................... 8
2.4 Motion with Constant Acceleration ................................................................................. 9
2.5 Falling Objects ............................................................................................................... 11
CHAPTER 3 - NEWTON'S LAWS OF MOTION ........................................................... 16
3.1 Force Concept ................................................................................................................ 16
3.2 Newton’s First Law of Motion ...................................................................................... 16
3.3 Mass Concept ................................................................................................................. 16
3.4 Newton’s Second Law of Motion .................................................................................. 17
3.5 Newton’s Third Law of Motion ..................................................................................... 17
3.6 Friction Force ................................................................................................................. 22
3.7 Applications of Newton’s Laws..................................................................................... 24
3.7.1 Atwood’s Machine .................................................................................................. 24
3.7.2 Two Boxes and a Pulley ......................................................................................... 25
CHAPTER 4 - WORK AND ENERGY ............................................................................ 28
4.1 Work .............................................................................................................................. 28
4.2 Kinetic Energy, and Work-Energy Theorem ................................................................. 29
4.3 Potential Energy ............................................................................................................. 31
4.4 Conservation of Energy ................................................................................................. 31
CHAPTER 5 - MECHANICAL PROPERTIES OF MATTER ........................................ 38
5.1 Density and Specific Gravity ......................................................................................... 38
5.2 Hooke’s Law .................................................................................................................. 40
5.3 Elastic Properties of Solids ............................................................................................ 41
5.3.1 Young’s modulus (Y): Elasticity in length ............................................................. 42
5.3.2 Shear Modulus (S): Elasticity of Shape .................................................................. 43
I
5.3.3 Bulk Modulus (B): Elasticity of Volume ................................................................ 45
CHAPTER 6 - THERMAL PROPERTIES OF MATTER ................................................. 50
6.1 Heat ................................................................................................................................ 50
6.2 Specific heat ................................................................................................................... 50
6.3 Heat transfer by conduction ........................................................................................... 52
6.4 Linear Thermal Expansion ............................................................................................. 54
6.5 Volume Expansion of Solids and Liquids ..................................................................... 55
CHAPTER 7 - HYDROSTATICS .................................................................................... 60
7.1 Pressure .......................................................................................................................... 60
7.2 Variation of Pressure with Depth ................................................................................... 61
7.3 Absolute and Gauge Pressure ........................................................................................ 62
7.4 Buoyant Forces and Archimedes’ Principle .................................................................. 63
7.5 Viscosity and Viscous Fluids ......................................................................................... 65
CHAPTER 8 - DIRECT CURRENT CIRCUITS .............................................................. 70
8.1 Introduction .................................................................................................................... 70
8.2 Electric Current .............................................................................................................. 70
8.3 Ohm’s Law..................................................................................................................... 71
8.4 Series and Parallel Circuits ............................................................................................ 72
8.4.1 Resistors in Series ................................................................................................... 72
8.4.2 Resistors in Parallel................................................................................................. 74
8.4.3 Current Divider ....................................................................................................... 75
Appendix A – Important Terminology Translations ......................................................... 78
Appendix B – Conversion Factors ..................................................................................... 78
Appendix C – Equation Sheet ............................................................................................ 79
Appendix D – Answer Key ................................................................................................ 80
II
CHAPTER 1 - UNITS AND DIMENSIONS
1.1 Introduction
In order to study the physical world, all the physical quantities should be measured and
expressed in numbers. The laws of physics are described in terms of a few fundamental quantities.
When we look out at the universe we see that it occupies space, within which we find matter, that
change with time. To measure the entire physical world we use the seven fundamental quantities
of length, mass, time, electric current, luminous intensity, amount of substance, and
temperature. These quantities are called the fundamental (basic) quantities. Other quantities are
called derived quantities.
1.2 Derived Quantities
Most of the quantities observed and used in Physics are derived, meaning that they can be
obtained from the fundamental quantities. Examples of these quantities are: speed (distance
travelled divided by time, that is length/time), volume (length times width times height, i.e.
length3), and density (mass divided by volume, i.e. mass/length3). All other quantities are derived
in the same way, in terms of the seven fundamental quantities mentioned earlier.
1.3 Systems of Units
The two major systems of units in use today are the SI system (International System) and
the British Engineering System. We will use the SI system (also known as the metric system)
because it is widespread in science and technology throughout the world. Table 1.1 lists the units
used in SI system for the fundamental quantities.
Table 1.1: SI Units
Fundamental Physical Quantity
International System (SI) Units
Length
Meter (m)
Mass
Kilogram (kg)
Time
Second (s)
Electric Current
Ampere (A)
Temperature
Kelvin
Luminous Intensity
Candela
1
Prefixes
However, quantities of the same nature (two distances, for example) can vary by a large
amount: the size of the atom and the size of a planet are difficult to express both in meters. Thus,
we can use prefixes in order to amplify or reduce unit. Since the metric system is a decimal system
(i.e. of base 10), larger or smaller values of any unit can be obtained by multiplying or dividing
the standard unit by factors of 10.
We know that (1 centimeter) is one hundredth of a meter, and that a kilometer is one
thousand meters. For example, the prefixes centi (which means 0.01) and kilo (which means
1000) are used to express a small and a large quantity. Similarly, to express the size of an atom
we will need a prefix of even smaller value (e.g. nano = 10-9). For large quantities, prefixes of
larger value are used (e.g. mega = 106 or giga = 109).
Table 1.2: Some prefixes for powers of ten
Prefix
Symbol
Value
femto
f
10-15
pico
p
10-12
nano
n
10-9
micro

10-6
milli
m
10-3
centi
c
10-2
deci
d
10-1
deka
da
10
hecto
h
102
kilo
k
103
mega
M
106
giga
G
109
tera
T
1012
1.4 Conversion of Units
Conversion is the method or operation that allows us to express the value of a quantity in
a different unit than the one it is given in. For example, we may wish to give the distance between
two cities in kilometers kilometres instead of meters or miles; we may want to express mass in
grams instead of kilograms; we could also convert a time interval from hours and minutes to
2
second; and so on.
In general, though not always, we will want to convert a given quantity's value to its
equivalent value in standard units, in order to use it in a formula or equation. For example, if a
speed in a problem is given in km/h, we will need to convert it to m/s before replacing it in our
equation, because that's the standard unit for velocities in physics. Likewise, a mass given in grams
will (generally) have to be converted to kilograms before being used in a calculation. (There are
exceptions to this general rule, we will encounter some of them in the problems of the following
chapters; we can simply say now that conversions will be needed unless two quantities of the same
units are being divided in an equation, in that case conversion is not necessary.)
The first we will have to always keep in mind is that conversion can only take place
between units of the same type. For example, we can convert between km/h and m/s, but not
between km/h and m; the same applies for the following example cases:
1. cm2 to m2 but not to m
2. g/cm3 to kg/m3 , but not to kg or kg/m
3. mm3 to m3 or cm3 , but not to m or m2 or km
1.4.1 Conversion Method
We are now going to explain a simple conversion method; this requires three operations:
1. Dividing and multiplying the initial unit by the final one;
2. Replacing prefixes (such as centi or kilo) by their numerical values using Table 1.2).
3. Cancelling any units left out from the divide-and-multiply operation and finally
multiplying all the numbers that one gets at the end.
This method has two great advantages: a) it works the same way in all cases, no matter
how complicated the units; b) the only thing that one needs to know is the value of the various
prefixes (or at least the main ones ) given in Table 1.2.
Example 1.1
Convert the distance: 30 km into m. The first step requires that we write:
30 km = 30
km
• 𝑚
𝑚
Then the second step requires us that we write: 30 km = 30
103 𝑚
𝑚
• 𝑚
In the final step, just cancel the units in the fraction (in this case: meter in the numerator and the
denominator) and multiply the numbers 30 and 103, to obtain the final result:
30 km = 30 (103 ) m = 30,000 m = 𝟑 × 𝟏𝟎𝟒 𝐦
3
Example 1.2
Convert 90 km/h into m/s.
We will now proceed with the two steps in sequence without stopping; just follow the following
operations:
km
3
3
km
h × m = 90 km ( s ) × (m) = 90 10 m s × m = 90 10 m = 𝟐𝟓 𝐦
90
= 90 m
h
s
h m
s
3600 s m s
3600 s
𝐬
s
Example 1.3
Now we'll convert units with power.
Say we have an area of 400 cm2 that we need in m2. We will proceed always in the same fashion;
in an intermediate step we must group the units in the fraction under the power before replacing
the value of the prefix (centi = 10-2). This is important; many students make mistakes at this level.
2
cm2
cm 2
10−2 m
400 cm = 400 2 × m2 = 400 ( ) × m2 = 400 (
) × m2
m
m
m
2
= 400(10−2 )2 × m2 = 400(10−4 ) × m2 = 𝟒 × 𝟏𝟎−𝟐 𝐦𝟐
Example 1.4
Convert 7.8 g/cm3 into kg/m3
g
g
3
kg
g m3
kg
g
m 3
kg
7.8 3 = 7.8 cm × 3 = 7.8 ( 3 ) × ( 3 ) = 7.8 3 ( ) × ( 3 )
kg
cm
m
kg cm
m
10 g cm
m
m3
3
g
m
kg
1
kg
𝐤𝐠
= 7.8 3 ( −2 ) × ( 3 ) = 7.8 3
( 3 ) = 𝟕𝟖𝟎𝟎 𝟑
−6
10 g 10 m
m
10 × 10
m
𝐦
1.5 Dimensions
Dimensions represent the types of units or base quantities being used. For example, the units
of an area are always a length squared (m2, cm2, etc...) whether it is for example a circle, rectangle,
or triangle. Likewise, velocity is always given as a distance over a time (m/s, km/h, etc...) whether
the velocity is constant or changing with time.
Dimensional analysis is a useful method for determining the units of a variable in an
equation. Also, with the use of dimensional analysis we can check the correctness of an
equation. Even a minor error in algebra can be detected because it will often result in an
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equation which is dimensionally incorrect.
The basic dimensions are mass (M), length (L), and time (T). As said before for the units,
the dimensions of derived quantities can be expressed in terms of basic quantities (length, time,
mass, etc...). Thus, we can say that area has a dimension of [L2], velocity has a dimension of
[L/T], the dimension of acceleration is [L/T2], and so on.
Dimensional analysis is useful in checking the correctness of a mathematical relationship. To
do so, two rules must be applied:
a) In any formula, we can only add or subtract quantities of the same dimensions;
b) In any formula, quantities on each side of the equation must have the same dimension.
Example 1.5:
Is the formula v = v0 + 1/2 at2 correct? Where v and v0 are velocities, a is the acceleration, and t
is time.
Applying the first rule, we check whether v0 and 1/2 at2 have the same dimension: v0 is
velocity, its dimension is [L/T], a is an acceleration, its dimension is [L/T2], and for t, it is [T].
Multiplying a and t2, the last term has a dimension of [L], so it is different from the dimension of
v0, and it cannot be added to it. The first rule is not satisfied, so the formula is incorrect (we
do not need to check the second rule).
Note that numbers, such as 1, 2, and  are dimensionless, and so we cannot check if the
numbers in a formula are correct or not. For example, a dimensional analysis of the formula V =
R3 gives a correct result from the point of view of dimensional analysis, even though there is no
object for which the volume is related to its radius in this way. So, our conclusion is that either
the formula is wrong or it is correct from the point of view of dimensional analysis.
5
PROBLEMS
1. Find the standard units of the following quantities:
a- area b- speed c- acceleration d- force e- energy
2. Convert the following quantities as specified:
a. 0.32 cm2
= _____________m2 = _____________mm2
b. 110 km/h
= _____________m/s
c. 13.6 g/cm3
= _____________kg/m3
d. 0.11 cal/cm.s.oC = _____________J/m.s.oC
(1 cal = 4.186 J)
3. Find the dimension of the following physical quantities
a- Force b- Pressure c- Density d- Energy
4. Apply the dimensional analysis method to check whether the following formulas
are correct:
a) t 2 − t1 = 2π√
L
g
b) x = vt 2 + 2at
1
c) x = v0 t + ( ) at 2
2
d) x = v0 t + 2at 2
Where x is the distance covered in time t, v and a represents the velocity and
acceleration respectively.
6
CHAPTER 2 - KINEMATICS IN ONE DIMENSION
Mechanics is the field concerned with the study of the motion of objects and the related
concepts such as force and energy. Mechanics is usually divided into two parts: dynamics, and
kinematics. Dynamics deals with the study of the motion of objects and the relation of this
motion to physical quantities such as force and mass. Kinematics is concerned with the
description of the motion using the concepts of space and time regardless of the causes of this
motion. In the following chapter, we will discuss the one-dimensional motion only.
2.1 Average Velocity
The most important aspect of the motion of an object is how fast it is moving. The
motion of the particle is completely known if its position in space is known at all times. To
discuss one-dimensional motion of an object in general, suppose that an object is moving along
x-axis from point A to point B as shown in Figure 2.1. Assume at some moment in time t1, the
object is at point A on the x-axis at position x1, and at some later time t2 suppose it is at point
B on the x axis at position x2.
x
B
x
x0
A
t
t0
t
Figure 2.1: Position time graph for a particle moving along x-axis
The displacement of the object is x = x - x0, and the time elapsed is t = t – t0. Then the
average velocity, v , is defined as the displacement divided by the elapsed time and can be
written as:
ν̄ =
x−x0
t−t0
=
Δx
(2.1)
Δt
Where v stands for velocity and the bar (-) over the v is a standard symbol meaning
“average”, and it has the standard unit of m/s. It should be noted that there is a distinction
between the two terms; velocity and speed. Velocity is used to identify both the magnitude
(numerical value) of how fast an object is moving and the direction in which it is moving.
7
Velocity is therefore a vector which has both magnitude and direction. On the other hand, speed
is used to identify only the magnitude of how fast an object is moving. There is a second
difference between velocity and speed: namely, the average velocity is defined in terms of
displacement, rather than total distance traveled.
Average speed and average velocity often have the same magnitude, but sometimes
they don’t. As an example, if a person walks 70 m towards the east then 30 m towards the west,
the total distance traveled is 70 m + 30 m = 100 m. But, the displacement is 40 m only. If this
walk took 70 s to complete, then the average speed is: distance / time = 100 m / 70 s = 1.4 m/s.
On the other hand, the magnitude of the average velocity is: displacement/ time = 40 m /70 s =
0.57 m/s.
Example 2.1
The position of a moving point as a function of time is plotted as moving along the x-axis of a
coordinate system. During a 4 s time interval, the point position changes from x 0 = 50 m to
x=30 m, as shown in the figure below. What is the point average velocity?
Solution
x
0
10
20
30
40
50
Average velocity is the displacement divided by the elapsed time:
v̅ =
x − x0 30 m − 50 m
−20 m
𝐦
=
=
= −𝟓
t − t0
4s
4s
𝐬
The displacement and the average velocity are negative, which tells us that the point is
moving to the left along the x-axis, as indicated by the arrow in the above figure. Thus we can
say that the point average velocity is 5 m/s to the left.
2.2 Instantaneous Velocity
Instantaneous velocity is defined as the velocity of a particle at any instant of time, or
the velocity of the particle on a position-time graph. Instantaneous velocity is important when
the average velocity in different time intervals is not constant.
2.3 Acceleration
An object whose velocity is changing with time is said to be accelerating. If a particle
undergoes this change in velocity in less time than another, it is said to have a greater
8
acceleration. That is, acceleration specifies how rapidly the velocity of an object is changing.
Suppose that a particle moving along x axis have the velocities v2 and v2 at times t2 and t1. The
average acceleration of the particle is defined as the velocity change v (v – v0) over the time
interval t (t2 - t1) and is written as:
ā =
ν−ν0
t−t0
=
Δv
(2.2)
Δt
The standard unit of acceleration is m/s2 and it is also a vector, but for one-dimensional
motion, we need only to use a plus or minus sign to indicate direction relative to a chosen
coordinate system. The instantaneous acceleration a can be defined in the same way as
previously stated for instantaneous velocity.
Example 2.2
A car accelerates along a straight road from rest to 75 km/h in 5 s. What is the magnitude of its
average acceleration?
Solution
The car starts from rest, so v0 = 0.
The final velocity is v = 72*1000/3600 = 20 m/s.
So, the average acceleration is:
m
m
v − v0 20 s − 0 s
𝐦
a̅ =
=
=𝟒 𝟐
t − t0
5s
𝐬
2.4 Motion with Constant Acceleration
Many practical situations occur in which the acceleration is constant. That is, the
acceleration doesn’t change over time. If the magnitude of the acceleration is constant and the
motion is in a straight line, the motion is said to be under uniform acceleration. In this situation,
the average acceleration is equal to the instantaneous acceleration, and a can be replaced by a.
To simplify the calculation, let us consider the initial time to be zero, i.e. t1 = 0, and let
t2 = t and let x0 and v0 represent the initial position x1 and the initial velocity v1 of the object.
At time t the position and velocity will be x and v (rather than x 2 and v2) and the average
velocity during the time t can be obtained using equation (2.2).
ν̄ =
x−xo
t−to
=
x−xo
(2.3)
t
So, the acceleration which is assumed to be constant in time can be calculated using
equation (2.2):
9
a=
v−νo
t−to
=
v−νo
(2.4)
t
Rearranging equation (2.4), we get:
v = νo + at
(2.5)
Next, let us see how to calculate the position of an object after a time t when it is
moving with constant acceleration. Solving equation (2.3) for x gives:
x = xo + ν̄ t
(2.6)
Because the velocity increases at a uniform rate, the average velocity, v, will be
midway between the initial and final velocities:
v̅ =
νo +ν
(2.7)
2
Combining the last two equations with Equation (2.5), we find:
x = xo + νo t +
1
2
at 2
(2.8)
We now can derive another equation which is useful in situations where the time t is
not known. Rearranging and manipulating the previous equations, it is easy to get the
following equation:
ν2 = ν2o + 2a (x-xo )
(2.9)
Equations 2.6, 2.7, 2.8 and 2.9 are the most useful equations relating positions,
velocity, acceleration, and time for motion with constant acceleration. In many cases we can
set x0 = 0, and this simplifies the above equations. Note that x represents position, not distance,
and x – x0 is the displacement.
Example 2.3
You are designing an airport for small planes. One kind of airplane that might use this airfield
must reach a speed of at least 27.8 m/s before take-off (100 km/h), and can accelerate at 2 m/s2.
(a) If the runway is 150 m long, can this airplane reach the proper speed to take off? (b) If not,
what minimum length must the runway have?
Solution
(a) The airplane’s acceleration is a = 2.00 m/s2, and we know the plane can travel a distance
of (x =150 m). We want to find its velocity at the end of the runway, to determine if it will be
at least 27.8 m/s. Using equation (2.9):
v 2 = v02 + 2a(x − x0 ) = 0 + (2) (2
m
m2
[150
)
m]
=
600
s2
s2
10
Therefore; v = √600
m2
s2
= 24.5
m
s
m
(< 27.8 s )
So, the runway’s length is not sufficient to reach the required velocity.
(b) Given v = 27.8 m/s and a = 2.0 m/s2, the minimum length is determined by calculating (xx0). In this case, equation (2.9) can be rewritten in the form:
2
x − x0 =
v −
2a
v02
m 2
m 2
(27.8 s ) − (0 s )
=
= 𝟏𝟗𝟑. 𝟐 𝐦
m
2 (2 2 )
s
Example 2.4
How long does it take a car to cross a 30 m wide interaction after the light turns green, if it
accelerates from rest at a constant acceleration of 2.00 m/s2?
Solution
Starting from rest means that v0=0, substituting in equation (2.8):
1
1
x = x0 + v0 t + at 2 = at 2 ,
2
2
∴t=√
2x
2(30 m)
=√
m = 𝟓. 𝟒𝟖 𝐬
a
2 2
s
Example 2.5
A boy on a skateboard is traveling up a hill. Given that his speed at the bottom of the hill is 12
m/s and he came to rest after traveling a distance of 24 m. find his constant acceleration.
Solution
vo = 12 m/s, v = 0, and assume xo = 0
m 2
m 2
(0 s ) − (12 s )
v 2 − v02
𝐦
a=
=
= −𝟑 𝟐
2(x − x0 )
2(24 m)
𝐬
Note that the boy is slowing down and his initial velocity is greater than his final velocity. So,
his acceleration in this case is negative. A negative acceleration is called deceleration.
2.5 Falling Objects
It was widely believed until the time of Galileo that heavier objects fall faster than
lighter objects and that the speed of fall is proportional to the weight of the object. For free
fall, Galileo’s postulated that all objects would fall with the same constant acceleration in the
absence of air resistance. He showed that this postulate predicts that for an object falling from
11
rest, the distance traveled will be proportional to the square of the time, i.e. d  t2. We can
conclude this relation from equation (2.8), but Galileo was the first to derive this mathematical
relation. In fact, among Galileo’s great contributions to science was to establish such
mathematical relations. Another great contribution of Galileo was to propose theories with
specific experimental consequences that could be quantitatively checked (such as d  t2).
Galileo also claimed that all objects, light or heavy, fall with the same acceleration, at
least in the absence of air. If you hold a piece of paper horizontally in one hand and a heavier
object as a baseball in the other, and release them at the same time, the heavier object will
reach the ground first. But if you repeat the experiment, this time crumpling the paper into a
small, you will find that the two objects will reach the floor at nearly the same time.
Galileo was sure that air acts as a resistance to very light objects that have a large
surface area. But in many ordinary circumstances this air resistance is negligible. In a chamber
from which the air has been removed, even light objects like a feather or a horizontally held
piece of paper will fall with the same acceleration as any other object. Such a demonstration
in vacuum was of course not possible in Galileo’s time, which makes Galileo’s achievement
greater. The acceleration of a falling object is called acceleration due to gravity on the Earth,
and it is assigned the symbol g, and its magnitude is approximately equal to 9.81 m/s2. In brief,
we can say that: At a given location on the Earth and in the absence of air resistance, all objects
fall with the same constant acceleration.
Example 2.6
Suppose that a ball is dropped from a tower 70 m high. How far will it have fallen after 1 s and
3 s? Assume y is positive downward. Neglect air resistance.
Solution
The acceleration, a = g = + 9.81 m/s2, which is positive because we have chosen downward as
positive. Since we want to find to distance fallen given the time, t, equation (2.8) can be
applied with v0 = 0 and y0 = 0. Then, after 1.00s, the position of the ball is:
y1 =
1 2 1
m
at = (9.81 2 ) (1 s)2 = 4.91 m
2
2
s
So the ball has fallen a distance of 4.91 m after 1 s. Similarly, after 3 s,
y3 =
1 2 1
m
at = (9.81 2 ) (3 s)2 = 44.145 m
2
2
s
12
Example 2.7
Suppose the ball in Example 2.6 is thrown downward with an initial velocity of 3.00 m/s,
instead of being dropped. (a) What then would be its position after 1s and 3 s? (b) What would
its speed be after 1 s and 3 s? Compare to the speeds of the dropped ball.
Solution
(a) We can approach this in the same way as Example 2.6, using equation (2.8), but this time
v0 is not zero but is v0 = 3 m/s. Thus, at t = 1 s, the position of the ball is:
1
m
1
m
y = v0 t + at 2 = (3 ) (1 s) + (9.81 2 ) (1 s)2 = 𝟕. 𝟗𝟏 𝐦
2
s
2
s
and at t = 3.00s,
1
m
1
m
y = v0 t + at 2 = (3 ) (3 s) + (9.81 2 ) (3 s)2 = 𝟓𝟑. 𝟏𝟓 𝐦
2
s
2
s
As expected, the ball falls farther each second than if it were dropped with vo = 0.
(b) The velocity is readily obtained from equation (2.5):
m
m
v = v0 + at = (3 ) + (9.81 2 ) (1 s) = 𝟏𝟐. 𝟖𝟏
s
s
m
m
v = v0 + at = (3 ) + (9.81 2 ) (3 s) = 𝟑𝟐. 𝟒𝟑
s
s
𝐦
𝐬
𝐦
𝐬
[at t = 1 s]
[at t = 3 s]
When the ball is dropped (v0 = 0), the first term in the above equations is zero, so,
m
m
m
v = v0 + at = (0 ) + (9.81 2 ) (1 s) = 9.81
[at t = 1 s]
s
s
s
m
m
m
v = v0 + at = (0 ) + (9.81 2 ) (3 s) = 29.43
[at t = 3 s]
s
s
s
We see that the speed of a dropped ball increases linearly in time. In Example 2.6, it is obvious
that the distance fallen increases as the square of the time. The downwardly thrown ball also
increases linearly in speed, but its speed at any moment is always 3.0 m/s higher than the falling
ball.
13
PROBLEMS
1. A car accelerates from 11 m/s to 24 m/s in 5.5 s. Find:
a) The acceleration
b) The distance traveled.
2. A car slows down from a speed of 20 m/s to rest in 5 s. How far did it travel in that
time?
3. A car slows down from 120 km/h to rest in a distance of 160 m. The car is
assumed to move with uniform acceleration. Calculate:
a) The uniform acceleration.
Rest
120 km/h
b) The time required.
160 m
4. An electron in a cathode ray tube accelerates uniformly from 3104 m/s to 6106 m/s
in a distance of 2 cm. (a) what is the acceleration of the electron in this region. (b) How
long is the electron in the region where it accelerates?
5. A stone is dropped from the top of a building. The stone hits the ground after 4
seconds. Find:
Stone
a) The height of the building (y).
b) The velocity of the stone just before hitting
the ground (v).
y

6. A ball is thrown vertically upward from the ground with an initial velocity of
15 m/s. Determine:
a) The time required to reach maximum height?
b) The maximum height?
ymax
c) The velocity of the ball after 1 second.
15 m/s
7. A ball thrown vertically upward is caught by the thrower after 3.2 seconds.
Find:
a) The initial velocity of the ball.
b) The maximum height it reaches.
14
MULTIPLE CHOICE QUESTIONS
1. 1 Newton is equal to:
a) kg.m/s
b) m/s2
c) kg.m/ s2
d) kg/m.s
2. The relation between velocity (v) and distance (x) for an object moving with constant
acceleration is given by:
a) v 2  v 02  ax
b) v 2  v 02  1 2 ax c) v 2  v 02  2ax
d) v 2  v 02  3ax
3. The relation between distance (x) and time (t) for an object moving with constant
acceleration is given by:
a) x  v 0t  1 2 at 2 b) x  v 0t  1 2 at c) x  v 0t  2at 2
d) x  v 0t 2  1 2 at
4. A car accelerates from rest (vo= 0) to (v= 36 km/h) in 5s. The car’s acceleration is:
a) 2 m/s2
b) 4 m/s2
c) -2 m/s2
d) -4 m/s2
5. A car slows down from (vo= 20 m/s) to (v= 10 m/s) in 5 seconds. The car acceleration
is:
a) 2 m/s2
b) 4 m/s2
c) -2 m/s2
d) -4 m/s2
6. If a ball is thrown upward, the velocity of the ball:
a) doubles
b) remains constant
c) decreases
d) increases
7. If a ball is thrown downward, the velocity of the ball:
a) doubles
b) remains constant
c) decreases
d) increases
8. A stone is dropped from the top of a building (vo= 0). It hits the ground after 2.0 s.
The height of the building is:
a) 9.8 m
b) 9.6 m
c) 4.9 m
d) 19.6 m
9. A ball is thrown vertically upward from the ground with an initial speed
(vo= 9.8 m/s). The maximum height of the ball is:
a) 9.8 m
b) 9.6 m
c) 4.9 m
d) 19.6 m
15
CHAPTER 3 - NEWTON'S LAWS OF MOTION
Chapter 2 explains how motion can be handled using velocity and acceleration.
In this chapter, we discuss why an object at rest begins to move as well as what causes
a body to accelerate or decelerate. Also, the relation between force and acceleration will
be presented.
3.1 Force Concept
The concept of force can be observed from everyday experiences. If one pushes or pulls
an object, he exerts a force on it. When a motor lifts an elevator, or a hammer hits a nail, a force
is being exerted. One may push very hard on a heavy table, and it may not move, so forces do
not always give rise to motion. When several forces act simultaneously on an object, the object
will accelerate only if the net force acting on it is not equal to zero. If the net force is zero, the
acceleration is zero and the velocity of the object remains constant. So, when the velocity of a
body is constant or if the body is at rest, it is said that the object is in equilibrium.
The fundamental forces in nature are: gravitational attraction between objects due to
their masses, electromagnetic forces between charges at rest or in motion and nuclear forces
between atomic particles. The force is a vector quantity which has direction and magnitude.
3.2 Newton’s First Law of Motion
Newton discovered that in the absence of external resistance, motions in a straight line
and at constant speed would continue. That is Newton's first law of motion:
An object at rest will remain at rest and an object in motion will continue in motion
with a constant velocity unless it experiences a net external or resultant force.
Simply, one can say that when the net force F on a body is zero, its acceleration (a) is
zero. The behavior of a body to maintain its state of rest or of uniform motion in a straight line
is called inertia. That is why Newton's first law is often called the law of inertia.
3.3 Mass Concept
The mass m is an inherent property of a body and it is a measure of the inertia of a
body. As the mass of the body increases, it is harder to change its state of motion. So, it is
harder to start it moving from rest, or to stop it when it is moving. In SI units, the unit of mass
16
is kilogram (kg) as mentioned in Chapter 1. It is important to distinguish between mass m and
weight mg. Mass is a property of a body and it is a measure of its inertia or quantity of matter.
Weight is the force of gravity acting on a body. So, if an object is taken to the Moon (gravity
on Moon is one sixth that one on Earth), the object weight is about one sixth as its weight on
Earth. However, the mass of the object is the same on both Moon and Earth.
3.4 Newton’s Second Law of Motion
Newton’s first law of motion states that if no net force is acting on a body, it remains
at rest, or if moving, it continues to move with constant speed in a straight line. If the net force
is not zero, Newton concluded that the velocity of the body will change. A change in speed or
velocity gives rise to acceleration. Let us assume that certain net force F is required to make
the block moves with certain acceleration a. So, when the applied force is doubled, the
acceleration doubles, and so on. From such observation, Newton concluded that the
acceleration of an object is directly proportional to the net force acting on it.
The acceleration of an object depends also on its mass m. In another way around, if you
apply a force F to a block, it will move with acceleration a. If the mass of the block is doubled,
the block will move with acceleration a/2 with the same applied force. Also, if the mass of the
block is tripled, the block will move with acceleration a/3 with the same applied force, and so
on. These observations are summarized in Newton's second law of motion, which states that:
The acceleration of an object is directly proportional to the net force acting on it
and is inversely proportional to its mass. The direction of the acceleration is in the
direction of the net force acting on the object.
In an equation form, Newton's second law of motion can be written as:
 F  ma
(3.1)
Where F stands for the net force, m for the mass, and a for the acceleration.
3.5 Newton’s Third Law of Motion
Newton's second law deals with a single object on which a force is exerted. The third
law of motion discovered by Newton explains what happens to the object that is exerting the
force. The third law can be summarized by stating that:
Whenever one object exerts a force on a second object, the second exerts an equal
and opposite force on the first.
17
This tells us that the object exerting the force feels a ``reaction'' force with the same
magnitude, but in the opposite direction from the force that it is exerting. As an example, we
will consider the following two cases to illustrate Newton's third law of motion. First, imagine
that you are pushing on a solid wall with a fairly large force, and assume you don't break
through. According to Newton's third law, the wall is exerting an equal and opposite force back
on you. How is this consistent with the second law, which says a force causes acceleration?
The answer lies in the fact that as you push on the wall, and the wall pushes back on you, the
friction between your shoes and the floor also exerts a force on you, which, if you are lucky, is
just enough to keep you from moving backwards away from the wall. If you push against a
wall while standing on ice, the frictional force is much less and you do in fact accelerate
backwards away from the wall.
Another example of the third law is a book lying on a table as shown in the figure below.
If the table were not there the book would fall under the force of gravity between the book and
the earth. The table is exerting a force on the book that is just enough to cancel the gravitational
force on the book. By Newton's third law, the book is also exerting an equal and opposite force
on the table. If the book were very heavy, its weight would cause the table to bend, or even
break. This is illustrated in the following figure.
N
Free-Body Diagram (FBD)
N
PHYSICS
PHYSICS
mg
mg
The forces acting on the book are: its weight mg (the force the earth exerts on the book)
and the upward reaction force N which the table exerts on the book. N is called the normal
force.
Example 3.1
A box of mass 5.0 kg is pulled vertically upwards by a force of 59 N applied to a rope
attached to the box. Find:
Direction of Motion
i.
The acceleration of the box.
ii.
The vertical velocity of the box after 4 seconds.
5 kg
18
a
Solution
Direction of Motion
T = 59 N
From Newton’s second Law:
i.
ΣF = ma
5 kg
a
T − mg = ma
m
T − mg 59 N − [(5 kg) (9.81 s2 )]
𝒎
a=
=
=𝟐 𝟐
m
5 kg
𝒔
ii.
mg
Since a is constant
v = v0 + at
m
m
𝐦
v = (0 ) + [(2 2 ) (4 s)] = 𝟖
s
s
𝐬
Example 3.2
What are the forces acting on a box if you push down
F
N
on it with a force F while it sits on a smooth,
horizontal surface as shown in the figure?
mg
Solution
Draw all the forces on the box, this is called a "free body diagram". The forces, for this case,
will be the "normal" force (the perpendicular force) exerted by the surface and is labeled N in
the diagram and the force F exerted by the hand. Gravity exerts a force downward, called the
weight and labeled mg. Just as in the previous example, we can write
ΣF = m a
Nothing happens in the horizontal direction. All of the forces have only vertical components,
taking up as positive, we have:
ΣFy = N - mg - F
Since the box sits on a surface and it does not accelerate, so a = 0. This means:
N - mg - F = 0
N = mg + F
Example 3.3
Two blocks of masses m1 and m2 are placed in
contact with each other on a smooth, horizontal plane
as shown. A constant horizontal force F is applied to
m1. What is the acceleration of each mass?
19
F
m1
m2
Solution
Simply, one can say that a force F is applied to an object whose mass is m = m1 + m2. So its
acceleration must be:
a=
F
(m1 + m2 )
Also, if we look at all the forces on m1 and make a free-body diagram of the forces acting on
m1. Of course the external force F pushes to the right on the mass m1. Gravity pulls down with
force m1g and the plane responds with a normal force N1. But the other mass, m2 exerts a force
on mass m1. This force is labeled P and points to the left. We can apply Newton's Second Law
to the y-component forces and find that N1 = mg1. But now there is an additional and unknown
force in the x-component of Newton's Second Law.
ΣF = m a
N1
F - P = m1 a
F
P
m1
m1 g
We need more information so we turn to the other mass, m2.
The first mass m1 exerts a force P on the mass,
m2. Applying the y-component of  F  ma ,
N2
P’
we find: N2 = m2g
m2
m2g
For the x-components, the only force acting on m2 is P’ so,
 F = P’ = m2 a
However, from Newton's third law, F12 = - F21, we know that P = P' so
F - P = m1 a
F - m2 a = m1 a
F = m1 a + m2 a
F = (m1 + m2) a
a=
F
(m1 + m2 )
That is the same answer we found so quickly earlier but this may provide a pattern to use for
more complex situations.
20
Example 3.4
A person weighs a fish using a spring scale attached to the ceiling of
an elevator as shown in the figure below. Show that if the elevator
accelerate or decelerate upwards or downwards, the spring scale reads
a weight different from the true weight of the fish.
Solution
The forces acting on the fish are shown in the free-body diagram. T is
the tension supplied by the scale; this is the value the scale reads. We
may call it the apparent weight of the fish. The net force on the fish is:
ΣF = m a
The net force is equal to the mass times the acceleration. This fish is
moving along with the elevator. In this diagram we have taken the
acceleration to be up so it is positive.
∑ F = ma = T − mg
T = mg + ma
T = m (g + a), i.e. T >mg (upward)
While the elevator accelerates upward, the apparent weight of the fish is greater than its
true weight, mg.
What happens as the elevator accelerates downward?
21
Solution
The forces on the fish are again shown in the free-body diagram,
ΣF = m a
The net force is equal to the mass times the acceleration. This fish is moving along with the
elevator. Now the acceleration is up so it is negative.
∑ F = ma = mg − T
T=mg-ma
T = m (g - a), i.e. T < mg (downward)
While the elevator accelerates downward, the apparent weight of the fish is less than its
true weight, mg.
3.6 Friction Force
If a body moves on a rough surface, there is a resistance to motion. This resistance to
motion is called force of friction, f. Experiments show that this friction force arises from the
roughness of the two surfaces. If a horizontal force F is applied to the right on a block on a
horizontal table as shown in figure. The block remains constant if F is not large enough to
move it. The force responsible of keeping the block from motion acts to the left and is called
force of static friction, fs.
N
f
m
F
mg
Increasing the value of F, the block starts to move with acceleration to the right. The
resistance force in this case is called the force of kinetic friction, fk. If F= fk , the block moves
to the right with constant speed. Both fs and fk are proportional to the normal force acting on
the block.
So, the force of static friction between any two surfaces in contact is opposite to the
applied force and can be expressed as:
fs  s N
(3.2)
Where the dimensionless constant, s is the coefficient of static friction and N is the normal
force.
Also, the force of kinetic friction acting on an object is opposite to the direction of motion of
the object and is given by:
22
fk  k N
(3.3)
The dimensionless constant, k is the coefficient of kinetic friction.
Example 3.5
A ball of mass 0.6 kg traveling on a certain surface at a speed of 9 m/s slows down to 3.0 m/s
over a distance of 20 m. Find:
i.
The frictional force acting on the ball.
ii.
The coefficient of kinetic friction between the ball and the surface.
N
fk
Direction of motion
mg
Solution
We can use the following kinematic equation since the acceleration is constant:
v 2 = v02 + 2ax
m 2
m 2
v 2 − v02 (3 s ) − (9 s )
𝐦
∴a=
=
= −𝟏. 𝟖 𝟐
2x
2(20 m)
𝐬
The negative sign means that the object is decelerating.
Applying Newton’s second law to the ball in x-direction:
ΣF = ma
−𝑓𝑘 = ma
m
i.
−𝑓𝑘 = (−0.6 kg) (−1.8 s2 ) = 𝟏. 𝟎𝟖 𝐍
ii.
Applying Newton’s second law to the ball in the y–direction:
N − mg = 0
N = mg = (0.6 kg) (9.81
𝑓𝑘 = 𝜇𝑘 𝑁,
∴ 𝜇𝑘 =
fk
N
𝑚
) = 5.886 𝑁
𝑠2
1.08 N
= 5.886 N = 𝟎. 𝟏𝟖𝟒
23
3.7 Applications of Newton’s Laws
3.7.1 Atwood’s Machine
Atwood’s Machine is an arrangement of two unequal masses hung vertically by a thin cord
over a light, frictionless pulley as shown in the figure below. Atwood’s Machine is sometimes
used in laboratory to measure the acceleration due to gravity. From the figure shown below,
we can determine the acceleration due to gravity, and the tension in the cord.
Free-Body Diagram for m1 & m2
T
a
a
m1
T
m2
m1
a
a
m2
m1 g
m2 g
Solution
The free-body diagrams for the two masses are shown. Assuming m1 is greater than m2 we can
determine the acceleration and the tension in the cord. Applying Newton's second law to m1:
ΣF = m a
T − m1 g = m1 a
(1)
Similarly for m2:
m2 g − T = m2 a
(2)
Adding equations (1) and (2), we get:
T − m1 g + m2 g − T = m1 a + m2 a
m2 g − m1 g = (m1 + m2 ) a
(3)
Solving for the acceleration (a), we get:
𝐦 −𝐦
𝐚 = (𝐦𝟐 +𝐦𝟏 ) 𝐠
𝟏
(4)
𝟐
Substitute equation (4) into (1), the tension in the cord can be determined:
m2 − m1
T − m1 g = m1 [(
) g]
m1 + m2
m2 − m1
T = m1 [(
) g] + m1 g
m1 + m2
m1 m2 −m21
T=(
m1 +m2
m1 m2 −m21 +[(m1 +m2 )(m1 )]
) g + m1 g= (
m1 +m2
)g
m1 m2 − m12 + m1 m2 + m12
𝟐𝐦𝟏 𝐦𝟐
T=(
)g = (
)𝐠
m1 + m2
𝐦𝟏 + 𝐦𝟐
24
3.7.2 Two Boxes and a Pulley
Consider two boxes of masses m1 and m2 that are connected by a thin cord running over a
pulley on a table as shown in the figure below. The coefficient of kinetic friction between box
1 and the table is k. Ignore the mass of the cord and the pulley and any friction in the pulley.
i.
Determine the group acceleration (a)
ii.
a
m1
Determine the tension in the cord (T)
m2
Solution
a
First let us draw the free-body diagram for each box.
N
fk
T
a
m1
m2
T
a
m2g
m1g
N = mg
fk = μk N = μk m1 g
(1)
Applying Newton's second law to m1:
T − fk = m1 a
(2)
Substituting from (1) in (2) gives:
T − μk m1 g = m1 a
(3)
Applying Newton's second law to m2:
m2 g − T = m2 a
(4)
Adding equations (3) and (4), we get:
T − μk m1 g + m2 g − T = m1 a + m2 a
m2 g − μk m1 g = (m1 + m2 )a
∴𝐚=
𝐦𝟐 𝐠−𝛍𝐤 𝐦𝟏 𝐠
(5)
𝐦𝟏 +𝐦𝟐
Substituting from (5) in (3), one can get:
 m g   k m1g 

T   k m1g  m1  2
m

m


1
2
So, the tension in the cord T is given by:
𝐦𝟐 𝐠 − 𝛍𝐤 𝐦𝟏 𝐠
𝐓 = 𝐦𝟏 (
) + 𝛍𝐤 𝐦𝟏 𝐠
𝐦𝟏 + 𝐦𝟐
25
PROBLEMS
1. A 4 kg particle starts from rest and moves a distance of 5 m in 3s under the action of
a constant force. Find the magnitude of that force.
2. What is the net force needed to
accelerate a 800 g object from rest to 16
m/s over a distance of 60 cm?
F=?
16 m/s
800 g
800 g
60 cm
3. How much tension must a rope withstand if it is used to accelerate a 2.500 kg body
vertically upward at 1.2 m/s2? Ignore friction.
4. The two masses shown in figure are each initially 1.8 above the ground, and the
massless frictionless pulley is 4.8 m above the ground. Determine:
a. The acceleration of the smaller mass.
b. The velocity of the smaller mass at the moment the larger one hits the ground.
c. The maximum height the smaller mass reaches.
2.2 kg
3.2 kg
4.8 m
1.8 m
5. Find the acceleration and the tension in the cord of an Atwood’s Machine when m1
= 3 kg and m2 = 5 kg.
6. A 4 kg block at rest lies on a horizontal surface (k = 0.50). This block is
connected by a thin string that passes over a pulley to a second block of mass 6
kg. Find:
v
a. The group acceleration.
4 kg
4 kg
b. The tension in the cord.
c. The velocity of the 4 kg block after 2 seconds.
x
d. The distance traveled by the 4 kg block after 2
seconds.
26
6 kg
MULTIPLE CHOICE QUESTIONS
1. A certain force produces an acceleration (a = 2 m/s2) when applied to an object of
mass m1. The same force produces an acceleration (a = 8 m/s2) when applied to another
object of mass m2. The ratio m1/m2 is:
a) 16
b) 4
c) 0.25
d) 0.5
2. A 1.0 kg mass is hung from a spring scale attached to the ceiling of an elevator. The
scale reads 10.8 N. The direction of the elevator is:
a) downward
b) upward
c) no motion
3. A 1.0 kg mass is hung from a spring scale attached to the ceiling of an elevator. The
scale reads 7.8 N. The direction of the elevator is:
a) downward
b) upward
c) no motion
4. A 1.0 kg mass is hung from a spring scale attached to the ceiling of an elevator. The
scale reads 9.81 N. The direction of the elevator is:
a) downward
b) upward
c) no motion
5. The range of values for the coefficient of friction () is:
a)  > 0
b)  < 0
c)  = 1
d) 0    1
6. If m1 = m2 in Atwood’s machine, the acceleration (a) will be:
a) a = g
b) a > g
c) a < g
27
d) a = 0
CHAPTER 4 - WORK AND ENERGY
In this chapter, we shall be concerned only with the mechanical form of energy. The
concepts of work and energy can be applied to the dynamic of a mechanical system without
resorting to Newton's laws. However, it is important to note that the work-energy concepts are
based on Newton's laws and therefore do not involve any new physical principles. In its
simplest form, energy can be defined as the ability of a body or system of bodies to perform
work. To describe the energy of a body or a system, we must first define the concept of work.
4.1 Work
Whenever a force is applied to an object, causing the object to move, work is done by
the force. If a force is applied but the object doesn't move, no work is done. When force is
applied and the object moves a distance x in a direction other than the direction of the force,
less work is done than if the object moves a distance x in the direction of the applied force.
Consider an object that undergoes a displacement x along a straight line under the action of a
constant force F, which makes an angle  with x as shown in Figure 4.1.
F
F
θ
θ
Figure 4.1: Work done by a constant force
The work done by the constant force is defined as the product of the component of the
force in the direction of the displacement and the magnitude of the displacement. Since the
component of F in the direction of x is F.cos, the work done W by the force F is given by:
W = F . cos () x
(4.1)
It should be noted that when the object does not move (x=0), the force does no work on
the object. For example, if a person pushes hard against a brick wall, a force is exerted on the
wall but the person does no work unless he actually moves the wall in the direction he is
pushing. Also, the work done by a force is zero when the force is perpendicular to the
displacement, since  = 90° and cos90° = 0. Finally, if the applied force F acts along the
direction of the displacement then  = 0 and cos () = 1. In this case, the work done by the
force has its maximum value and is given by:
W = F. x
(4.2)
28
Work is a scalar quantity and its unit can be derived from its definition; it is the unit of
force times the unit of distance, or Newton-meters (N.m). This combination of units is given
the name of Joules (J). Work can be either positive or negative: if the force has a component in
the same direction as the displacement of the object, the force is doing positive work. If the
force has a component in the direction opposite to the displacement, the force does negative
work. As an example, suppose you are trying to stop a car that is rolling towards you by pushing
against it. Clearly, if you push hard enough, you will succeed in slowing the car down, but in
this process you will move backwards as the car rolls towards you. In this case the force you
exert is in the opposite direction of the motion, and is negative. By Newton's third law, the car
is exerting an equal and opposite force on you, and is consequently doing positive work on
you. So when you do negative work on an object, that object in reality is doing positive work
on you.
Example 4.1
A force of 15 N is acting at an angle of 25 to the horizontal is used to pull a box a distance of
5 m across a floor. How much work is done?
Solution
The work done is found by using equation (4.1):
W = F cos x= (l5.0 N)(cos 25˚) (5 m) = 68 N.m = 68 J
Example 4.2
A force of 867 N is applied to a car to bring it to rest in a distance of 350 m. How much work
is done in stopping the car?
Solution
W = F cos x = (867 N)( cos 180˚)(350 m) = - 3.035×105 J
Note that, cos (180˚) = -1, and hence, the work done is negative.
4.2 Kinetic Energy, and Work-Energy Theorem
Energy is the ability to do work and it is a scalar quantity and has the same units as
work (i.e. Joule, J). There are various types of energy, such as kinetic energy, gravitational
potential energy, heat energy, electrical energy and chemical energy. The kinetic energy of a
body is the energy that a body possesses by its motion. The kinetic energy (KE) of an object
of mass m that is moving with velocity v is given by:
29
1
KE = 2 mν2
(4.3)
The work done on an object by a net force equals the change in the kinetic energy of
the object. This relationship is called the work-energy theorem. To prove this theorem,
assume a constant force F acts on a body of mass m causing it to be displaced a distance x in
the same direction. The work done W by the force is given by:
W= Fx
(4.4)
From Newton’s second law, F = ma. So equation (4.4) becomes:
W= (ma) x
(4.5)
From the third equation of motion:
 2   o2
a
2x
(4.6)
Substituting from equation (4.6) into equation (4.5) gives:
1
W = m(ν2 − ν2o )
2
W = KE − KEo
(4.7)
Thus, the work required to cause an object of mass, m, to move a distance x is equal to the
change in its kinetic energy.
Example 4.3
If a car doubles its speed, what happens to its kinetic energy?
Solution
KE0 =
1
1
1
1
mv02 = m(2v0 )2 = m[4v02 ] = 4 [ mv02 ] = 𝟒 𝐊𝐄𝟎
2
2
2
2
The Kinetic Energy increases by a factor of 4.
Example 4.4
How much work is required to accelerate a 1200 kg car from 10 m/s to 30 m/s?
Solution
1
1
1
1
W = KE − KEo = mν2 − mν2o = (1200kg)(30m/s)2 − (1200kg)(10m/s)2
2
2
2
2
W = 𝟒. 𝟖 × 𝟏𝟎𝟓 𝐉
30
4.3 Potential Energy
We have illustrated that an object is said to have energy by virtue of its motion, which
we call kinetic energy. The energy associated with force that depends on the position or
configuration of a body is called potential energy PE. The most common example of potential
energy is gravitational potential energy. Gravitational potential energy is defined as the energy
that a body possesses by virtue of its position. If a block of mass m is lifted to height h above
some reference level such as the ground, then that block would have potential energy in that
raised position. The potential energy of the block is equal to the work done to put the block
into that particular position. Thus the gravitational potential energy PE is expressed as:
PE = W=Fh cos0= mgh
(4.8)
4.4 Conservation of Energy
There are many forms of energy: mechanical, chemical, electrostatic, heat, nuclear. In
any isolated system, energy can be transformed from one kind to another, but the total amount
of energy is constant (conserved). As an example a battery contains chemical energy and can
be used to produce mechanical energy. Also, kinetic friction generally transforms kinetic
energy into thermal energy. The law of conservation of energy states that energy cannot be
created or destroyed; it can merely be changed from one form of energy to another. When we
say that something is conserved, we mean that this quantity is constant and does not change
with time.
The work a conservative force does on an object in moving it from a certain point A to
another point B is path independent - it depends only on the end points of the motion. Thus the
force of gravity and the spring force are conservative forces. For a non-conservative (or
dissipative) force, the work done in going from A to B depends on the path taken as friction
and air resistance forces. For conservative forces, the total energy of the body remains constant
during the motion. Kinetic friction, on the other hand, is a non-conservative force, because it
reduces the mechanical energy in a system. Note that non-conservative forces do not always
reduce the mechanical energy; a non-conservative force changes the mechanical energy, so a
force that increases the total mechanical energy, like the force provided by a motor or engine,
is also a non-conservative force.
The analysis of a system whose energy is conserved leads us to the law of conservation
of energy. In any closed system, that is an isolated system, the total energy of the system
remains constant. The law of conservation of energy applies to mechanical energy. Mechanical
31
energy is the sum of the potential and kinetic energies in a system. The principle of the
conservation of mechanical energy states that the total mechanical energy in a system (i.e., the
sum of the potential energy and the kinetic energy) remains constant as long as the only forces
acting are conservative forces.
Let us illustrate the law of conservation of energy with the following two examples. For
an object falling, on its way down the object loses height and gains speed; it thus loses
gravitational potential energy and gains kinetic energy. So, the gravitational potential energy
is transformed to kinetic energy as the object falls. On the other hand, if an object is thrown
upwards, it gains height and loses speed, thus increasing its gravitational potential energy and
decreasing its kinetic energy. In this case it is the kinetic energy which is being transformed
into gravitational potential energy. However, in either case, at any given point in time one finds
that the sum of the kinetic and the gravitational potential energy remains the same.
The second example is the roller coaster car which operates on the same principle of
energy conservation. Work is initially done on a roller coaster car to lift it to its initial position.
Once lifted to the top, the roller coaster car has a high potential energy and virtually no kinetic
energy (the car is almost at rest). Assuming that no external forces are doing work upon the car
as it travels from the initial point to the end of the track, then the total mechanical energy of
the car is conserved. As the car descends hills as shown in figure, its potential energy is
transformed into kinetic energy as the car speeds up. As the car climbs up hills and loops, its
kinetic energy is transformed into potential energy as the car slows down. In the absence of
external forces doing work, the total mechanical energy of the car is conserved. Conservation
of energy on a roller coaster car means that the total amount of mechanical energy is the same
at every location along the track.
Figure 4.2: Conservation of energy in a roller coaster car
It should be noted that for the roller coaster car, it is assumed that there were no external
forces doing work, which is not true. The roller coaster car experiences the force of friction
and the force of air resistance during its motion. Friction and air resistance are both external
forces and would do work on moving objects. The presence of friction and air resistance would
32
do negative work and cause the total mechanical energy to decrease during the motion. So,
assuming that mechanical energy is conserved is a useful approximation which assists in the
analysis of other complex motions.
Example 4.5
A 250 grams ball is thrown vertically upward with an initial velocity of 20 m/s. Find:
(a) The total energy of the ball (ETotal)
(b) The maximum height of the ball (hC)
(c) The velocity of the ball at a height of 15 meters (vB)
Solution
This problem can be solved using kinematics equations. To illustrate the principle of energy
conservation, we will solve it using that principle. The total energy (ETotal = kinetic energy +
potential energy) of the ball is constant at any position.
VC = 0 m/s, KEC = 0
C
c) vB =?
B
b) hC = ?
vA = 20 m/s,
PEA = 0
hB = 15 m
A
1
1
m 2
(a) ETotal = ETotal,A = KEA + PEA = 2 mvA2 + 0 = 2 (0.25kg) (20 s ) = 𝟓𝟎 𝐉
(b) At maximum height hC, the ball’s velocity C = 0, applying the conservation law of
energy at position C:
ETotal,A = ETotal,C = KEC + PEC =
50 J = 0 + (0.25kg) (9.81
hC =
50 J
(0.25 kg) (9.81
1
mv 2 + mghC
2 C
m
)h
s2 C
m = 𝟐𝟎. 𝟑𝟗 𝐦
)
s2
(c) Applying the conservation law of energy at position B:
1
ETotal,A = ETotal,B = KEB + PEB = mvB2 + mghB = 50 J
2
33
1
m
(0.25 kg)(vB2 ) + (0.25 kg) (9.81 2 ) (15 m) = 50 J
2
s
m
) (15 m)
𝐦
s2
= 𝟏𝟎. 𝟐𝟖
1
𝐬
(0.25 kg)
2
50J − (0.25 kg) (9.81
∴ vB = √
Example 4.6
A roller coaster starts from rest at the top of a 30 meter hill. Determine:
a) The speed of the roller coaster car at the bottom of the hill (vB).
b) The height (hC) when the roller coaster speed is 8 m/s.
A
vc=8 m/s
30 m
vB=?
B
C
hC=?
Solution
a) Applying conservation law of energy between points 1 and 2:
ETotal, A = ETotal, B = constant
1
1
mvA2 + mghA = mvB2 + mghB
2
2
m
1
0 + m (9.81 2 ) (30 m) = mvB2 + 0
s
2
m
) (30 m)
𝐦
s2
= 𝟐𝟒. 𝟐𝟔
1
𝐬
2m
m (9.81
vB = √
b) Applying conservation law of energy between points A and C we have:
1
1
mvA2 + mghA = mvC2 + mghC
2
2
m
1
m 2
m
0 + m (9.81 2 ) (30 m) = m (8 ) + m (9.81 2 ) hC
s
2
s
s
m
1
m 2 m [(9.81 m ) (30 m) − 1 (8 m)2 ]
(30
m (9.81 2 )
m) − 2 m (8 s )
2
s
s2
s
hC =
=
m
m
m (9.81 2 )
m (9.81 2 )
s
s
hC = 𝟐𝟔. 𝟕𝟑 𝐦
34
4.5 Power
The work done on an object is independent of the time taken to travel the distance
covered. For example, suppose one lifts a 200 N weight off the floor a distance of 2 m. The
work done in this case would be the product of the force exerted (the 200 N) times the distance
covered (the 2 m), independent of how fast the weight was lifted. This is somewhat against our
everyday idea of what work is. One might say that one does more “work” if one lifts the weight
faster. However, it is true that the same amount of work is done independently of the time
taken; rather, it is the power P in this case that is different. The power exerted by a force is
defined as the work done over time, i.e. the rate at which work is done. Therefore power is:
P=
W
t
=
Fx
t
= Fν
(4.9)
So, exerting a force at a larger velocity results in a greater power output. The units of
power can be derived from its definition; they are the units of work (Joules) divided by the
units of time (seconds). This combination of units (J/s) is called a Watt (W).
Example 4.7
A force of 6000 N keeps a car moving at 72 km/h. How much power is needed by the car?
Solution
P=
W Fx
1000 m
=
= Fv = (6000 N) (72 ×
) = 𝟏. 𝟐 × 𝟏𝟎𝟓 𝐖
t
t
3600 s
Example 4.8
A person pulls a block with a force of 15 N at an angle of 35 with the horizontal. How much
power is exerted if the block is moved 5 m in the horizontal direction in 5 seconds?
Solution
W F cosθ x (15 N)(cos 35° )(5 m)
P= =
=
= 𝟏𝟐. 𝟐𝟗 𝐖
t
t
5s
Example 4.9
An 80 kg person climbs a set of stairs in 20 seconds. The vertical height of the stairs is 3 meters.
Find the person’s output power.
Solution
m
W mgy (80 kg) (9.81 s 2 ) (3 m)
P= =
=
= 𝟏𝟏𝟕. 𝟕𝟐 𝐖
t
t
20 s
35
PROBLEMS
1. A force of 80.0 N is used to pull a box through a distance of 60 m. If the force makes
an angle of 40° with the horizontal, how much work is done?
2. How much work is required to accelerate a 1000 kg car from 20 m/s to 45 m/s?
3. An object of mass 2 kg traveling at 5.0 m/s enters a region of ice where the coefficient
of kinetic friction is 0.35. Use the work energy theorem to find the distance the object
travels before coming to rest.
4. A person pushes a 40 kg crate at a constant speed of 1.40 m/s across a horizontal
floor (k = 0.25) for 5 s. Determine: (a) the work done on the crate
(b) the energy dissipated by the frictional force?
5. A 2000 kg car is at rest on a frictionless track. A constant force acts on it for 3 s, after
which the car is moving with a speed of 0.20 m/s. Find: (a) the magnitude of the force,
(b) the kinetic energy of the car, (c) the work done on the car, and (d) the displacement
of the car.
6. A 0.4 kg ball is thrown vertically upward with a velocity of 30 m/s. Find:
a) The total energy of the ball.
b) The maximum height of the ball (hc).
C
c) The velocity of the ball at a height of 16 m.
Note: Solve the problem first using conservation
law of mechanical energy and then using the
kinematics equations and compare the results
hC
B
vA = 30 m/s
m = 0.4 kg
hB = 16 m
A
7. The roller coaster shown in the figure below is pulled up to point A where it is
released from rest. Assuming no friction, calculate:
A
a) The car speed at points B and C
b) The height at which the car speed
is 12.5 m/s
32 m
C
B
6m
8. A certain automobile engine delivers 2.24 x 104 W to its wheels when moving at a
constant speed of 27 m/s. What is the resistive force acting on the automobile?
36
MULTIPLE CHOICE QUESTIONS
1. The potential energy (PE) of an object thrown vertically upward is:
a) increasing
b) constant
c) decreasing
d) minimum at the end
2. The total energy of an object thrown vertically up or downward is:
a) increasing
b) constant
c) decreasing
d) maximum at the end
3. The kinetic energy (KE) of an object thrown vertically upward is:
a) increasing
b) constant
c) decreasing
d) maximum at the end
4. Conservation law of mechanical energy states that:
a) PE = KE
b) PE > KE
c) PE < KE
d) PE + KE = constant
5. If a car’s velocity (v) is doubled (2v), then its kinetic energy (KE) will:
a) decrease by a factor of 2
b) increase by a factor of 4
c) remain constant
d) decrease by a factor of 4
6. The power (P) needed to lift a 150 kg load for 10 m in 20 s is:
a) 294000 W
b) 735 W
c) 73.5 W
d) 75 W
7. A pump is used to lift 10 kg of load per minute through a height of 4 m. The output
power (P) of the pump motor is:
a) 6.54 W
b) 65.33 W
c) 392 W
d) 39.2 W
8. The time (t) required for a 1225 W motor to lift a 25 kg object to a height of 10 m is:
a) 16 s
b) 8 s
c) 4 s
d) 2 s
37
CHAPTER 5 - MECHANICAL PROPERTIES OF MATTER
Materials usually exist in one of three states: solid, liquid, and gas as shown in Figure
5.1. The atoms of a solid material are closely packed, a solid has fixed shape which cannot be
changed easily (Figure 5.1, a). In a liquid, the atoms or molecules are close together but they
can move freely relative to each other (Figure 5.1, b). Liquids have no definite shape. Rather,
they take the shape of the container they placed in. A gas is made of widely spaced atoms or
molecules moving freely in all directions (Figure 5.1, c). Like liquids, gases take the shape of
the container and it will expand to fill its container. Gases are easily compressed (density
changes as the pressure changes), while liquids compress with great difficulty. Since liquids
and gases do not maintain a fixed shape, they have the ability to flow; they are thus often
referred to as fluids.
(a) Solid
(b) Liquid
(c) Gas
Figure 5.1: Material states
5.1 Density and Specific Gravity
It is sometimes said that iron is heavier than wood. This cannot really be true, since a
large log clearly weighs more than an iron nail. What we should say is that iron is denser than
wood. The density of a substance is defined as the amount of mass in a unit volume of that
substance.

m
V
(5.1)
The SI unit of density is (kg/m3). Densities of solids and liquids are very nearly constant
but the densities of gases vary greatly with temperature and pressure. Table 5.1 presents a list
of densities for various materials.
38
Table 5.1: Densities of Various Substances
Substance
Density  (kg/m3)
Aluminum
Brass
Copper
Gold
Iron (Steel)
Lead
Silver
Concrete
Wood
Ice
Benzene
Blood
Alcohol, ethyl
Glycerin
Mercury
Water (4oC)
Sea Water
Helium
Air (0oC, 1 atm pressure )
Carbon dioxide (0oC, 1 atm pressure )
2700
8600
8900
19300
7800
11300
10500
2300
300 - 900
917
879
1050
790
810
13600
1000
1025
0.179
1.29
1.98
Material State
Solids
Liquids
Gases
The specific gravity of a substance is defined as the ratio of density of a substance to the
density of water (at 4.0C) which is 1000 kg/m3, the specific gravity of any substance will be:
Specific gravity =
 material
 water
(5.2)
Example 5.1
A lead sphere has a radius of 5 cm. The density of lead is 11.3  103 kg/m3, determine:
a) The mass of the sphere
b) The specific gravity of lead
Solution
a) The volume of the sphere is:
Vsphere =
4 3 4
πr = π(0.05 m)3 = 5.23 × 10−4 m3
3
3
So we have from Eq. (5.1)
m = ρV = (11.3 × 103
b) Specific Gravity =
ρmaterial
ρwater
kg
) (5.23 × 10−4 m3 ) = 𝟓. 𝟗𝟏 𝐤𝐠
m3
=
kg
m3
kg
1000 3
m
11.3×103
39
= 𝟏𝟏. 𝟑
Example 5.2
Calculate the height of an aluminum cylinder with a diameter of 2 cm and a mass of 27 g. The
density of aluminum is 2.7g/cm3.
Solution
𝑑
2
= r = 1 cm = 0.01 m, m = 27/1000=0.027 kg
g
  (2.7
cm
ρAlum. =
)  (2.7 
3
10-3 kg
10
-6
m
3
)  2700 kg / m3
m
m
= 2
V πr h
Solving for h, we obtain:
h=
m
πr 2 ρAlum.
=
0.027 kg
π(0.01m)2 (2700
kg
)
m3
= 𝟎. 𝟎𝟑𝟏𝟖 𝐦
Example 5.3
A person would like to design a water bed. If the size of the bed is 2.20 m long, 1.8 m wide,
and 0.3 m deep, what mass of water is needed to fill the bed?
Solution
m = ρV = (103
kg
) (2.20 m)(1.80 m)(0.30 m) = 𝟏𝟏𝟖𝟖 𝐤𝐠
m3
5.2 Hooke’s Law
If a force is exerted on an object, such as the vertically suspended metal bar or a spring
as shown in Figure 5.2, the length of the object changes.
L
x
F
Figure 5.2: Hooke’s Law
If the amount of elongation (x) is small compared to the length of the object,
experiment shows that (x) is proportional to the force exerted on the object; this was first
observed by Robert Hooke (1635-1703). This proportionality can be written as an equation and
is called Hooke's law:
F = k Δx
(5.3)
40
Where ‘F’ represents the force (or weight) pulling on the object, ‘x’ is the change in
length or sometimes referred to as the elongation, and ‘k’ is the proportionality constant which
is known as force constant.
The elastic properties of solids will be discussed in terms of the concepts of stress and
strain. Stress is a quantity that is proportional to the force causing a deformation. Strain is a
measure of the degree of deformation.
For a small stresses, the stress is proportional to the strain and the constant of
proportionality depends on the material. This proportionality constant is defined as elastic
modulus which is given by:
Stress
Elastic Modulus = Strain
(5.4)
Figure 5.3 shows a typical stress-strain curve. Up to a point called the “Elastic limit”,
the curve is a straight line and obeys Hook’s law (equations 5.3). Up to the elastic region, the
object will return to its original state if the applied force is removed. If the elastic limit is
exceeded by applying a larger stress, the object enters the plastic region where the curve
deviates from a straight line and no linear relationship exists between stress and strain. The
maximum stress applied is called the ultimate stress. If the stress is increased even further, the
material will break (breaking point).
Stress
Ultimate stress
Breaking point
Strain
Figure 5.3: Stress - Strain Curve
5.3 Elastic Properties of Solids
There are three types of elastic modulus:
1. Young’s modulus (Y): measures the resistance of a solid to a change in its length.
2. Shear modulus (S): measures the resistance of a solid to a change in its shape.
3. Bulk modulus (B): measures the resistance of a solid, or a liquid, to changes in their
volume.
41
5.3.1 Young’s modulus (Y): Elasticity in length
If a force (F) is applied perpendicular to the cross sectional area (A) of a bar with length
(L), the bar will stretch or compress by a distance (L). The ratio of the elongation (L) in the
bar length to the original length of the bar (L) is called the tensile strain:
change of length L

original length
L
Tensile Strain 
(5.5)
The ratio of the magnitude of the applied force to the cross- sectional area of the wire is called
the tensile stress acting on the wire.
F
Tensile Stress = A
(5.6)
The ratio of tensile stress to tensile strain is defined as Young's modulus (Y):
Y
Tensile Stress
F/A

Tensile Strain L / L
(5.7)
Its value depends only on the material. The value of Young's modulus for various materials is
given in Table 5.2.
Table 5.2: Elastic Constants for Various Substances
Substance
Young's Modulus, Y
(×1010 N/m2)
Bulk Modulus, B
(×1010 N/m2)
Shear Modulus, S
(×1010 N/m2)
Aluminum
7.0
7.0
2.5
Brass
9.1
6.0
3.5
Copper
11
14.0
4.2
Iron
9.1
9.0
7.0
Steel
21
0.77
0.56
Lead
1.6
16.0
8.4
Concrete
2.0
Marble
5.0
7.0
Granite
4.5
4.5
Wood
1.0
Nylon
0.5
Water
0.2
Alcohol
0.1
Mercury
0.25
42
Example 5.4
A 10 kg mass is hung from a 1.5 m steel wire with a diameter of 1.0 mm. (a) what is the stress
on the wire? (b) What is the strain? (c) How much will the wire stretch?

Young's modulus of steel is Ysteel = 2.1×1011 N/m2
Solution
r = 0.0005 m, A = πr2 = π (0.0005 m)2 = 7.85×10-7 m2
The force of weight stretching the wire is: F = mg = (10 kg)(9.81 m/s2 ) = 98.1 N
a) The stress acting on the wire is:
Stress =
F mg
98.1 N
𝐍
𝟖
= 2=
=
𝟏.
𝟐𝟓
×
𝟏𝟎
A πr
7.85 × 10−7 m2
𝐦𝟐
b) The strain of the wire is:
8 N
stress 1.25 × 10 m2
Strain =
=
= 𝟓. 𝟗𝟓 × 𝟏𝟎−𝟒
N
Y
2.1 × 1011 2
m
c) The elongation of the wire is given by:
∆L = strain × L = (5.95 × 10−4 )(1.5 m) = 𝟖. 𝟗𝟑 × 𝟏𝟎−𝟒 𝐦
Example 5.5
A load of 102 kg is supported by a wire of length 2m and cross-sectional area 0.1 cm2. The
wire is stretched by 0.22 cm. Find: (a) the tensile stress, (b) tensile strain, and (c) Young's
modulus of the wire
Solution
m
F mg (102 kg) (9.81 s 2 )
𝐍
a) Tensile Stress = =
=
= 𝟏. 𝟎 × 𝟏𝟎𝟖 𝟐
−4
2
A
A
0.1 × 10 m
𝐦
∆L 0.22 × 10−2 m
b) Tensile Strain =
=
= 𝟏. 𝟏 × 𝟏𝟎−𝟑
L
2m
8 N
Tensile Stress 1.0 × 10 m2
𝐍
𝟏𝟎
c) Y =
=
=
𝟗.
𝟏
×
𝟏𝟎
Tensile Strain
1.1 × 10−3
𝐦𝟐
5.3.2 Shear Modulus (S): Elasticity of Shape
In addition to being stretched or compressed, a body can be deformed by changing the
shape of the body. If the body returns to its original shape when the distorting stress is removed,
the body exhibits the property of elasticity of the shape, called shear.
43
As an example, consider the cube fixed to the surface in Figure 5.4. A tangential force
(F) is applied at the top surface of the cube with side (h). The magnitude of this force (F) times
the height (h) of the cube would normally cause a torque acts on the cube to rotate it. However,
since the cube is not free to rotate, the body instead becomes deformed and changes its shape.
F
x
A
F
h
h

Figure 5.4: Shear Force
The tangential force thus causes a change in the shape of the body that is measured by
the angle (), called the angle of shear. We can also relate () to the linear change from the
original position of the body (x) by noting from Figure 5.4 that:
tan  =
∆x
(5.8)
h
Because the deformations are usually quite small, tan () can be replaced by the angle () itself,
expressed in radian. Thus,

x
h
(5.9)
Equation (5.9) represents the shear strain of the body. The shear stress is defined as:
F
Shear Stress = A
(5.10)
The shear stress is directly proportional to the shear strain. The constant of proportionality is
called the shear modulus (S):
S
shear stress F / A

shear strain

(5.11)
Values of (S) for various materials are given in Table 5.2. Note that the shear modulus
is smaller than Young's modulus. This implies that it is easier to slide layers of molecules over
each other than compressing or stretching them.
Example 5.6
A tangential force of 7.5107 N is applied on the top surface of a sheet of copper (1.5 m length,
1.0 m width, and 10 cm thickness). The shear modulus for copper is 4.21010 N/m2. Find (a)
the shearing stress, (b) the shearing strain, and (c) the displacement.
44
Solution
a) The area that the tangential force is acting on is:
A = length  width = (1.5 m) (1 m) = 1.5 m2
Shear Stress =
F 7.5 × 107 N
𝐍
=
= 𝟓. 𝟎 × 𝟏𝟎𝟕 𝟐
2
A
1.5 m
𝐦
b) The shear strain, found from equation (5.11), is:
N
5.0 × 107 2
F
m
ϕ=
=
= 𝟕. 𝟗 × 𝟏𝟎−𝟒 𝐫𝐚𝐝
AS (1.5 m2 ) (4.2 × 1010 N )
m2
c) The linear displacement (x), found from equation (5.9):
∆x = hϕ = (0.1 m)(7.9 × 10−4 rad) = 𝟕. 𝟗 × 𝟏𝟎−𝟓 𝐦
Example 5.7
A brass cube with 5.0 cm sides is subjected to a tangential force, if the angle of shear is 0.010
rad, what is the magnitude of the tangential force? The shear modulus of brass is SBrass =
3.61010 N/m2.
Solution
From equation (5.11) we get:
F
A
= Sϕ
N
F = SAϕ = (3.6 × 1010 m2 ) (5 × 10−2 m)2 (0.010 rad) = 𝟗. 𝟎 × 𝟏𝟎𝟓 𝐍
Example 5.8
A copper block (7.50 cm) on a side, is subjected to a tangential force of 3.5103 N. Find the
angle of shear.
Solution
F
3.5 × 103 N
ϕ=
=
= 𝟏. 𝟒𝟖 × 𝟏𝟎−𝟓 𝐫𝐚𝐝
SA (4.2 × 1010 N ) (7.50 × 10−2 m)2
m2
5.3.3 Bulk Modulus (B): Elasticity of Volume
If a uniform force is exerted on all sides of an object, as in Figure 5.5, such as a block
under water, each side of the block is compressed.
45
F
F
F
F
F
F
Figure 5.5: Compression Force
Thus, the entire volume of the block decreases. The compression stress is defined as:
F
Volume Stress = A
(5.12)
F represents the magnitude of the normal force acting on the cross-sectional area (A) of the
block. The strain is the change in the volume per unit volume, that is:
Volume Strain =
∆V
(5.13)
V
The volume stress is directly proportional to the volume strain. The constant of
proportionality is called the Bulk modulus (B):
Volume Stress
F/A
B = Volume Strain = − ∆V/V
(5.14)
The minus sign is introduced in equation (5.14) because an increase in stress (F/A)
causes a decrease in the volume, leaving (V) negative. For fluids, the volume stress is the
pressure (P), and equation (5.14) becomes:
Volume Stress
P
B = Volume Strain = − ∆V/V
(5.15)
The bulk modulus is a measure of how difficult it is to compress a substance. The
reciprocal of the bulk modulus B, is called the compressibility (K) which is a measure of how
easy it is to compress the substance:
1
k=B=−
∆V/V
(5.16)
F/A
Values of the Bulk Modulus (B) for various materials are given in Table 5.2.
Example 5.9
A solid copper sphere of 0.5 m3 volume is placed 100 m below the ocean surface where the
pressure is 3.0×105 N/m2. What is the change in the volume of the sphere? The bulk modulus
for copper is BCopper = 1.4×1011 N/m2.
Solution
The change of the volume is:
∆V = −
VP
B
N
=
(0.5 m3 )(3.0×105 2 )
m
N
14×1010 2
m
= −𝟏. 𝟎𝟕 × 𝟏𝟎−𝟔 𝐦𝟑 (The -ve sign indicates compression)
46
Example 5.10
A cube of lead with 15.0 cm sides is subjected to a uniform pressure of 5.0105 N/m2. By how
much does the volume of the cube change? The Bulk modulus of lead is to BLead 1.6×1011 N/m2.
Solution
Using equation (5.15) to calculate the change of the volume:
5 N
−2
3
P V (5.0 × 10 m2 ) (15.0 × 10 m)
∆V = −
=
= −𝟏. 𝟎𝟓𝟓 × 𝟏𝟎−𝟖 𝐦𝟑
N
B
1.6 × 1011 2
m
Example 5.11
A liter of glycerin decreases by 0.21cm3 when subjected to a pressure 9.8105 N/m2. Calculate
(a) the bulk modulus and (b) the compressibility of glycerin.
Solution
a) The magnitude of the bulk modulus is:
V
1000 cm3
N
𝐍
5
𝟗
B=−
p=−
(9.8
×
10
)
=
𝟒.
𝟔𝟕
×
𝟏𝟎
∆V
−0.21 cm3
m2
𝐦𝟐
b) The compressibility (k) is:
k=
1
1
𝐦𝟐
=
= 𝟐. 𝟏𝟒 × 𝟏𝟎−𝟏𝟎
B 0.46 × 1010 N
𝐍
m2
Example 5.12
A solid steel sphere of volume 0.5 m3 is lowered to a depth in the ocean where the water
pressure is equal to 2.0×107 N/m2. The bulk modulus of steel is equal to 7.7 × 109 N/m2. What
is the change in volume of the sphere?
Solution
From the definition of bulk modulus, we have:
p
B=
∆V
V
pV
∆V =
B
∆V =
N
)(0.5 m3 )
m2
N
7.7×109 2
m
(2.0×107
= −𝟏. 𝟑 × 𝟏𝟎−𝟑 𝐦𝟑 (The -ve sign indicates compression)
47
PROBLEMS
1. Calculate the mass of a solid iron sphere that has a diameter of 3.0 cm.
2. Calculate the mass of a solid iron cylinder that has a height of 2 m and a diameter of
3.0 cm.
3. A solid cube of 5.0 cm side has a mass of 1.31 kg. What is its material?
4. A 1.50 m long aluminum wire has a diameter of 0.750 mm. If a force of 60.0 N is
suspended from the wire. Find: (a) The stress, (b) the strain, and (c) the elongation of
the wire. Young’s modulus of aluminum is YAlum. = 7.0×1010N/m2.
5. The figure on the right shows a 1.0 m long steel wire
that has a diameter of 0.750 mm. When an unknown
weight is suspended from the wire it stretches by 0.2
mm. What was the load placed on the wire?
[Young's modulus of steel is 2.1×1011 N/m2].
d = 0.750 mm
Length = 1.0 m
Stretch = 0.2 mm
Weight =?
6. A mass of 2 kg is supported by a copper wire of length 4m and a diameter of 4mm.
Determine: (a) the stress on the wire and (b) the elongation of the wire.
[Young's modulus of copper is 1.1×1011 N/m2]
7. A copper cylinder, 7.50 cm high, and 7.50 cm in diameter, is subjected to a tangential
force of 3.5×103 N. Find the shear angle, where the shear modulus of copper is 4.2×1010
N/m2.
8. A shearing force is applied to the top edge of a 12.5 cm thick steel block, causing it
to distort by 5.20 mm. The area of its top surface is 2.50×10-4 m2. Calculate: (a) the
shearing angle, and (b) the shearing force?
[Shear modulus of steel is 5.6×109 N/m2].
9. A pressure of 1.013×107 N/m2 is applied to a volume of 15.0 m3 of water. If the bulk
modulus of water is 2.0×108 N/m2, (a) by how much will the water be compressed? (b)
What is the compressibility of water?
10. A total force of 5.72 × 105 N acts uniformly on the surfaces of a 1.5 m aluminum
cube. Calculate: (a) the stress on the cube, (b) the strain, (c) the change in volume, and
(d) the compressibility of aluminum.
[Bulk modulus of aluminum is 7.0×1010 N/m2]
11. A solid copper cube has a side length of 65.5 cm. How much pressure must be
applied to the cube to reduce the side length to 65 cm? The bulk modulus of copper is
1.4×1011 N/m2
48
MULTIPLE CHOICE QUESTIONS
1. The SI unit of density is:
a) kg. m3
b) kg.m2
c) kg/ m2
d) kg/ m3
2. The density of mercury () in a certain unit is 13.6 g/cm3. So, the density of mercury
in SI unit is:
a) 136 kg/m3
b) 13600 kg/m3
c) 1360 kg/m3
d) 1.36 kg/m3
3. The specific gravity (G) of aluminum is 2.7. An aluminum volume (V= 1000 cm3) will
have the mass:
a) 0.027 kg
b) 0.27 kg
c) 2.7 kg
d) 27 kg
4. The SI unit of Shear Modulus (S) is:
a) kg/m2
b) kg.m2
c) N/ m2
d) N. m3
5. The SI unit of Shear Strain is:
a) m
b) kg
c) degree
d) rad
6. The SI unit of Bulk Modulus (B) is:
a) kg/m2
b) kg.m2
c) N/ m2
d) N. m3
7. The compressibility (k) is defined as:
a) 1/B
b) 1/ B2
c) 1/ S2
d) 1/S2
8. Which of the following is a material constant?
a) stress
b) shear modulus c) strain
d) length
9. Bulk modulus (B) is a measure for the change in:
a) area
b) length
c) volume
d) mass
10. A certain spring obeys Hooke's Law. A force (F) of 10 N stretches the spring 3 m.
A force of 30 N will stretch the spring by:
a) 1 m
b) 9 m
c) 6 m
d) 12 m
49
CHAPTER 6 - THERMAL PROPERTIES OF MATTER
6.1 Heat
A solid body is composed of a large number of atoms or molecules arranged in a lattice
structure. These molecules have potential energy and vibrational kinetic energy. The sum of the
potential and kinetic energy of all these molecules is called the internal energy of the body. Heat
is the amount of internal energy flowing from a body at a higher temperature to another body at a
lower temperature, i.e. the transfer of internal energy between two bodies as a result of temperature
difference between them. Hence, when the body cools, its internal energy decreases and when it
is heated, its internal energy increases. The traditional unit of heat is the kilocalorie (kcal) which
is defined as the quantity of heat (Q) required to raise the temperature of 1 kg of water by 1C.
The unit of mechanical energy is the Joule. The equivalence between mechanical energy and heat
energy is called the mechanical equivalent of heat and is given by:
1 cal = 4.186 J
(6.1)
6.2 Specific heat
Various substances require different amount of heat energy to raise the temperature of a
given mass of the substance by the same amount. For example 1 kcal of energy is required to raise
the temperature of 1 kg of water by 1C, while a 0.093kcal is required to raise the temperature of
1 kg of copper by 1 C. Specific heat (c) is defined as the amount of heat energy required to change
the temperature of 1 kg of the material by 1 C. Thus,
Q
c = m ∆T
(6.2)
Q = heat energy absorbed/added or lost/removed, J
m = mass of substance, kg
T = Tf -Ti = temperature change, C
Ti and Tf = initial and final temperatures.
The SI unit of specific heat is J/ (kg.C). Specific heats for some substances are given in
Table 6.1 below.
50
Table 6.1: Specific Heats of Various Substances
Specific Heat, c
Material
J/(kg . C)
Air
1009
Alcohol (Ethyl)
2430
Aluminium
900
Concrete
2900
Copper
385
Glass
837
Gold
130
Granite
800
Human Body
3470
Iron (Steel)
452
Lead
130
Mercury
140
Platinum
134
Silver
238
Steam
2013
Water
4186
Zinc
389
Example 6.1
Convert: a) 150 cal to kcal b) 150 cal to Joules
c) 600 J to cal
Solution
a) 150 cal = 150/1000 kcal = 0.150 kcal
b) 150 cal = 150  (4.186) = 627.9 J
c) 600 J = 600/4.186 = 143.34 cal
Example 6.2
Find (a) the heat energy required to raise the temperature of 100 g of water from a temperature of
25.0 C to 55.0 C. (b) What is the heat energy required to raise the temperature of 100 g of iron
through the same temperature interval?
51
Solution
Mass of water (mw) = 100 g = 0.100 kg; Mass of iron (mi) = 100 g = 0.100 kg
Initial temperature (Ti) = 25.0 ℃; Final temperature (Tf) = 55.0 ℃
Specific heat of: water: cw = 4186 J/ (kg. ℃), iron: ci = 452 J/ (kg. ℃)
a) Heat energy needed to raise the temperature of water is:
Qw = cw mw (Tf − Ti ) = (4186)(0.1)[55.0 − 25.0 ] = 𝟏. 𝟐𝟔 × 𝟏𝟎𝟒 𝐉
b) Heat energy required to raise the temperature of iron (Qi) is given by:
Qi = ci mi (Tf − Ti ) = (452)(0.1)[55.0 − 25.0 ] = 𝟏. 𝟑𝟔 × 𝟏𝟎𝟑 𝐉
Example 6.3
A 300 g steel ball at 20 ℃ is placed in a pan of boiling water. How much thermal energy is
absorbed by the ball?
Solution
Since water boils at 100 ℃, T2 = 100 ℃
Q = c m (Tf − Ti ) = (452)(0.3)[100 − 20 ] = 𝟏. 𝟎𝟖𝟓 × 𝟏𝟎𝟒 𝐉
6.3 Heat transfer by conduction
When heat energy is absorbed by or lost from a system because of the difference in
temperature between the system and its surroundings, we refer to this process as heat transfer.
Thermal energy can be transferred by one or all of the following three mechanisms: conduction,
convection, and radiation. Conduction is the transfer of thermal energy through the exchange
of kinetic energy between the molecules without any motion of the medium. Heat transfer by
conduction can occur in solids, liquids, and gases, but it is most important in solids.
The thermal energy (Q) transferred by
T2
T1
conduction through a slab of material (Figure 6.1)
is directly proportional to its cross-sectional area
(A), the temperature difference (T1-T2) between
the hot and cold surfaces, the time (t). The thermal
A
Q
Q
Q
L
energy is also found to be inversely proportional
Figure 6.1: Heat Conduction in a Slab
to the thickness of the slab (L).
T1 −T2
Q ∝ A(
L
)t
(6.3)
52
Introducing a proportional constant (k), equation (6.3) becomes:
T1 −T2
Q = kA(
L
)t
(6.4)
The proportionality constant (k) is called the thermal conductivity of the material. If (k)
is large, then a large amount of thermal energy will flow through the slab, and the material is called
a good conductor of heat. On the other hand, if (k) is small, then a small amount of thermal energy
will flow through the slab, and the material is called a poor conductor or a good insulator of heat.
The units of k is (J/m.s.℃).
The ratio (
T1 −T2
L
) is the temperature gradient. If the temperatures at both surfaces (T1
and T2) became constant in time, this is called steady state condition. In this case, both the
T1 −T2
temperature gradient (
L
Q
) and the heat flow rate ( t ) are constant. The thermal conductivity
of various materials is given in Table 6.2.
Table 6.2: Thermal Conductivity for Various Materials
Material
Thermal Conductivity, k
J / (m. s.℃ )
Aluminium
234
Copper
402
Iron ( Steel )
87.9
Gold
313
Lead
35.6
Platinum
71.2
Silver
427
Zinc
117
Glass
1.4
Concrete
1.30
Brick
0.65
Plaster
0.72
Helium
0.15
Air
0.023
53
Example 6.4
Find the amount of thermal energy that flows per hour through a steel wall 4.00 cm thick, 10 cm
long, and 8 cm high, if the temperature of the inside wall is 20 C while the temperature of the
outside wall is 70C .
Solution
L = 4 cm = 0.04 m, A = 8 cm×10 cm = 80 cm2 = 0.008 m2
T = T1 - T2= (70-20) = 50C, k =87.9 J/m.s. C, t = 60×60 = 3600 s
T1 − T2
50
Q = kA(
) t = (87.9)(0.008) (
) (3600) = 𝟑. 𝟏𝟔 × 𝟏𝟎𝟔 𝐉
L
0.04
Example 6.5
A rod made of Aluminium is 2 m long and 4 cm in diameter. If one end of the rod is maintained
at 90 C and the other end is maintained at 20 C. Find:
a) The temperature gradient.
b) The heat rate transferred through it.
c) The temperature at a distance of (0.5 m) from the hot end.
Solution
T1 = 90 C, T2 = 20 C, L = 2 m, r = 0.04/2=0.02 m,
A   r2 =
 (0.02) 2 =1.25610-3 m2
a) Temperature gradient =
b) Heat rate = (Q/t) = k A
T1 −T2
L
T1 −T2
L
=
90−20
2
= 𝟑𝟓
℃
𝐦
= 234 (1.256  10-3 ) (35) = 10.29 J/s
c) At steady state, the temperature gradient is constant, then
T1 − T2 T1 − T ′ 90 − 20 90 − T ′
=
→
=
,
L
L′
2
0.5
T1 −T2
L
=constant,
Solving for T ′ → 𝐓 ′ = 𝟕𝟐. 𝟓℃
6.4 Linear Thermal Expansion
It is a well-known fact that most materials expand when heated. This expansion is called thermal
expansion. If a solid thin rod of initial length Li, at initial temperature (Ti) is heated to a final
temperature (Tf), then the rod expands by a small length (L) as shown in Figure 6.2.
Li
Figure 6.2: Linear Expansion
54
L
The change in length (L) is given by:
L   Li T
(6.5)
T is the temperature change (Tf - Ti),  is a constant called the coefficient of linear
expansion and has the unit (1/ C) or (C -1). The value of () depends on the type of material. The
coefficient of linear expansion for selected materials is given in Table 6.3.
Table 6.3: Coefficients of Linear & Volume Thermal Expansion
Coefficient of …
Material
Linear Expansion ()
Volume Expansion ()
(×10-5 oC-1 )
(×10-4 oC-1)
Aluminium
2.5
Copper
1.7
Brass
1.9
Steel
1.2
Lead
2.9
Zinc
2.6
Glass (ordinary)
0.9
Glass (pyrex)
0.32
Ethyl alcohol
11
Water
2.1
Mercury
1.8
Glycerin
5.0
6.5 Volume Expansion of Solids and Liquids
Volume also changes with temperature. The change in volume (V) at constant pressure
is proportional to the initial volume (Vi) and to the change in temperature (T) according to the
relation:
V   Vi T
(6.6)
 is called the coefficient of volume expansion. The value of  depends on the type of
material and it has the same unit of  [C-1]. It is tabulated for common materials in Table 6.3.
Equation (6.6) applies to any body volume, and for an isotropic solid:
  3
(6.7)
55
An isotropic solid is one in which the coefficient of linear expansion is the same in the
three directions. For liquids and gases,  has no meaning and  should be determined
experimentally.
Example 6.6
A steel railroad track is 100 m long and 5 cm in diameter. The track is initially at a temperature of
20 C. Find:
a) The change in length of the track when the temperature rises to 50 C.
b) The force necessary to compress the rail back to its initial length.
Solution
steel = 1.2×10-5 C -1 , Ysteel = 2.1×1011 N/m2
r = 0.05/2=0.025 m, A =  (.025)2 =1.963 ×10-3 m2
T = 50 - 20 = 30 C, Li = 100 m
a) ∆L = α Li ∆T = (1.2 × 10−5 ℃−1 )(100 m)(30℃) = 𝟎. 𝟎𝟑𝟔 𝐦
∆L
b) F = (Ysteel ) (A) ( L )
F = (Ysteel ) (A) (αsteel ) (∆T)
= (2.1 × 1011
N
) (1.963 × 10−3 m2 )(1.2 × 10−5 ℃−1 )(30℃)
m2
= 𝟏. 𝟒𝟖 × 𝟏𝟎𝟓 𝐍
Example 6.7
What is the coefficient of linear expansion of a solid rod if a 100 ℃ temperature change causes a
2.00 m long rod to expand by 4.80 mm?
Solution
L = 4.8 mm = 4.8×10-3 m, Li = 2.0 m, T = 100 C
∆L
4.8 × 10−3 m
α=
=
= 𝟐. 𝟒 × 𝟏𝟎−𝟓 °𝐂 −𝟏
(2.0
(100
Li ∆T
m)
℃)
Example 6.8
What is the change in volume of 2.0 L of glycerin when it is heated from 20.0 C to 70.0 C? [
for glycerin is 5.1×10-4 C –1]
56
Solution
 = 5.1×10-4 C –1 , Vi = 210-3 m3 , T = 70 - 20 = 50 C ,
∆V = β Vi ∆T = (5.1 × 10−4 ℃−1 )(2.0 × 10−3 m3 )(50 ℃) = 𝟓. 𝟏 × 𝟏𝟎−𝟓 𝐦𝟑
Example 6.9
A 2.0 L Pyrex beaker is filled with alcohol at 20.0 C. How much alcohol will overflow if the
beaker and alcohol are heated to 70.0C?
The Volume expansion coefficients for alcohol and Pyrex are:

βalcohol = 1.1×10-3 ℃-1

βpyrex = 9.6×10-6 ℃-1
Solution
The amount of alcohol that overflows is equal to the difference between the volume changes of
alcohol and Pyrex:
For alcohol:
Vi,alcohol = 2.0 10-3 m3, T = 50 C, a = 1.1×10-3 C –1
∆Va = βa Vi,a ∆T = (1.1 × 10−3 )(2.0 × 10−3 )(50) = 1.1 × 10−4 m3
For Pyrex:
Vi,pyrex = 2.0 10-3 m3, T = 50 C, p = 9.6×10-6 C -1,
∆Vp = βp Vi,p ∆T = (9.6 × 10−6 )(2.0 × 10−3 )(50) = 9.6 × 10−7 m3
Therefore the overflow is equal to:
∆V = ∆Va − ∆Vp = (1.1 × 10−4 m3 ) − (9.6 × 10−7 m3 )
∆V = 𝟏. 𝟎𝟗 × 𝟏𝟎−𝟒 𝐦𝟑
57
PROBLEMS
A) Specific Heat
1. A 250 g glass (c = 837 J/kg℃) is taken from a freezer at -23.0 ℃ and placed into beaker
of boiling water. How much thermal energy is absorbed?
2. A 590 g sample of an unknown substance cools from 85.5 ℃ to 25.5 ℃ when 1.6×104
J is removed from it. What is the specific heat of the unknown substance?
B) Heat Transfer by Conduction
3. A certain material with a surface area of 1.56 m2 and a thickness of 5.25 mm loses
2.59×105 kcal in 215 s when the inside temperature is 158 ℃, and the outside temperature
is 15.2 ℃. What is the thermal conductivity of the material?
4. How much thermal energy flows through a glass window 0.35 cm thick, 1.2 m high and
0.80 m wide, in 12.0 hr, if the temperature on the outside of the window is -8.00 ℃ and
the temperature on the inside of the window is 20.0 ℃?
• Thermal conductivity (kglass) = 1.4 J/m.s.C
5. A (3.0 m  7.0 m) brick wall is 10.0 cm thick. Find the rate of heat energy flow through
the wall when its outer surface is at -5.0 C and its inner surface is at 30.0 ℃.
• Thermal conductivity (kbrick) = 0.65 J /m.s.℃
6. One end of a copper rod is maintained at a temperature of 100 ℃, where the other
end is maintained at a temperature of 25.0 ℃. The rod is 1.5 m long and 3.0 cm in
diameter. Find:
a) The amount of thermal energy that
3.0 cm
Kcopper = 402 J/ms℃
flows through the rod in 5.0 min.
b) The temperature of the rod at a distance Q
of 60 cm from the hot end.
100 ℃
25 ℃
L = 1.5 m
• Thermal conductivity (kcopper) = 402 J/m.s C
C) Thermal Expansion
7. A metal rod is 60.0 cm long and 4 cm in diameter. A 75 ℃ temperature increase causes
the rod to expand 0.810 mm. Find (a) the coefficient of linear expansion of the metal rod.
(b) The force necessary to compress it back to its initial length. Ymetal = 1.11011N.m-2.
8. What is the volume expansion coefficient of a solid, if a 50.0 ℃ temperature change
causes 1 m3 of the solid to expand by 1×10-3 m3?
9. An open Pyrex tube is filled to the top with 3000 cm3 of mercury at an initial
temperature of 20 ℃. How much mercury will overflow if the tube and mercury are
heated to 100 ℃? The coefficient of volume expansions are:
• βpyrex = 9.6×10-6 ℃-1, βMercury = 9.6×10-4 ℃-1
58
MULTIPLE CHOICE QUESTIONS
1. The SI unit of specific heat (c) is:
a) J.kg/℃
b) J/(kg℃)
c) kg℃/J
d) J.kg. ℃
2. The amount of heat required to change the temperature of a unit mass of the substance
by one degree is defined as:
a) heat capacity
b) thermal energy c) thermal conductivity
d) specific heat
3. The SI unit of thermal conductivity (k) is:
a) J.kg/m℃
b) Jm/kg℃
c) mkg℃/J
d) J/ms℃
4. A 500 g aluminium block (c=900 J/kg℃) at -10.0 C is placed in an oven at 190 ℃.
The thermal energy absorbed by the block is:
a) 81000 J
b) 90000 J
c) 9000 J
d) 8100 J
5. A 3.0 m copper rod is heated from 20.0 ℃ to 80.0 ℃. The rod length is increased
by (L = 3 mm). The coefficient of linear expansion of copper () is:
a) 1.67×10-5 ℃-1
b) 1.67×10-3 ℃-1
c) 1.67×10-2 ℃-1
d) 1.67×10-4 ℃-1
6. If the thickness of a wall (L) with uniform properties is doubled (2L), the heat passes
through the wall (Q) will be:
a) doubled
b) tripled
c) halved
d) constant
7. The volume expansion coefficient () of an isotropic solid is equal to:
a) β = α
b) β = 2α
c) β =3α
d) β = 4α
59
CHAPTER 7 - HYDROSTATICS
The study of fluids is usually treated from two different approaches. Firstly, fluids at rest
will be considered. This portion of the study of fluids is called Fluid statics or hydrostatics.
Second, we will study the behavior of the fluids when they are in motion. This part of the study
is called fluid dynamics or hydrodynamics. We start the study of the fluids by defining and
analyzing the macroscopic variables.
7.1 Pressure
The forces exerted by a fluid at rest are called hydrostatic forces. Pressure is defined as
the magnitude of the normal hydrostatic force acting per unit surface area. The pressure is thus
a scalar quantity. We write this mathematically as:
F
P=A
(7.1)
The SI unit for pressure is (N/m2), which is also named Pascal (Pa). Hence, 1 Pa = 1
N/m2. There are several units to describe pressure:
Metric units:
1 Pa = 1 N/m2
Other units:
1 torr = 1 mm of Hg (mercury) at 0 ℃.
1 torr = 133.3 Pa
1 bar = 1.0×105 N/m2 = 1.0×105 Pa
1 standard atmosphere (atm) =14.7 psi=76 cm Hg= 1.013×105 Pa
Example 7.1
A glass with straight sides contains 0.05 kg of water. The inside diameter of the glass is 2.2
cm. What pressure is exerted by the weight of the water on the bottom of the glass?
Solution
The area of the bottom of the glass is:
A = πr2 = π (1.1×10-2 m) 2 = 3.81×10-4 m2
Weight of the water: F = mg = (0.05 kg) (9.81 ms-2) = 0.491 N
The bottom of the glass must support the weight of the water. The pressure is:
P=
F mg
0.49 N
𝐍
𝟑
= 2=
=
𝟏.
𝟐𝟗
×
𝟏𝟎
A πr
3.81 × 10−4 m2
𝐦𝟐
60
Example 7.2
A man weighs 100 kg. At one particular moment when he walks, his right heel is the only part
of his body that touches the ground. If the heel of his shoes measures 7 cm  6 cm, what
pressure does the man exert on the ground?
Solution
The pressure that the man exerts on the ground is:
m
F mg (100 kg) (9.81 s 2 )
𝐍
P= =
=
= 𝟐. 𝟑𝟑𝟔 × 𝟏𝟎𝟓 𝟐
(0.07 m) (0.06 m)
A
A
𝐦
Example 7.3
A rectangular swimming pool measures 7m × 12m at the bottom. Calculate the pressure at the
bottom of the pool, where the mass of water is 160000 kg.
Solution
m
F mg (160000 kg) (9.81 s 2 )
P= =
=
= 𝟏. 𝟖𝟔𝟗 × 𝟏𝟎𝟒 𝐏𝐚
(7 m) (12 m)
A
A
7.2 Variation of Pressure with Depth
Consider a point at a depth (h) below the surface of the liquid (that is, the surface is at
height h above this point), as shown in Figure (7.1). The pressure of the liquid at this depth (h)
is due to the weight of the column of liquid above it. Thus the force acting on the area is:
F = mg = (V) g= A h g
Where (Ah) is the volume of the column,  is the density of the liquid (assumed to be
constant), and g is the acceleration due to gravity. The pressure p is then:
F
P=A=
mg
A
=
ρAhg
A
= ρhg
(7.2)
Thus the pressure is directly proportional to the density of the liquid, and to the depth
within the liquid. In general, the pressure at equal depths within a uniform liquid is the same.
h
A
Figure 7.1: Pressure Variation with Depth
61
Example 7.4
Find the water pressure at a depth of 3.0 m in a swimming pool.
Solution
The density of the water is 1000 kg/m3, and the water pressure is:
p = ρhg = (1000
kg
m
(3m)
)
(9.81
) = 𝟐. 𝟗𝟒 × 𝟏𝟎𝟒 𝐏𝐚
m3
s2
The pressure is 2.94×104 Pa at every point at a depth of 3m in the pool and exerts the same
force in every direction at that depth.
7.3 Absolute and Gauge Pressure
We have discussed atmospheric pressure, it is obvious that the total or absolute pressure
exerted at a depth (h) for example in a pool of water must be greater than the value of the
pressure at the pool surface. The air above the pool is exerting an atmospheric pressure on the
surface of the pool. Hence the total or absolute pressure (pa) at the depth (h) in the pool is the
sum of the atmospheric pressure (po) plus the pressure of the water itself, that is:
Pa = P0 + ρhg
(7.3)
The pressure of the liquid (hg), which is the difference between the absolute pressure
and atmospheric pressure, is called gauge pressure (pg) . The open tube manometer (Figure 7.2)
is a device used to measure the gauge pressure.
Atmospheric Pressure
Gauge
Pressure
h
Figure 7.2: Open Tube Manometer
It consists of a U-shaped tube containing a liquid usually water or mercury. When both
ends of the tube are open to the atmospheric, the level of the liquid in both columns is the same.
However, if one end of the tube is connected to a system whose gauge pressure we want to
measure, then the liquid in the other end rises until the pressures are equal. If (h) is the
difference in heights of the fluid columns, then the gauge pressure is pg =  h g and equation
(5.3) becomes:
Pa = Po + Pg
(7.4)
62
Example 7.5
What is the absolute pressure at a depth of 3.00 m in a swimming pool?
Solution
The density of the water is 1000 kg/m3, the absolute pressure is:
pa = po + ρ h g = (1.013 × 105
N
kg
m
(3m)
)
+
[(1000
)
(9.81
)] = 𝟏. 𝟑𝟎𝟕 × 𝟏𝟎𝟓 𝐏𝐚
m2
m3
s2
Example 7.6
Suppose that you use an air pressure gauge to measure the pressure in your car’s tires. If the
pressure of one tire is 2.2 atm, what is the absolute pressure of the air in the tire?
Solution
The absolute pressure in the tire is:
pa = po + pg = 1 atm + 2.2 atm = 𝟑. 𝟐 𝐚𝐭𝐦 = 3.2×1.013×105 Pa = 𝟑. 𝟐42×105 Pa
7.4 Buoyant Forces and Archimedes’ Principle
The variation of pressure with depth has a significant consequence. It allows the fluid
to exert a buoyant force on the bodies immersed in the fluid. If this buoyant force is equal to
the weight of the body, the body floats in the fluid. This result was stated by Archimedes and
is called Archimedes’ Principle. It can be stated as follows: "A body completely or partially
submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the
body". To prove that the buoyant force is equal in magnitude to the weight of the displaced
fluid, we take an object of height (h) immersed in a fluid as shown in Figure 7.3.
F2
h2
h1
h1 - h2
F1
Figure 7.3: Buoyant Force
The pressure at the bottom of the body (p1) at depth (h1) is:
p1 = f h1 g
(7.5)
where (f) is the density of the fluid. Similarly, the top of the body is at a depth (h2) below the
surface of the fluid, and the fluid pressure (p2) is given by:
p2 = f h2 g
(7.6)
63
Because the force due to the pressure acts equally in all direction, there is an upward force on
the bottom of the body and is given by:
F1 = p1 A
The force acting on the body downward is given by:
F2 = p2 A
The net force (Fb) acts upward since F1 > F2 on the body and is given by:
Fb = F1 - F2 = p1 A - p2 A
Replacing the pressures p1, and p2 from equations (7.5) and (7.6) then:
Fb = f h1 g A - f h2 g A = f g A (h1 - h2)
(7.7)
But: A (h1 - h2) = VO = Volume of the object. Therefore equation (7.7) becomes:
Fb = f VO g
(7.8)
Where Fb is the buoyant force exerted by the fluid, f is the density of the fluid, and VO
is the object volume (also the volume of the fluid displaced by the object).Since the mass of
the displaced fluid is (mf = f Vo), we see that:
Fb = mf g
(7.9)
where (mf g) is the weight of the displaced fluid. The apparent weight (W’) is the difference
between the actual weight (W) and the buoyant force (Fb), i.e.
W’=W- Fb
(7.10)
Example 7.7
Calculate (a) the buoyant force acting on a copper cube of 0.2 m side when it is completely
submerged in water. (b) What is the result if the cube is made of steel?
The density of copper is (8600 kg/m3), and the density of steel is (7850 kg/m3).
Solution
The buoyant force is given by equation (7.8)
f (water) =1000 kg/m3, VO= (0.2 m) 3 = 0.008 m3
For copper:
Fb = ρf Vo g = (1000
kg
m
) (0.008 m3 ) (9.81 2 ) = 𝟕𝟖. 𝟒𝟖 𝐍
3
m
s
For steel:
Fb = ρf Vo g = (1000
kg
m
) (0.008 𝑚3 ) (9.81 2 ) = 𝟕𝟖. 𝟒𝟖 𝐍
3
m
s
Note: the buoyant force depends only on the volume of the object not its kind.
64
Example 7.8
A wooden cube (10 cm) on side floats on oil surface. If the density of the wood is 700 kg/m3,
and the density of oil is 850 kg/m3. How deep does the cube sink in oil?
Solution
L = 0.1 m, f = 850 kg/m3
Since the cube floats, it is in equilibrium, so:
Fb = W
(W= weight of the object)
f Vo g = mwood g
f L2 h g = wood L3 g
kg
ρwood L3 ρwood L (0.1 m) (700 m3 )
h=
=
=
= 𝟎. 𝟎𝟖𝟐 𝐦
kg
ρf L2
ρf
850 3
m
7.5 Viscosity and Viscous Fluids
In deriving Bernoulli’s equation from the work-energy relationship, the friction has
been ignored. Bernoulli’s equation works well for fluids that flow easily over short distances
through large-diameter pipes and it still works in systems that have fluid friction. There are two
types of fluid flow. When layers of fluid slip over each other without eddy currents, the fluid
is said to have laminar flow (Figure 7.4a). When eddy currents (whirling movement) are
present, the fluid is said to have turbulent flow (Figure 7.4b). Turbulent flow is usually
difficult to analyze.
Figure 7.4 (a): Laminar Flow
Figure 7.4 (b): Turbulent Flow
Laminar flow involves shear in a viscous fluid. Fluids offer some resistance to sharing
motion. This resistance is a form of internal friction which is called viscosity. The term
viscosity can be explained with the aid of the following example. Consider a thin layer of fluid
between two parallel plates as shown in Figure 7.5. The lower plate is fixed, while the upper
one is moving with speed (v) under the action of an external force (F). The velocity of
successive layers increases linearly from 0 to v as one moves from a layer adjacent to the fixed
plate to a layer adjacent to the moving plate.
65
Moving plate
v
F
L
Fluid
Fixed plate
Figure 7.5: A layer of fluid between two parallel plates
In a time interval (t), the fluid adjacent to the upper plate moves a distance (x). The shear
stress on the fluid is equal to (F/A), while the shear strain is equal to (x/L). So, the shear strain
per unit time is equal to (v/L). The term (v/L) is also defined as the velocity gradient. The shear
stress (F/A) is proportional to the shear strain per unit time (v/L).
The constant of
proportionality is defined as the coefficient of viscosity () and is given by:
η=
F/A
(7.11)
v/L
It is important to mention that equation (7.11) is only valid if the fluid velocity varies
linearly with position, i.e. the velocity gradient is uniform. If the velocity gradient is not
uniform we have to replace v/L by dv/dy.
The basic unit of viscosity in SI is the poiseuille (1N.s/m2) or pascal.second (Pa.s).
Most engineering handbooks use a smaller unit called poise. A poise is still an inconveniently
large unit for some applications. Often viscosity is expressed in centipoise.
1 poiseuille (PI) = 1 N.s/m2 = Pa.s
1 poise (p) = 10-1 PI
1 centipoise (cp) = 10-2 p = 10-3 N.s/m2
The coefficient of viscosity of liquids usually decreases with increasing temperature,
while gases behave differently. Hot gases have higher viscosities than cool gases. Table 7.1
gives typical values of coefficient of viscosity for some common fluids.
66
Table 7.1: Coefficient of Viscosity of Various Fluids
Temperature
Coefficient of Viscosity, 
(C)
(10-3 Pa.s)
0
1.8
20
1.0
100
0.3
Whole Blood
37
4
Blood Plasma
37
1.5
20
1.2
30
1.0
20
1500
30
630
30
200
0
0.017
40
0.019
0
0.009
80
0.0098
Fluid
Water
Ethyl alcohol
Glycerin
Engine Oil
Air
Hydrogen
Example 7.9
A metal plate of area 0.05 m2 is connected to an 8g mass via a string that passes over an ideal
pulley (massless and frictionless). A lubricant with a thickness of 0.3 mm is placed between
the plate and the surface. When released, the plate is observed to move to the right with a
constant speed of 0.085 m/s. Find the coefficient of the viscosity of the lubricant.
Film
8g
Solution
The force (F) that makes the plate to move is equal to the tension force (T) of the string, which
equals the suspended weight (W), therefore:
F = T =W = mg = (8×10-3 kg) (9.81 m/s2) = 7.85×10-2 N
The lubricant in contact with the horizontal surface is at rest, while the layer in contact with
the plate moves at the speed of the plate. Assuming the velocity gradient is uniform, we have:
F/A FL (7.85 × 10−2 N)(0.3 × 10−3 m)
𝐍. 𝐬
−𝟑
η=
=
=
=
𝟓.
𝟓𝟒
×
𝟏𝟎
m
v/L Av
𝐦𝟐
(0.05 m2 ) (0.085 )
s
From the previous example, can you explain why we put lubricant oil inside a car’s engine?
67
PROBLEMS
A) Pressure
1. Suppose that you use an air pressure gauge to measure the pressure in your car’s tires.
If the pressure of one tire is 1.9 atm, what is the absolute pressure of the air in the tire?
2. At what depth in a lake is the absolute pressure equal to three times atmospheric
pressure?
3. A 0.5 m layer of oil floats on the surface of water in a 2.5 m deep container. The
density of oil is 800 kg/m3, and atmospheric pressure is 1.01105 Pa. What is the
pressure at the bottom of the container?
4. A rectangular swimming pool is 2 m deep and measures (7 m  12 m) at the bottom.
If the atmospheric pressure is 1.01105 Pa and the pool is completely filled with water,
calculate:
a) The pressure at the bottom of the pool?
b) The total force on the pool’s bottom surface?
B. Archimedes’ Principle
5. What is the density of a fluid if a buoyant force of 23 N is exerted on a 2510-4 m3
object which is completely submerged in it?
6. A 70 kg rock lies at the bottom of a lake. Its volume is 3.010-3 m3. How much force
is needed to lift it?
7. A 144.06 N crown has an apparent weight of 136.612 N when submerged in water.
What is the crown made of? (Refer to Table 5.1 for material densities)
8. A block of wood 0.12 m high, 0.06 m wide, and 0.1 m long having a density of 700
kg/m3 floats on water.
a) How much of it will be above the water surface?
b) How much weight has to be placed on top of the block so that its top is just level
with the water?
68
MULTIPLE CHOICE QUESTIONS
1. The SI unit of pressure is:
a) N/m
b) N/m3
c) atmosphere
d) Pascal
2. The relation between absolute pressure (pa), atmospheric pressure (po) and gauge
pressure (pg) is given by:
a) pa = po - pg
b) pa = po + pg
c) pa = pg – pa
d) pa = 2po + 2pg
3. A 2.0 kg cubic block with 10 cm on a side is placed on a table. The pressure exerted
by the block on the table is:
a) 0.02 Pascal
b) 0.2 Pascal
c) 1960 Pascal
d) 0.196 Pascal
4. The atmospheric pressure (po) is equal to 1.01105 Pa. The absolute pressure (pa) at
the bottom of a lake that is 5 m deep is given by:
a) 1.01106 Pa
b) 1.5105 Pa
c) 1.5103 Pa
d) 1.01105 Pa
5. The buoyancy force (Fb) is given by:
a) Fb = o Vf g
b) Fb = o Vo g
c) Fb = f Vf g
d) Fb = f Vo g
6. The buoyant force (Fb ) acting on a block 10 cm long by 12 cm wide by 15 cm
high when completely submerged in mercury (f = 13600 kg/m3 ) is:
a) 239.9 N
b) 2.399 N
c) 23.99 N
d) 0.2399 N
69
CHAPTER 8 - DIRECT CURRENT CIRCUITS
8.1 Introduction
Alessandro Volta invented the electric battery in 1800 which is considered as one of
the most important practical discoveries in science. The electric battery represented the basis
for a wide range of subsequent development in electrical technology. There are many different
kinds of batteries in use today. One of the most common types of batteries is referred to as a
dry cell. Such battery consists of zinc case serving as the negative terminal, and a carbon rod
to serve as the positive terminal. The space between the two terminals contains a paste-like
mixture of manganese dioxide, ammonium chloride, and carbon. When these materials are
assembled in this way, two chemical reactions take place; one occurs at the zinc case, the other
at the manganese dioxide layer surrounding the carbon rod. Positive charged zinc ions leave
the case and enter the ammonium chloride paste, where they combine with chloride ions. As
each zinc ion is removed from the case, it leaves behind two electrons. As additional zinc ions
leave the case, more electrons accumulate, leaving the zinc case with a net negative charge.
When a chloride ion breaks free from the ammonium chloride molecule, the remaining portion
of the molecule becomes singly ionized. This positively charged ion is neutralized by the
manganese dioxide, which supplies the needed electrons. As a result, the carbon rod surrounded
by its manganese dioxide layer ends up with a net positive charge.
Direct current or DC electricity is the continuous movement of electrons from an area
of negative (-) charges to an area of positive (+) charges through a conducting material such as
a metal wire. Whereas static electricity sparks consist of the sudden movement of electrons
from a negative to positive surface. A DC circuit is necessary to allow the current or the flow
of electrons to flow. Such a circuit consists of a source of electrical energy (such as a battery)
and a conducting wire running from the positive end of the source to the negative terminal. The
flow of DC electricity is similar to the flow of water through a hose.
8.2 Electric Current
When a continuous conducting path is connected between the terminals of a battery,
charges can flow from one terminal of the battery to the other. This flow of charges is called
an electric current. The electric current in a conductor is defined as the net amount of charge
that passes through the conductor per unit time at any point. The current is a measure of the
flow or quantity of electrons that pass a given point in one second. Thus, the electric current
70
(I) is expressed as:
I=
ΔQ
Δt
(8.1)
Q is the amount of charge in coulombs (C) that passes through the conductor at any point
during the time interval t.
Electric current is measured in units of amperes (A), which are equivalent to coulombs
per second (i.e. 1A = 1C/s). The direction of an electric current is by convention the direction
in which a positive charge would move. Thus, the current in the external circuit is directed
away from the positive terminal and toward the negative terminal of the battery. Electrons
would actually move through the wires in the opposite direction. Knowing that the actual
charge carriers in wires are negatively-charged electrons may make this convention seem a bit
odd and misleading; however this is the convention which is used worldwide and one must
become accustomed to.
8.3 Ohm’s Law
A battery is an electrical device whose terminals maintain a potential difference. An
electric field will be generated when the battery terminals are connected electrically. The
potential difference of the battery causes the flow of charges or simply an electric current.
Ohm's law says that the value of electric current in a piece of conductor is directly proportional
to the potential difference applied to it. The proportional constant is called the electrical
resistance and it depends on the conductor material, conductor shape and size, but does not
depend on the applied potential. Ohm's law is valid for many, but not all electrical devices. For
example, semiconductor elements do not obey Ohm's law at all and these materials are called
non-Ohmic materials, while elements which obey Ohm's law are called Ohmic materials or
resistors. G.S. Ohm was the first to establish experimentally that the electric current (I) in a
metal conductor is proportional to the potential difference (V) applied to its ends.
R
I
_
+
V
IαV
(8.2)
The potential difference (V) is produced by a battery or a power supply to the ends of
71
the conductor. It was found that the amount of electric current that flows in a wire depends not
only on the potential difference (V) but also on the resistance (R) of the wire offered to the
flow of electrons. This means that electrons are slowed down because of interactions with the
atoms of the wire. For a given voltage, higher resistance results the less current. So the current
in the wire (conductor) is inversely proportional to its resistance at a given voltage;
Iα
1
(8.3)
R
Combining the above two equations, we obtain Ohm’s law:
I =
V
(8.4)
R
Ohm's law states that the electric current (I) through a metal conductor is proportional
to the applied voltage (V) and (R) is a constant of the metal conductor independent of (V) called
electrical resistance. The electrical resistance (R) can be expressed as:
R =
V
(8.5)
I
The unit of the electrical resistance is called Ohm (Ω) and is defined as:
1Ω =
1 volt
1 Ampere
8.4 Series and Parallel Circuits
A series circuit is a circuit in which resistors are arranged in a chain, so the current has
only one path to take. The current is the same through each resistor. The total resistance of the
circuit is found by simply adding up the resistance values of the individual resistors. Parallel
circuit is a circuit in which the resistors are arranged with their heads connected together, and
their tails connected together. The current in a parallel circuit breaks up, with some flowing
along each parallel branch and re-combining when the branches meet again. The voltage across
each resistor in parallel connection is the same. The total resistance of a set of resistors in
parallel is found by adding up the reciprocals of the resistance values, and then taking the
reciprocal of the total.
8.4.1 Resistors in Series
Resistors can be connected in series; that is, the current flows through them one after
another. The circuit in Figure 8.1 shows three resistors connected in series, and the direction of
current is indicated by the arrow.
72
_
+
R1
A
+
I
I
+
V
I
_
V
_
I
R2
R3
Figure 8.1: Resistors in Series
There is only one path for the current to travel, so the current through each of the resistors is
the same.
I = I1 = I2 = I3
(8.6)
Also, the voltage drops across the resistors must add up to the total voltage supplied by the
battery:
V = V1 + V2 + V3
(8.7)
Since V = I R, then
V = I R1 + I R2 + I R3 =I (R1 + R2 + R3)
(8.8)
But Ohm's Law must also be satisfied for the complete circuit:
V = I Re
(8.9)
where Re is the equivalent resistance.
Setting equations (8.8) and (8.9) equal, we get:
I Re = I (R1 + R2 + R3)
(8.10)
So the currents cancel on both sides of equation (8.10), and we arrive at an expression for the
equivalent resistance for resistors connected in series:
Re = R1 + R2 + R3
(8.11)
In general, the equivalent resistance of resistors connected in series is the sum of their
resistances. That is,
Re = ∑ Ri
Example 8.1:
The current flowing in a circuit containing four resistors connected in series is I = 1.0 A. The
potential drops across the first, second and third resistors are, respectively: 5 V, 8 V, and 7 V.
The equivalent resistance of the circuit is Re = 30. Find the resistance of each resistor in the
circuit and the total voltage supplied by the battery.
73
Solution
R1
Because the resistors are connected in
series, then the same current flows
I
V
through each one. Using Ohm's Law,
I
R3
R4
we can find the resistances of the first,
second and third resistors.
R2
I
I
V1 5V
V2 8V
V3 7V
=
= 𝟓𝛀, R 2 =
=
= 𝟖𝛀, R 3 =
=
= 𝟕𝛀
I
1A
I
1A
I
1A
R 4 = R e − (R1 + R 2 + R 3 ) = 30Ω − (5Ω + 8Ω + 7Ω) = 10Ω
R1 =
V4 = I ∙ R 4 = 1A(10Ω) = 10V
V = V1 +V2 +V3 +V4 = 5V + 8 V + 7V + 10 V = 30 V
Or
V = IRe = 1A (30Ω) = 30 V
8.4.2 Resistors in Parallel
Resistors can be connected such that they branch out from a single point (known as a
node), and join up again somewhere else in the circuit as shown in Figure 8.2. Each of the
three resistors in Figure 2 is another path for current to travel between points A and B.
-
+
A
I
+
V
V
I1
R1
I2
-
R2
I3
R3
Figure 8.2: Resistors in Parallel
Note that the node does not have to physically be a single point; as long as the current has
several alternate paths to follow, then that part of the circuit is considered to be parallel. At
point A the potential must be the same for each resistor. Similarly, at point B the potential must
also be the same for each resistor. So, between points A and B, the potential difference is the
same. That is, each of the three resistors in the parallel circuit must have the same voltage.
V = V1 = V2 = V3
(8.13)
Also, the current splits as it travels from A to B. So, the sum of the currents through the
three branches is the same as the current at A and at B (where the currents from the branch
reunite).
I = I1 + I2 + I3
(8.14)
74
By Ohm's Law, equation (8.14) is equivalent to:
V
Re
V
V
V
1
2
3
=R +R +R
(8.15)
So the V's cancel out in equation (8.15) and we are left with:
1
Re
1
1
1
=R +R +R
1
2
(8.16)
3
This result can be generalized to any number of resistors connected in parallel.
1
Re
1
= ∑R
(8.17)
i
For two resistors connected in parallel:
R R
R e = R 1+R2
1
(8.18)
2
Example 8.2:
Find the total current for the circuit shown in Figure (8.2) knowing that:
R1 = 2, R2 = 3 , R3 = 6, and Vt = 12 V.
Solution
The potential drop across all the branches of the circuit is equal to 12 V.
V1 = V2 = V3 = 12 V
I1 =
V
12V
V
12V
V
12V
=
= 6A, I2 =
=
= 4A, I3 =
=
= 2A
R1
2Ω
R1
3Ω
R1
6Ω
I = I1 + I1 + I1 = 6A + 4A + 2A = 𝟏𝟐𝐀
Or
1
1
1
1
1 1 1 6+4+2
1
=
+
+
= + + =
= 1Ω → R e =
= 1Ω
R e R1 R 2 R 3 2 3 6
12
1Ω
I=
V
12V
=
= 𝟏𝟐𝐀
Re
1Ω
8.4.3 Current Divider
To obtain I1 and I2, which shows two resistors connected in parallel (Figure 8.3):
V1 = V2 = V
I
I1 R1 = I2 R 2 = I R e
V
R1 R 2
I1 R1 = I2 R 2 = I
R1 + R 2
∴ I1 = I
∴ I2 = I
R2
R1 +R2
R1
R1 +R2
R1
I1
R2
I2
(8.19)
Figure 8.3: Current Divider
(8.20)
75
PROBLEMS
1. Find the equivalent resistance of the circuit shown below.
3Ω
2Ω
A
18 Ω
9Ω
1Ω
6Ω
B
2. Find the equivalent resistance between A and B for the circuit shown below.
6Ω
3.4 Ω
A
2Ω
5Ω
B
3. For the circuit shown to the right, find:
a) The total current in the circuit (IT)
b) The current in the 4 resistor
c) The voltage drop across the 3 resistor
d) The voltage drop across the 5 resistor
4. For the circuit shown to the right, find:
a) The total current in the circuit (IT)
b) The current in the 3 resistor (I1)
c) The voltage drop across the 6
resistor (V6)
5. For the circuit shown below, find:
a) The total current (IT)
b) The current in the 2 resistor
c) The voltage drop across the 3
resistor
12 V
IT
4Ω
5
6Ω
I2
I1=?
IT =?
6Ω
3Ω
24 V
V6Ω = ?
2
IT =?
6Ω
6. For the circuit shown below, find:
a) The total current (IT)
b) The current in the 6 resistor
c) The voltage drop across the 3
and 5 resistors
2
3Ω
40 V
3Ω
6Ω
10 Ω
6Ω
8Ω
3Ω
5Ω
1Ω
IT =?
11 V
76
MULTIPLE CHOICE QUESTIONS
1. A potential difference of 10 V is applied across a 5 resistor. The current (I) passes
through the resistor is:
a) I= ½ A
b) I = 2 A
c) I = 50 A
d) I = 5 A
2. Three resistors (2, 4 and 6) are connected in series. The equivalent resistance
(Re) should be:
a) Re < 2 
b) Re < 4 
c) Re < 6 
d) Re > 6 
3. Three resistors (2, 4 and 6) are connected in parallel. The equivalent resistance
(Re) should be:
a) Re > 6 
b) Re > 4 
c) Re >2 
d) Re < 2 
4. Three resistors (16, 16 and 8) are connected in parallel. The equivalent
resistance (Re) is:
a) 4 
b) 40 
c) 8 
d) 16 
5. Three resistors (R1, R2 and R3) are connected in series. The current through R1 is (I1=
2 A). So, the current through R3 is:
a) I3= 1 A
b) I3= 2 A
c) I3= 3 A
d) I3= 4 A
6. A circuit consists of a battery and three equal resistors (R1=R2=R3) connected in
parallel. If the total current (It) in the circuit is 9 A, so the current (I) passes in each
resistor is:
a) 3 A
b) 27 A
c) 9 A
d) 6 A
77
Appendix A – Important Terminology Translations
‫الفصل‬
‫عربي‬
‫أنجليزي‬
‫الرمز‬
‫الوحدة الدولية‬
Chapter
Arabic
English
Unit
Length
Width
Height
Mass
Volume
Density
Area
Time
Distance
Velocity
Acceleration
Gravity
Force
Friction
Weight
Work
Energy
Radius
Diameter
Thermal Energy
Thermal Expansion
Coefficient
Pressure
Atmospheric
Pressure
Apparent Weight
Voltage
Current
Resistance
Symbol
SI Unit
L
w
h
m
V
ρ
A
t
x,y
v
a
g
F
fk
W
W
E
r
d
Q
m
m
m
kg
m3
kg/m3
m2
s
m
m/s
m/s2
m/s2
N
N
J
J
m
m
J
α,β
˚C-1
p
Pa
Po
Pa
1
2
3
4
5
6
‫الوحدة‬
‫الطول‬
‫العرض‬
‫االرتفاع‬
‫الكتلة‬
‫الحجم‬
‫الكثافة‬
‫المساحة‬
‫الزمن‬
‫المسافة‬
‫السرعة‬
‫تسارع‬
‫الجاذبية‬
‫القوة‬
‫احتكاك‬
‫الوزن‬
‫العمل‬
‫الطاقة‬
‫نصف القطر‬
‫القطر‬
‫الطاقة الحرارية‬
‫ثابت التمدد‬
‫الحراري‬
‫الضغط‬
7
‫الضغط الجوى‬
8
W’
‫الوزن الظاهري‬
V
‫الجهد الكهربى‬
I
‫شدة التيار‬
R
‫المقاومة‬
Appendix B – Conversion Factors
Length
1 meter (m) = 100 cm = 39.4 in.
= 3.28 ft = 6.21×10-4 mi
1 centimeter (cm) = 10-2 m = 10 mm
= 0.394 in.
1 inch (in.) = 2.54 cm = 0.0254 m
N
V
A
Ω
Power
1 watt (W) = 1 J/s = 2.39×10-4 kcal/s
1 kilowatt (kW) = 1000 W
Density
1 kg/m3 = 1×10-3 g/m3
78
= 0.083 ft
1 foot (ft) = 0.305 m = 30.5 cm = 12 in
1 mile (mi) = 1610 m = 1.61 km
= 5280 ft
1 kilometer (km) = 1000 m = 0.621 mi
1 nanometer (nm) = 10-9 m = 10-7 cm
1 micron (μ) = 10-6 m = 1 μ m = 10-4 cm
1 g/cm3 = 1000 kg/m3
Area
1 m2 = 104 cm2 = 1.55×103 in2
1 cm2 = 10-4 m2 = in2
1 km2 = 106 m2
Acceleration
1 m/s2 = 3.281 ft/s2
Velocity
1 m/s = 3.28 ft/s = 2.24 mi/hr
= 3.60 km/hr
1 km/hr = 0.278 m/s = 0.621 mph
= 0.91 ft/s
Pressure
1 N/m2 = 1 pascal (Pa) = 9.87×10-6 atm
1 atmosphere (atm) = 1.013×105 N/m2
Volume
1m3 = 103 L
Electricity
1 volt (V) = 1 joule/coulomb =J/C
1 ampere (A) = 1 C/s
1 ohm (Ω) = 1 volt/ampere
Mass
1 kg = 1000 g
Weight
1 lb = 4.45 N
1 N = 0.225 lb
Energy, Work, Thermal Energy
1 joule = 2.39×10-4 kcal = 0.239 cal
1 kilocalorie (kcal) = 4186 J
1 calorie = 4.186 J
Appendix C – Equation Sheet
Chapter 2: Kinematics in One Dimension
Chapter 3: NEWTON'S LAWS OF MOTION
Horizontal Motion:
1. v = v0 + at
1
2. x = v0 t + at 2
2
3. v 2 = vo2 + 2ax
Newton’s 2nd Law of Motion:
ΣF=ma
Vertical Motion:
1. v = v0 + gt
1
2. y = v0 t + gt 2
2
3. v 2 = vo2 + 2gy
Chapter 4: Work & Energy
Work (W) & Power (P)”
W = Fx
1

K. E = mv 2
2
P. E = mgh
1
} ETotal = P. E + K. E = mv 2 + mgh = constant
2
79
P=
W Fx
=
= Fv
t
t
Chapter 5: Mechanical Properties of Matter
Density(ρ), Mass(m), Volume(V):
m
ρ=
Specific Gravity (G)
ρ
G = material
Volumes (V):
Hooke’s Law:
F = kx
V
Bulk Modulus (B):
F
stress
FV
A
B=
=−
=−
∆V
strain
A∆V
V
fluids
PV
B =
⏞ −
∆V
Compressibility (k):
1
K =
B
Shear Modulus (S):
F
shear stress A
F
S=
= =
shear strain ϕ Aϕ
∆x
ϕ=
h
ρwater
Young’s Modulus (Y):
4
Vsphere = πr 3
3
Vcube = L3
F
stress
FL
A
Y=
=
=
strain ∆L A∆L
L
Vcylinder = πr 2 h
ϕ
Chapter 6: Thermal Properties of Matter
Specific Heat (c)
Q
c=
m(∆T)
Q > 0 (𝐻𝑒𝑎𝑡 𝐺𝑎𝑖𝑛)
Q < 0 (𝐻𝑒𝑎𝑡 𝐿𝑜𝑠𝑠)
∆T = Tf − Ti
Thermal Conductivity (k):
𝑄
𝑘=
𝑇 − 𝑇2
𝐴( 1
)𝑡
𝐿
T −T
 temp. gradient = 1 2 = constant SS

q (heat rate) =
Q
t
L
= 𝑘𝐴 (
𝑇1 −𝑇2
𝐿
) = const.
Linear (α) and Volume (β) Thermal Expansion
∆L
α=
L(∆T)
β = 3α (isotropic materials)
∆V
β=
V(∆T) }
∆T = Tf − Ti
Chapter 7: Hydrostatics
Absolute Pressure (Pa)
Pa = Po + ∆P = Po + ρhg
Buoyancy Force (Fb)
Fb = ρf Vo g
Body Completely Submerged
W ′ = W − Fb
W
G=
W − W′
Body Partially Submerged
hρ
h1 = o
h
ρf
h2 = h − h1
Fb
h2
h1
mg
Chapter 8: Direct Current Circuits
Ohm’s Law
V = IR
Series connection
R eq = R1 + R 2 + R 3
Current Divider
R2
I1 = I (
)
R1 + R 2
R1
I2 = I (
)
R1 + R 2
Parallel connection
1
1
1
1
=
+
+
R eq R1 R 2 R 3
or
I
R1
I1
R2
I = I1 + I2
Appendix D – Answer Key
Chapter
1
Answers to Problems
1a) m2 1b) m/s 1c) m/s2 1d) N 1e) J
2a) 0.000032 m2, 32 mm2 2b) 30.6 m/s
2c) 13600 kg/m3 2d) 46.046 J/m.s.oC
3a) N 3b) N/m2 3c) kg/m3 3d) J
4a) incorrect 4b) incorrect 4c) correct
4d) correct
80
Answers to
Multiple Choice Questions
I2
2
3
4
5
6
7
8
1a) 2.36 m/s2 1b) 96.195 m
2) 50 m 3a) -3.47 m/s2 3b) 9.6 s
4a) 9×1014 4b) 6.63×10-9 s
5a) 78.48 m 5b) 34.34 m/s
6a) 1.53 s 6b) 11.47 m 6c) 5.19 m/s
7a) 15.7 m/s 7b) 12.55 m
1) 2.22 N 2) 170.67 N 3) 27.53 N
4a) 1.82 m/s2 4b) 2.5 m/s 4c) 3.6 m
5) 2.80 m/s2 6a) 3.92 m/s2 6b) 35.32 N
6c) 7.84 m/s 6d) 7.84 m
1) 3677.01 J 2) 812500 J 3) 3.64 m
4a) 686.7 J 4b) 686.7 J
5a) 133.3 N 5b) 40 J 5c) 40 J 5d) 0.3 m
6a) 180 J 6b) 45.87 m 6c) 24.21 m/s
7a) vB = 25.06 m/s, vC = 22.59 m/s
7b) 24.00 m 8) 829.63 N
1) 0.11 kg 2) 11.03 kg 3) Silver
4a) 1.36×108 N/m2 4b) 1.94×10-3
4c) 2.91×10-3 m 5) 18.56 N
6a) 1.56×106 N/m2 6b) 5.68×10-5 m
7) 1.89×10-5 rad 8a) 0.0416 rad
8b) 58240 N 9a) 0.76 m3 9b) 5×10-9 m2/N
10a) 42370 N/m2 10b) 6.05×10-7
10c) 2.04×10-6 m3 10d) 1.43×10-11 m2/N
11) 3.18×109 N/m2
1) 25,738 J 2) 452 J/kgoC
3) 118.8 J/msoC 4) 4.64×108 J 5) 4777.5 J/s
6a) 4262 J 6b) 70˚C 7a) 1.8×10-5 ˚C-1
7b) 104968 N 8) 2×10-5 ˚C-1
9) 2.28×10-4 m3
1) 2.9 atm 2) 2P0/ρg 3) 124544 Pa
4a) 120620 Pa 4b) 10132080 N
5) 937.8 kg/m3 6) 657.27 N 7) Gold
8) 0.036 m
1) 12Ω 2) 10Ω 3a) 2A 3b) 2A 3c) 6V
3d) 10V 4a) 6A 4b) 4A 4c) 2A 5a) 10A
5b) 6A 5c) 12V 6a) 1A 6b) 0.167A
6c) 1V, 5V
81
1) c 2) c 3) a 4) a
5) c 6) c 7) d 8) d
9) c
1) b 2) b 3) a 4) c
5) d 6) d
1) a 2) b 3) c 4) d
5) b 6) b 7) a 8) d
1) d 2) b 3) c 4) c
5) d 6) c 7) a 8) b
9) c 10) b
1) b 2) d 3) d
5) a 6) c 7) c
4) b
1) d
5) d
2) b
6) a
3) c 4) b
1) b
5) b
2) d
6) a
3) d
4) a