MODULE IN
PHYSICS FOR ENGINEERS
ENGGPHYS
MECHANICAL AND MECHATRONICS ENGINEERING
SCHOOL OF ENGINEERING AND ARCHITECTURE
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TABLE OF CONTENTS
Contents
Page
Title Page
i
Course Overview
1
Course Study Guide and House Rules
3
Study Schedule
5
Study Schedule Table
5
Short Term Academic Calendar
10
Assessment and Evaluation Guide
11
General Requirements
11
Formative Assessment Guide
11
Evaluative Assessment Guide
12
Technological Tools
13
Grading System
13
Course References
14
Facilitator Contact Details
14
Annexes
A. Rubrics for Evaluation
15
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2
ENGGPHYS
COURSE LEARNING OUTCOMES
PHYSICS FOR ENGINEERS
As a result of their educational experiences
in the subject EnggPhys, graduates should
be able to:
CLO 1:
Demonstrate knowledge of
physics concepts and principles by
describing everyday phenomena and
analyze problems on vectors, one- and
two-dimensional motion and Newton’s
laws.
CLO 2:
Demonstrate knowledge of
physics concepts and principles by
describing everyday phenomena and
analyze problems on dynamics, work,
energy, energy and power, impulse
and momentum.
CLO 3:
Demonstrate knowledge of
physics concepts and principles by
describing everyday phenomena and
analyze problems on heat and
calorimetry, simple harmonic motion,
mechanical waves, and electricity.
CLO 4: Demonstrate ability to use
mathematical tools, including calculus
in solving problems involving physics
concepts and principles.
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COURSE INTRODUCTION
This course covers the following: vectors, kinematics, dynamics, work, energy and
power, impulse and momentum, rotation, dynamics of rotation, elasticity,
oscillation, fluid statics and kinematics, thermal expansion, thermal stress, heat
transfer, calorimetry, waves, electrostatics, electricity, magnetism, optics, image
formation by plane and curved mirrors, and image formation by thin lenses.
Module and Unit Topics
To ensure that you will demonstrate the above-cited course learning outcome at the
end of the semester, this course designed to be delivered in 54 contact hours was
structured into fourteen modules. Each module contains lecture notes and example
problems for each topic. Each topic is designed using the 5E constructivist model of
learning, developed by Rodger Bybee, that encourages students to engage,
explore, explain, elaborate, and evaluate their knowledge of topics covered therein.
It means that at the end of each unit, each module, and the course as a whole, you
will be assessed on your progress in attaining the course learning outcomes.
Outcomes-based education dictates that only when you can demonstrate the
course learning outcomes by the end of this course, can you be given a passing
mark. The modules that form the building blocks to help you attain the course
learning outcomes are as follows:
Module 1: Vectors



Introduction to vector quantities
Unit conversion
Vector Operations (Addition, Subtraction, Multiplication)
Module 2: Kinematics

Motion along a straight line
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
Motion in two dimensions
Module 3: Newton’s Law of Motion



Newton’s 1st Law of Motion
Newton’s 2nd Law of Motion
Newton’s 3rd Law of Motion
Module 4: Work, Energy and Power



Work and Energy
Law of Conservation of Energy
Power
Module 5: Impulse and Momentum


Impulse-Momentum relation
Law of Conservation of Momentum
Module 6: Rotational Motion, Dynamics of Rotation, Elasticity





Concepts on Angular displacement,
acceleration
Torque
Rotation of Rigid Bodies
Hooke’s Law
Young’s Modulus of Elasticity
Angular
velocity
and
angular
Module 7: Oscillations



Simple Harmonic Motion
Simple Pendulum
Spring-mass system
Module 8: Fluids


Properties of Fluids
Archimedes’ Principle
Module 9: Heat Transfer



Conduction
Convection
Radiation
Module 10: Waves


Properties of Waves
Transverse Waves and Longitudinal Waves
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
Mathematical representation of a wave
Module 11: Electrostatics



Basic concepts
Electrostatic force
Electric Field
Module 12: Electricity



Current, Voltage, Resistance
Ohm’s Law
Series and Parallel Circuits
Module 13: Magnetism



Magnetic Field
Magnetic Field Intensity
Magnetic Flux
Module 14: Optics



Nature of Light
Refraction
Reflection
Course Study Guide
The key to successfully finish this course lies in your hands. This module was prepared
for you to learn diligently, intelligently, and independently. Doing this will help and
prepare you, as this serves as a foundation for your higher Mechanical Engineering
courses. Aside from meeting the content and performance standards of this course
in accomplishing the given activities, you will be able to learn other invaluable
learning skills which you will be very proud of as a responsible learner
1. Schedule and manage your time to read and understand every part of the
module. Read it over and over until you understand the point.
2. Study how you can manage to do the activities of this module in
consideration of your other modules from other courses. Be very conscious
with the study schedule. Post it on a conspicuous place so that you can
always see. Do not ask about questions that are already answered in the
guide.
3. If you did not understand the readings and other tasks, re-read. Focus. If this
will not work, engage all possible resources. You may ask other family
members to help you. If this will not work again, email me first so that I can
provide the necessary assistance.
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4. Do not procrastinate. Remember, it is not others who will be short-changed
if you will not do your work on time. It will be you.
5. Before you start doing your tasks, read and understand the assessment tools
provided. Do not settle with the low standards, target the highest standards
in doing your assigned tasks. I know you can.
6. You are free to browse and read the different units of the module even prior
to doing the tasks in each unit. However, you need to ensure that you will
not miss any part of the module and you will not miss to accomplish every
activity in every unit as scheduled.
7. Before the end of each term, you will be tasked to send back through
correspondence the accomplished and scheduled modules for the term.
Make sure that you will follow it up with me through email or any other media
available for you.
8. While waiting for my feedback of your accomplished modules, continue
doing the task in the succeeding units of the module that are scheduled for
the next term.
9. If needed, do not hesitate to keep in touch with me through email.
Remember, if there is a will, there is a way.
10. In answering all the assessment and evaluation activities, write legibly. It will
help if you will not write your answers in the module if you are not yet sure of
your answers. You must remember that all activities in the module are
academic activities, which mean that the relevant academic conventions
apply. Think before you write.
a. Your answers should be composed of complete and grammatically
correct sentences. Do not use abbreviations and acronyms unless
these are introduced in the readings, and do not write in text-speak.
Avoid writing in all caps and cursive.
b. In the self-processed discussions, write appropriate and well-thought
arguments and judgements. Avoid merely approving or
disapproving with what is expressed in the material. You need to
support your inputs in the discussions from reliable information or from
empirical observation. Do not write uninformed opinions.
c. Do not write lengthy answers. Stick to the point. Be clear what your
main point is and express it as concisely as possible. Do not let your
discussion stray. Make use of the spaces in the module as your guide.
d. Quote your sources if there are in answering all the activities.
11. Lastly, you are the learner; hence, you do the module on your own. Your
family members and friends at home will support you but the activities must
be done by you. As Louisan, we always need to demonstrate our core
values of competence, creativity, social involvement and Christian spirit.
Additional Guidelines for Offline Students:
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


If you are a student opting for the offline mode of distance learning,
you will be tasked to send back the accomplished requirements at
given stages of the course through express mail correspondence on or
before the scheduled date to me. Make sure you will follow it up with
me through text or any other media available for you.
While waiting for my feedback of your accomplished requirements,
continue doing the task in the succeeding units of the module.
If needed, do not hesitate to keep in touch with me through any
available means. Remember, if there is a will, there is a way.
Study Schedule
Below are details in the conduct of this course arranged in chronological order visà-vis the topic learning outcomes and activities designed for you to undergo the five
stages of the 5E constructivist learning model.
Week
Topic Learning Outcome
Module 1: Vectors
1. Analyze and solve problems
involving unit conversion.
2. Perform mathematical
operations (addition,
subtraction, and
multiplication); on vector
quantities and solve
corresponding application
problems
Activities
Engage:




Introduce measurement
and unit conversion
Differentiate vector
quantities from scalar
quantities.
Recall trigonometric
functions of right
triangles.
Recall Sine and Cosine
Laws
Explore:


Identify basic conversion
factors.
Identify the different
mathematical operation
of vectors.
Explain:


Methods of vector
addition
Vector Multiplication (Dot
and Cross Product)
Elaborate:

Apply vector operation in
solving actual
engineering problems.
Evaluate:

Answer given problems
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Assignment no.1
Module 2: Kinematics
1. Analyze and solve problems
on Rectilinear Motion with
constant acceleration.
2. Analyze and solve problems
on Free-falling bodies.
3. Analyze and solve problems
on Projectile Motion.
Engage:

Introduce rectilinear
motion, free-falling
bodies and projectile
motion.
Explore:

Discuss the formulas on
Rectilinear Motion and
the concept of projectile.
Explain:

Demonstrate by solving
example problems on
rectilinear motion, freefalling bodies and
projectile motion.
Elaborate:

Apply concepts of
rectilinear motion and
projectile motion in
solving problems.
Evaluate:

Answer given problems
Assignment no.2
QUIZ #1 AND SUBMISSION OF ASSIGNMENTS 1 AND 2
Module 3: Newton’s Law of Motion
Apply Newton’s Laws of
Motion to analyze and solve
problems involving a body in
equilibrium or a body in
acceleration.
Engage:

State Newton’s three
laws of motion.
Explore:


Discuss the concepts and
procedure in solving
problems for systems in
equilibrium.
Discuss the concepts and
procedure in solving
problems for systems with
acceleration.
Explain:

Demonstrate by solving
sample problems for
systems under equilibrium.
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
Demonstrate by solving
sample problems for
systems with acceleration.
Elaborate

Analyze and solve
problems on Newton’s
laws of motion
Evaluate:

Answer given problems.
Assignment no.3
Module 4: Work, Energy and Power
Solve problems on work
done by a constant or by
a varying force, as well as
problems in mechanics,
applying the concepts of
gravitational potential
energy, kinetic energy,
work-energy theorem, and
mechanical power.
Engage:

Recall the different types
of mechanical energies
and the Law of
Conservation of Energy.
Explore:


Discuss formulas used in
determining mechanical
energies possessed by a
body.
Discuss the Law of
Conservation of Energy
Explain:


Solve sample problems in
determining the
mechanical energy of a
body.
Solve sample problems on
Law of Conservation of
Energy.
Elaborate:

Analyze and solve systems
applying the “law of
conservation of energy”.
Evaluate:

Answer given problem set.
Assignment no.4
QUIZ #2 AND SUBMISSION OF ASSIGNMENTS 3 AND 4
Module 5: Impulse and Momentum
Solve problems related to
momentum, impulse, and
conservation of
momentum.
Engage:


Define Impulse and
momentum.
Introduce the “Law of
conservation of
momentum”.
Explore:
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

Introduce the formula in
determining Impulse and
Momentum of a body.
Analysis on the Law of
Conservation of
Momentum
Explain:


Solve sample problems on
Impulse and Momentum.
Solve sample problems on
Law of conservation of
Momentum.
Elaborate:

Analyze and solve systems
applying the “law of
conservation of
Momentum”.
Evaluate:

Answer given problem set.
Assignment no.5
Module 6: Rotational Motion, Dynamics of Rotation, Elasticity
Solve problems on rotational
motion, dynamics of
rotation, and elasticity.
Engage:

Define angular
displacement, angular
velocity, angular
acceleration, torque and
Elasticity.
Explore:



Discuss the equations
involved for rotating
bodies.
Discuss the concept of
Torque
Discuss Elasticity
Explain:

Solve sample problems on
rotating bodies, torque
and elasticity.
Elaborate:

Analyze and solve
problems on rating
bodies.
Evaluate:

Answer given problem set.
Assignment no.6
Module 7: Oscillations
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Solve problems on SHM
involving horizontal spring
system, vertical spring
system, and simple
pendulum.
Engage:

Illustrate bodies
undergoing simple
harmonic motion (SHM).
Explore:

Discuss the concept and
equations used in solving
problems involving SHM.
Explain:


Solve sample problems on
systems involving
horizontal and vertical
springs.
Solve sample problems on
pendulum systems.
Elaborate:

Analyze and solve systems
problems applying the
concept of SHM.
Evaluate:

Answer given problem set.
Assignment no7.
MIDTERM EXAMINATION AND SUBMISSION OF
ASSIGNMENTS 5, 6 AND 7
Module 8: Fluids
Solve problems on the
application of Archimedes’
Principle
Engage:

Identify and define basic
properties of fluids.
Explore:


Discuss the equations in
determining basic
properties of fluid such as
Density, specific weight,
specific volume..etc.
Discuss Archimides’
Principle
Explain:

Solve sample problems on
Archimedes’ Principle
Elaborate:

Analyze and solve actual
Engineering problems
applying Archimedes’
Principle.
Evaluate:

Answer given problem set
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Assignment no.8
Module 9: Heat Transfer
Solve problems on heat
transfer
Engage:

Identify and define the
three modes of heat
transfer.
Explore:

Discuss the equations in
determining the rate of
heat transferred by
Conduction, Convection
and Radiation.
Explain:

Solve sample problems on
Heat transfer.
Elaborate:

Analyze and solve actual
Engineering problems
applying heat transfer by
conduction, convection
and radiation.
Evaluate:

Answer given problem set.
Assignment no.9
Module 10: Waves
Solve problems on the
mathematical
representation of a wave
and problems related to the
modes of mechanical
waves
Engage:

Differentiate Transverse
Waves and Longitudinal
waves.
Explore:


Discuss the Properties of
waves
Discuss Mathematical
representation of waves.
Explain:



Solve sample problems on
the mathematical
representation of a wave
and problems related to
the modes of mechanical
waves
Elaborate:
Analyze and solve actual
Engineering problems
involving waves.
Evaluate:
Answer given problem set.
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Assignment no.10
QUIZ #3 AND SUBMISSION OF ASSIGNMENTS 8,9 AND 10
Module 11: Electrostatics
Solve problems involving
the different methods of
charging, electric force,
and electric field
Engage:

Define Electric Force and
Electric Field.
Explore:


Discuss the concept of
Electrostatic Force/s.
Discuss charges located
on an electric field.
Explain:

Solve sample problems on
Electrostatics.
Elaborate:

Analyze and solve actual
Engineering problems
applying electric field and
electric force.
Evaluate:
Answer given problem set
Assignment no.11
Module 12: Electricity
Solve basic problems
involving current,
resistance, and voltage in
circuits that contain DC
sources and resistors in series
and/or parallel
Engage:


Discuss Electricity and
ElectricCircuits
Define Current, Voltage
and Resistance.
Explore:


Discuss OHM’s Law.
Discuss procedure and
concept in solving Series
and Parallel Circuits.
Explain:


Solve sample problems on
Ohm’s Law.
Solve sample problems on
Series and Parallel Circuits
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Elaborate:

Analyze and solve actual
Engineering problems on
Electric Circuits.
Evaluate:
Answer given problem set
Assignment no.12
QUIZ #4 AND SUBMISSION OF ASSIGNMENTS 11 AND 12
Module 13: Magnetism
Solve problems involving
magnetic field, magnetic
flux intensity, and magnetic
flux
Engage:


Discuss Magnets and
Magnetism
Define Magnetic field,
Magnetic Field Intensity,
Magnetic Flux
Explore:

Discuss concepts and
equations used to
determine Magnetic field
intensity and Magnetic
flux.
Explain:



Solve sample problems on
Magnetic field, Magnetic
Field Intensity and
Magnetic Flux
Elaborate:
Analyze and solve actual
Engineering problems on
Magnetic field, Magnetic
Field Intensity and
Magnetic Flux
Evaluate:

Answer given problem set
Assignment no.13
Module 14: Optics
Solve problems involving
the nature of light,
refraction, and reflection.
Engage:


Introduce the Nature of
Light
Define refraction, and
reflection
Explore:

Discuss concepts and
equations for reflection
and refraction.
Explain:
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

Solve sample problems on
reflection from mirrors.
Solve sample problems on
refraction
Elaborate:

Analyze and solve actual
Engineering problems on
reflection and refraction
Evaluate:

Answer given problem set
Assignment no.14
FINAL EXAMS AND SUBMISSION AOF ASSIGNMENTS 13 AND 14
Evaluation
The course modules rely on formative and summative assessments to determine the
progress of your learning in each module. To obtain a passing grade in this course,
you must:
1. Read all course readings and answer the pre-assessment quizzes, selfassessment activities, and reflection questions.
2. Submit all assignments and graded quizzes
3. Take the Midterm Examination.
4. Take the Final Examination.
 If you are a student under the offline DL mode, accomplish all print-based
and electronically saved discussion activities and requirements, and
submit them on time via express mail correspondence.
Formative Assessment Activities
Formative assessments for this course are applied to ungraded activities that are
used to monitor your learning experience and provide feedback to improve both
your learning approach as well as my instructional approach.
• You are required to answer the pre-assessment quizzes, self-assessment
activities, and reflection questions but your scores in activities will not be
included in the computation of your final grade.
• The reflection questions are designed to help you to critically analyze the
course readings for better understanding while the pre-assessment quizzes
and self-assessment activities are designed as a review management tool
to prepare you for the graded quizzes and examinations.
• Successfully answering formative activity questions and requirements will
serve as prompts to tell you if you need to study further or if you may
already move forward to the next unit of the module.
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• The completeness of your answers to the pre-assessment quizzes, selfassessment activities, and reflection questions will still be checked and will
still be part of your course completion. Hence, all pre-assessment quizzes,
self-assessment activities, and reflection questions must be answered.
• In doing your formative assessment activities, you can always ask the help
of your family and friends.
• The pre-assessment quizzes, self-assessment activities, and reflection
questions are required so you can take it anytime within the scheduled
days assigned for each module.
Summative Assessment Activities
The evaluative assessments are graded activities designed to determine if your
acquisition of learning and performance in tests is at par with standards set at certain
milestones in this course.
A. Quizzes, Examinations, and Assignments
Graded quizzes, examinations, and assignments are essential to determine
whether your performance as a student is at par with standards/goals that need
to be achieved in this course. The scores obtained from each of the graded
activities will contribute to your final grade, the weights of which are presented
in the grading system described in the succeeding sections of this text. Direct
scoring can be used on straightforward requirements like short answers and
multiple-choice responses, while scoring rubrics will be provided for answers that
are typically lengthy and involve a more complex level of thinking on your part.
B. Final Course Requirement
To achieve the course learning outcome, a final design submission of all
experiments is required. You are going to accomplish this in groups and
present learning outputs as scheduled in the study plan. For online students,
a live presentation will be scheduled on Google Meet. For offline students a
recorded and saved presentation will be accommodated for submission on
a USB flash drive. A separate rubric will be used for the write up and the
presentation.
Technological Tools
To be able to accomplish all the tasks in this course, you will need a computer or a
laptop with the following software applications: Word Processing, Presentation, and
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Publication for requirements that do not require online access. A smart phone with
video recording and editing features will also be used for activities that will require
you to record videos for saving and submission.
If you are a student online, access to the institutional Google Classroom will be
provided through your institutional account. An invitation to join the Google
Classroom will be sent to you through the SLU Student Portal and your institutional
email account, so make sure to activate your institutional email account. It is equally
important that you check your SLU Student Portal account at least twice a week and
turn your Gmail Notifications on in your mobile phone and computer.
If you are a student offline, the delivery of instructions and requirements will be
primarily through express mail correspondence of printed modules and saved digital
content on a USB flash drive. Feedback and clarifications will be facilitated through
text messaging and voice calls; hence, you need to have regular access to a cell
phone. If you need to call, or you want to talk to me, send me a message first and
wait for me to respond. Do not give my CP number to anybody. I will not entertain
messages or calls from numbers that are not registered in my phone. Hence, use only
the CP number you submitted to me.
Contact Information of the Facilitator
Engr. Ferdinand B. Itliong
ENGGPHYS Course Facilitator
Cellphone
SLU local extension number
Institutional email address
:
: Mechanical Engineering, loc. 273
:
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18
MODULE 1:
PHYSICAL QUANTITIES, UNITS AND VECTORS
MODULE 1: PHYSICAL QUANTITIES, UNITS AND VECTORS
1.1 Physics and its importance
The word physics comes from Greek, meaning “of nature” or “natural
philosophy”. Physics is concerned with the description of nature—that is, the
description and explanation of natural phenomena. In other words, physics is
concerned with how and why things work or behave the way they do.
Physics is an experimental science. Everything we know about the physical
world and about the principles that govern its behavior has been learned through
experiment, that is, through observations of the phenomena of nature. The ultimate
test of any physical theory is its agreement with experimental observations. These
observations usually involve measurements; thus physics is inherently a science of
experiment and measurements.
1.2 Physical Quantities
The study of Physics involves dealing with a lot of physical quantities. In
mechanics, we have the basic or fundamental quantities like mass, length and time.
All others are considered as derived quantities because they are obtained or
defined by simple relations between the fundamental quantities. The fundamental
quantities combined to form the derived quantity are sometimes called the
dimensions of the derived quantity.
Basic Quantities and Units
Unit
Meter
Symbol
m
Kilogram
kg
Second
s
Description
Measures
length
Measures
mass
Measures
time
Definition
Distance light travel in 1/299792458 second.
Mass of a special platinum-iridium cylinder in
Paris.
9192631770 oscillations of a special light
emitted by cesium-133 atoms.
Combinations of these basic quantities and units give various derived quantities
and units.
Example:
Force:1newton (N) = 1 kg-m/s2
In the proper expression of physical quantities, there should at least be a
number (to indicate how large or how small the quantity is) and the unit (to indicate
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19
the nature and type of the quantity). An expression that does not have one of these
two is meaningless.
1.3 Standards and Units
A standard is that quantity (usually in physical form i.e. an object) to which
other quantities being measured are compared. The measured quantity is then
expressed in terms of the standard, which now becomes the unit of the quantity.
Example: When we say the height of the building is 10 meters, it means
that the measured quantity (height of the building) is expressed in terms of
the length of an object (standard), which is considered to be one meter
long. Thus the "meter" is the unit for the height of a building.
1.4 Systems of Units:
There are two systems of units in common use: The English or British system
and the Metric system. A refinement of the old metric system was introduced in
1960 and is officially known as the International System of units or SI units. It is now
modern practice to use this system.
The English system is also known as the foot-pound-second (fps) system. The SI
system may be classified into the meter-kilogram-second (mks) system and the
Gaussian or centimeter-gram-second (cgs) system.
It is therefore necessary to look at the more salient aspects of the SI system:
1. In the SI system, the standard units for the different basic quantities are well
defined, clear and precise.
Example: For Length: 1 meter is defined as the distance traveled by
light in 1/299792458 sec
 The student is advised to look at the SI definitions for the other basic units.
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2. In the SI system of units, larger or smaller variations of these units are
obtained by attaching the proper prefix.
Prefixes in the Metric System or SI system
Prefix
Abbreviatio Power of 10
n
Prefix
Abbreviatio
n
Power of 10
exa
E
1018
deci
d
10-1
peta
P
1015
centi
c
10-2
tera
T
1012
milli
m
10-3
giga
G
109
micro

10-6
mega
M
106
nano
n
10-9
kilo
k
103
pico
p
10-12
hecto
h
102
femto
f
10-15
deka
da
101
atto
a
10-18
1.5 Unit Consistency and Unit Conversion
Unit consistency means that in a physical equation, each side of the
expression should have the same units otherwise the equation is an error.
Unit conversion is the process of changing the unit of a quantity to another
one within the same system or into another system. In physical computations, this is
usually done to attain unit consistency.
The process of unit conversion may be relatively easy but it has to be done in
an orderly manner to avoid errors.
One should also have considerable
knowledge of the needed conversion factors to be able to do it successfully.
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Steps in Performing Unit Conversion:
Example problems on Conversion of units:
1.
Convert 120 km/hr to mi/hr.
Solution:
120 𝑘𝑚
1𝑚𝑖
𝒎𝒊
×
= 𝟕𝟒. 𝟓𝟕
1 ℎ𝑟
1.6093 𝑘𝑚
𝒉𝒓
Note that the unit km cancels out due to division.
2.
As an astute observer walking around on continental crust (granite), you
might decide to test the hypothesis that the Earth is made entirely of granite.
You weigh a 1.00 ft3 piece of granite on your home scale and find that it
weighs 171 lbs. Thus you determine that the granite has a density of 171 lb/ft3.
Convert your granite's density to g/cm3.
Solution:
171 𝑙𝑏 1000𝑔
1𝑓𝑡 3
1𝑖𝑛 3
𝒈
𝑥
𝑥
(
)
𝑥
(
)
=
𝟐.
𝟕𝟒
𝑓𝑡 3
2.205𝑙𝑏 12𝑖𝑛
2.54𝑐𝑚
𝒄𝒎𝟑
Note that the units lb, ft3 and in3 cancels out due to division
Formative Problems: Practice conversion of units by solving the following problems.
1. The density of propane is 36.28 lb/ft3. Convert this to kg/m3. (Ans..581.67)
2. A box measures 3.12 ft in length, 0.0455 yd in width and 7.87 inches in height.
What is its volume in cubic centimeters? (Ans.. 7.91 x 103 cm3)
3. A block occupies 0.2587 ft3 . What is its volume in mm3 ? (Ans.. 7.326x106 mm3)
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1.6 Vectors and Vector Operations
Many physical quantities have magnitudes only but no direction. These are
called scalars. Examples are mass, time, density, temperature, etc. There are
however, many physical quantities such as force, velocity, displacement, etc.
which have directions as well as magnitude and these aspects always have to be
indicated when expressing these quantities. They are called vectors.
In physical computation and analyses, we have to be aware of the
difference between vectors and scalars because the mathematical treatments are
not the same. For example, we add scalars arithmetically but we cannot do the
same to vectors. Special methods are used.
1.6.1 BASIC ASPECTS ABOUT VECTORS
1. Vector Representation
a. Graphical representation - Vectors are represented by arrows.
b. Vector notations – Vectors are usually denoted with capital letters
written in boldface or with special markings. (A or A, B or B, etc.)
2. Indicating Directions of (coplanar) vectors:
METHOD 1: Using the angle θ that the vector makes with the “zero
reference line (usually the positive x-axis) ” measured going Counterclockwise.Illustration:
i.
Vector A = 3 units at 35o is a vector having a magnitude of 3 units,
and whose direction θA is 35o from the positive x-axis measured
going counter-clockwise.
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ii.
Vector A = 3 units at 1250 is a vector having a magnitude of 3 units,
and whose direction θA is 125o from the positive x-axis measured
going counter-clockwise
iii.
Vector A = 3 units at 2250 is a vector having a magnitude of 3 units,
and whose direction θA is 225o from the positive x-axis measured
going counter-clockwise.
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METHOD 2: Using Geographic Directions
Illustration: Let us assume that the figure below shows vector A = 3 units, θA = 25o
and vector B = 3 units, θB = 30o
The figure above means that:
i.
ii.
iii.
iv.
Vector A = 3 units 25o EN (East of North)
Vector A = 3 units 65o NE
Vector B = 3 units 30o SW
Vector B = 3 units 60o WS
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1.6.2 VECTOR OPERATIONS
1. VECTOR ADDITION AND SUBTRACTION:
Vector addition is the process of combining two or more vectors into
one. The combination is called the RESULTANT (R) of the vectors. Vector subtraction
is just like addition. In vector subtraction, the negative of one vector is added to the
other. For example, if two vectors A and B are to be added, the operation is
indicated as A + B. However, if vector B is to be subtracted from vector A, the
operation is indicated as A – B which is the same as A + (-B).

NOTE: The negative of a vector is another vector whose magnitude is the
same as the original vector but in the opposite direction.
METHODS OF VECTOR ADDITION
i.
The Algebraic method (for co-linear vectors only). Co-linear vectors are
vectors which lie along the same line.
Example: For the vectors shown in the diagram, determine a) their resultant; b) C A-D
E = 60 m
B = 20 m
C = 30 m
D = 25 m
A = 50 m
Solution:
For convenience we assign all vectors directed towards the right as
positive while all vectors directed towards the left are negative.
 Since vectors are co-linear simple arithmetic is applied
a) Resultant: R = A+B+C+D+E = 50m+(–20m)+(-30m)+ 25m+(-60m) = - 35 m, this
implies that the magnitude of the resultant vector has a magnitude of 35 m
and directed towards the left (negative sign)
b) C-A-D = C+(-A)+(-D) = -30m + (-50m ) + (-25m) = -101m, this implies that the
magnitude of C-A-D is 101m and directed towards the left (negative sign)
c) The Parallelogram Method
The procedure of "the parallelogram of vectors addition method" is

a.
b.
c.
d.
draw vector 1 using appropriate scale and in the direction of its action
from the tail of vector 1 draw vector 2 using the same scale in the direction of its action
complete the parallelogram by using vector 1 and 2 as sides of the parallelogram
the resulting vector R is represented in both magnitude and direction by the diagonal of the
parallelogram
A
A
A
R
B
B
B
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e.
Solve the resultant using sine law and cosine law
d) The Polygon method (Graphical method in determining the magnitude and
direction of the Resultant R)
Many vectors can be added together in this way by drawing the
successive vectors in a tip-to-tail fashion, as shown on the example below.
Scale: 1 cm = 1 unit
e) The Triangle method is similar to the Parallelogram Method but with the two
vectors connected from tip-to-tail.
Procedure:
a.
b.
c.
Construct the vector triangle by drawing the two vectors tip-to-tail. The vector that closes the
triangle is the resultant.
The resultant vector R of the two coplanar vectors can be calculated by trigonometry using
"the cosine law" for a non-right-angled triangle.
The angle between the vector and the resultant vector can be calculated using "the sine law"
for a non-right-angled triangle.
f) The Component Method
Ay
Many vector operations and analyses are carried
out using their
components. These are two or more vectors which when combined
or
A
added will give the original vector. For coplanar vectors (assumed to be on
the xy-plane) it is usually convenient to use two components which are
perpendicular to each other: one along the x-axis which is then called the xcomponent and the other one along the y-axis which is then called the ycomponent. These two components are collectively called the rectangular
components of the vector.
Determining the Components of a Vector
1. F1x is the magnitude of the x-component of vector F1.
2. The sign of F1x is positive if it points in the positive x-direction,
negative if it points in the negative x-direction.
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3. F1y is the magnitude of the y-component of vector F1.
4. The sign of F1y is positive if it points in the positive y-direction,
negative if it points in the negative y-direction.
UNIT VECTORS (3 – dimensional vectors)
Let
Vectors having a magnitude of unity with no units. Its purpose is to
describe a direction in space.
𝑖̂= unit vector pointing in the x – axis
𝑗̂ = unit vector pointing in the y – axis
𝑘̂ = unit vector pointing in the z – axis
y
𝑗̂
𝑖̂
x
𝑘̂
z
If A and B are in terms of their components:
A = Ax𝑖̂+ Ay𝑗̂+ Az𝑘̂
Addition:
and
B = Bx𝑖̂ + By𝑗̂ + Bz𝑘̂
A + B = (Ax + Bx)𝑖̂ +(Ay + By)𝑗̂+ (Az + Bz)𝑘̂
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PRODUCT OF VECTORS:
A) Scalar Product (Dot Product)
o Results to scalar quantity i.e. magnitude only, no direction.
A ∙ B = AB cos θ
B
Where A is the magnitude of vector A, and B is
the magnitude of vector B, and θ is the angle
between them.
θ
A

If θ = 90o, A∙B = AB cos 90o, cos 90o = 0, A∙B = 0; scalar product of 2
perpendicular vectors is always 0.

Using the unit vector computation:
A∙B = (Ax𝑖̂+ Ay𝑗̂+ Az𝑘̂) ∙ (Bx𝑖̂+ By𝑗̂+ Bz𝑘̂)
A∙B = AxBx + AyBy + AzBz
B) Vector Product (Cross Product)
o Vector quantity with a direction perpendicular to the plane of the
vector and a magnitude given by:


A x B = AB sin θ
If A and B are parallel, θ = 0 or 180o then A x B = 0 since, sin 0 & sin 180o =
0.
There are always two directions perpendicular to a given plane. Use the
right hand rule.
USING VECTOR REPRESENTATION:
A x B = (Ax𝑖̂+ Ay𝑗̂+ Az𝑘̂) x (Bx𝑖̂+ By𝑗̂+ Bz𝑘̂)
𝑖̂
Where:
𝑖̂ x 𝑖̂= 0
𝑖̂ x 𝑗̂= 𝑘̂
𝑘̂ x 𝑗̂= -𝑖̂
𝑗̂x̂𝑗 = 0
𝑗̂ x 𝑖̂ = - 𝑘̂
𝑖̂ x 𝑘̂ = -𝑗̂
𝑘̂ x 𝑘̂ = 0 𝑗̂x ̂𝑘 = 𝑖̂
𝑘̂
𝑘̂ x 𝑖̂ = 𝑗̂
+
𝑗̂
vector product of two parallel vectors is always zero.
̂
A x B = (AyBz – AzBy)𝒊̂ + (AzBx – AxBz)𝒋̂ + (AxBy = AyBx)𝒌
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SAMPLE PROBLEMS FOR VECTOR ADDITION:
1. Given are the following vector quantities:
A = 80 m due N
B = 40 m 300 NW
C = 60 m 150 NE
D = 60 m SE
Determine:
a. Magnitude and direction of the resultant of vectors A and B (using
Parallelogram Method)
b. Magnitude and direction of the resultant of vectors A and B (using
unit vectors)
c. Magnitude and direction of the Resultant of the four given vectors
using component method.
The given vectors drawn in the Cartesian-plane:
N
A=80m
B=40m
C=60m
300
150
W
45
E
0
D=60m
S
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31
a) Using PARALLELLOGRAM METHOD:
Let vector RAB represent the resultant of vectors A and B
N
RAB
A=80m
A=80m
1200
B=40m

300
W
E
By isolating the lower half of the parallelogram, our analysis in determining the
magnitude and direction of RAB can be determined by applying Sine and Cosine
Laws.
By cosine law:
S
RAB = √𝟖𝟎𝟐 + 𝟒𝟎𝟐 − 𝟐(𝟖𝟎)𝟒𝟎 𝐜𝐨𝐬 𝟏𝟐𝟎 = 𝟏𝟎𝟓. 𝟖𝟑𝒎
By sine law:
sin 𝛷 sin 120
=
105.83
80
Angle Φ = 40.89o
Direction of the resultant = 300 + 40.89o = 70.89o
THEREFORE: RAB = 105.83m, 70.89 NW (or North of West)

Note that the Triangle method will have the same solution as the
parallelogram method. Connect the two vectors from tip-to-tail. The
Resultant is a vector drawn from the tail of the first vector to the tip of the
second vector.
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B=40m
N
RAB
A=80m
W
E
S
b) USING COMPONENT
N
MET
A=80m
W
E
S
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X-COMPONENT: AX =0
Y-COMPONENT: AY=80m
N
By
B=40m
300
W
E
Bx
X-COMPONENT: BX = 40 cos 30 = 34.64m
Y-COMPONENT: BY = 40 sin 30 = 20m
S
By adding all x components of the vector: (vectors to the right “+”; vectors to the
left “-“), we have:
RABX = AX +BX = 0 +(-34.64) = - 34.64m
The negative sign means that the x-component of RAB is towards the left.
By adding all y-components of the vector: (vectors upward “+”; vectors
downward“-“), we have:
RABY = AY + BY = 80 + 20 = 100m
Since the value of the y-component of the resultant is positive, it means that it is
directed upwards.
The x and y components of the resultant can now be drawn as:
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N
RAB
RABY = 100m
AB
W
E
RABX = 34.64m
S
By Pythagorean Theorem:
RAB = √𝟑𝟒. 𝟔𝟒𝟐 + 𝟏𝟎𝟎𝟐 = 𝟏𝟎𝟓. 𝟖𝟑𝒎
From trigonometric functions of Right Triangles we have:
𝟏𝟎𝟎
Tan θAB = 𝟑𝟒.𝟔𝟒
θAB = 70.89O
THEREFORE: RAB = 105.83m, 70.89 NW (or North of West)
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c). Magnitude and direction of the Resultant of the four given vectors
From Previous solution:
The Components of vector A are:
X-COMPONENT: AX =0
Y-COMPONENT: AY=80m
The Components of vector B are:
X-COMPONENT: BX = 40 cos 30 = 34.64m
Y-COMPONENT: BY = 40 sin 30 = 20m
For Vector C, the components are:
N
C=60m
Cy
150
W
E
Cx
S
X-COMPONENT: CX = 60 cos 15 = 57.96m
Y-COMPONENT: CY = 60 sin 15 = 15.53m
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For Vector D, the components are:
N
W
E
45
0
D=60m
Dy
Dx
S
X-COMPONENT: DX = 60 cos 45o = 42.43m
Y-COMPONENT: DY = 60 sin 15o = 42.43m
x-component of the Resultant:
RX = AX + BX + CX + DX = 0 + (-34.64m) + 57.96m + 42.43m = 65.75m (to the right)
x-component of the Resultant:
RY = AY + BY + CY + DY = 80m + 20m + 15.53m + (-42.43m) = 73.1m (upward)
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N
Ry
R
θ
W
E
Rx
S
RX = √𝟔𝟓. 𝟕𝟓𝟐 + 𝟕𝟑. 𝟏𝟐 = 𝟗𝟖. 𝟑𝟐𝒎
𝟕𝟑.𝟏
Tan θ = 𝟔𝟓.𝟕𝟓
θAB = 48.03O
THEREFORE: R = 98.32m, 48.03 NE (or North of EAST)
Formative Problems:
1. A = 1km due south, B = 2km due west. Determine the resultant. (ans..2.24km,
63.4o W of S)
2. A = 72.4 m, 32.0° east of north, B = 57.3 m, 36.0° south of west, C = 17.8 m due
south. Determine the resultant. (ans..12.7m, 39o W of N)
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SAMPLE PROBLEMS FOR DOT PRODUCT:
SAMPLE PROBLEMS FOR CROSS PRODUCT:
Assignment # 1 (due on June 24, 2020)
1. Do the following conversions: (a) 15 m to ft, (b) 12 in to cm, (c) 30 days to sec
2. A football field is 300 ft long and 160 ft wide. What are the field’s dimensions
in meters and its area in square centimeters?
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3. In the Bible, Noah is instructed to build an ark 300 cubits long, 50 cubits wide
and 30 cubits high. (A cubit was a unit of length based on the length of the
forearm and equal to half of a yard.) What would the dimensions of the ark
be in meters? What would its volume be in cubic meters? (Assume that the
ark was rectangular.)
4. Which is longer and by how many centimeters, a 100-m dash or a 100-yd
dash?
5. Vector is A =2.80 cm long and is above the x-axis in the first quadrant. Vector
is 1.90 cm long and is below the x-axis in the fourth quadrant.
Find using triangle method: a) A + B, b) B-A, c) A-B
6. Three ropes pull on a large stone stuck in the ground, producing the vector
forces as shown in. Find the magnitude and direction of a fourth force on the
stone that will make the vector sum of the four forces zero.
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MODULE 2:
KINEMATICS
MOTION ALONG A STRAIGHT LINE (Rectilinear Motion)
Motion is apparent in widely ranging phenomena. Historically, motion was
one of the first phenomena to be studied carefully. Some progress was made in the
understanding of motion in ancient times, particularly by the philosophers of
classical Greece, but it was not until the Renaissance that the basic laws of motion
were discovered. Many individuals made important contributions, but two stand
above the rest: Galileo Galilei (1564-1642) and Isaac Newton (1642-1727).
Motion is defined as the continuous change in the position of a body. The
study of the motion of a body irrespective of the causes is the branch of mechanics
called Kinematics, which comes from a Greek word meaning “motion”.
In our study of motion we will be using a particle (an object whose
dimensions are negligible for the problem at hand and whose position is
represented by a mathematical point) as a model, which is actually a very small
body. However, larger bodies like a car or a sled can be represented by the
particle if all parts of it can be considered to be moving in the same way or it does
not rotate or change its shape while moving.
We start our study with the simplest type of motion a body can undergo. This
is called rectilinear motion or motion along a straight line. For the analysis, we will
be considering the line of motion as a coordinate axis, i.e. the x-axis if the line of
motion is horizontal or the y-axis if the line of motion is vertical.
2.1 BASIC CONCEPTS:
1. POSITION (x) of the body. This is to indicate the location of the body at any time as
it moves. It is the distance from a given reference point along the path at any time.


The reference point should not be confused with the starting point although
sometimes they are considered the same for convenience.
In cases where the line of motion is the y-axis, position is denoted by (y)
2. DISPLACEMENT (X)
certain length of time.
- this is the change in the position of the body during a
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41

the Greek letter “delta” () stands for “change in” a quantity.
X = X2 - X1
where
x1 - is the initial position or position at time t1
x2 - is the final position or position at time t2

Displacement is different from distance traveled in the sense that
displacement is a vector quantity directed from the initial to the final
position. However in rectilinear motion, the magnitude of the displacement
is the same as the distance travelled.
3. Time instant (t) and Time Interval (t)
Time instant is a point in time, i.e. at the time 5 seconds after starting or time 2
seconds before it stops, etc.
Time interval is a length of time, i.e. during the first 10 seconds or during the
time from t1 = 5 seconds to t2 = 10 seconds etc.
4. Velocity of the body. Generally, the velocity is the rate of change in the position
of the body. From this, it can be seen how fast a body is moving including its
direction of motion.
Average Velocity (vav) is the velocity of a body taken during a time interval
𝒗𝒂𝒗 =
𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕
𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍
=
∆𝒙
∆𝒕
or between two points along its path.
Instantaneous Velocity (v) is the velocity of a body at a particular time instant
or point along its path.
5. Acceleration of the body is the rate of change in the velocity of the body. A
body is said to be accelerating when the velocity is changing. This can happen in
the form of a change in the magnitude of the velocity i.e. the body moves faster
and faster or slower and slower, or in the form of a change in the direction. In
rectilinear motion however, acceleration can happen mostly in the form of a
change in magnitude (except at the instant when the body reverses its direction).
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Average acceleration (aav) is the acceleration of the body taken during a
time interval or during a certain displacement.
𝒂𝒂𝒗 =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 ∆𝒗
=
∆𝒕
𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍
Instantaneous acceleration (a) is the acceleration at a given time instant or
point along its path.
2.2 Uniformly Accelerated Rectilinear Motion or UARM
UARM is one common type of rectilinear motion. In this type of motion, the
change in velocity for consecutive equal time intervals is constant. For example, if
the change in velocity during the first 5 sec is 20 m/s, the change in velocity during
the second 5 sec will also be 20 m/s, so also during the third 5 sec and so on. The
body therefore can be observed to be moving faster or slower at a uniform rate.
If the acceleration is in the same direction as the velocity the body moves faster.
In this case the acceleration is considered to be positive relative to the velocity. If
the acceleration is opposite to the velocity the body moves slower. In this case the
acceleration is said to be negative relative to the velocity.
Basic Equations used for Analyzing UARM:
To simplify the equations, it will be assumed here that initially (at t = 0), the
position X = 0, Thus the time interval t will be the same as time instant t because ti =
0 and tf = t.
The displacement X will also become same as position X because Xi = 0 and
Xf = X at time instant t.
Therefore vav = X/t = X/t and since vav = (vi + vf)/2 then
𝐗=
(𝐯𝐢 + 𝐯𝐟 )𝐭
𝟐
In UARM, the average acceleration is equal to the instantaneous acceleration
therefore
𝑎 = 𝐚𝐚𝐯 =
𝐯𝐟 − 𝐯𝐢
𝐭
The two equations above are considered as the basic equations for analyzing
UARM. Any problem involving UARM can already be solved or analyzed by just
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using these 2 equations. However, in many instances, the two will always be used
together to solve even only a single quantity. Therefore there is a need for
additional equations that will enable us to solve for a quantity directly using only
one equation. To obtain these additional equations, we simply solve the two basic
equations simultaneously by elimination of a particular unknown quantity. That’s
why these additional equations are called derived equations.
If vf is eliminated from the basic equations,
X = vi(t) + (1/2)at2
If vi is eliminated,
X = vf (t) - (1/2)at2
If t is eliminated
vf2 = vi2 + 2aX
Thus, there are 5 equations for analyzing UARM, each one having its own particular
application.
** note that the subscript “i” refer to initial property while the subscript “f” are final
properties.
Examples:
1. Determine the average speed of a car that travels 80 km/hr for 2 hr, 100 km/hr
for 1 hr and at 30 km/hr for 0.5 hr?
Illustration of path and behavior of motion:
0
1
2
3
Solution:
vave =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
=
𝛥𝑋
𝛥𝑡
Where:
ΔX = X01+X12+X23
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X01 = (80 km/hr)(2hr) = 160 km ---- displacement after the first 2 hours
X12 = (100 km/hr)(1hr) = 100 km – displacement during the next hour
X23 = (30 km/hr)(o.5hr) = 15 km – displacement during the last half hour
Therefore: ΔX = 160km + 100km + 15 km = 275km
Total time interval: Δt = 2 + 1 + 0.5 = 3.5 hr
Vave = 275 km/3.5 hr = 78.5714 km/hr
2. You normally drive on a freeway at an average speed of 105 km/hr, and the
trip takes 2 hr and 20 min. On a Friday afternoon, however, heavy traffic slows
you down and you drive the same distance at an average speed of only 70
km/hr. How much longer does the trip take?
Solution:
Time difference = time during rainy days – time during normal day
1 ℎ𝑟
Let: tn = time on a normal day = 2 hr + (20min x 60 𝑚𝑖𝑛) = 2.3333hrs
Solving for the distance travelled:
X = (105 km/hr)(2.3333hr) = 245 km
Solving for time of travel during rainy day (tR)
tR = 245 km/(70 km/hr) = 3.5 hrs
Therefore:
Time difference = tR –tN = 3.5hr – 2.3333hr = 1.1667 hr
The trip during a rainy day takes 1.1667 hrs longer compared to that on a
normal day.
3. A car moving with constant acceleration covers the distance between two
points 60 m apart in 6s. Its velocity as it passes the second point is 15 m/s. (a)
what is its velocity at the first point? (b) What is the acceleration?
X = 60m ; t = 6s; a =?
1
2
V1 =?
V2 = 15 m/s
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a. Solving for initial velocity:
X = (vi + vf) t / 2
60m = (v1 + 15 m/s)(6s)/2
v1 = 5 m/s
b. Solving for acceleration:
X = vf (t) - (1/2)at2
60m = (15m/s)(6s) – ½(a)(6s)2
a = 1.6667 m/s2
2.3 Free Falling Bodies:
An excellent example of motion with (nearly) constant acceleration is a free
falling body. A body, which is dropped from a roof of a building, is a freely falling
body if there is no air resistance. However, there is a more accurate meaning of a
free falling body. It is a body such that the only force acting on it is the pull of gravity.
This implies that there is no air resistance or air friction, for air friction is considered a
force acting on a moving body.
The motion of falling bodies has been studied with great precision. When the
effects of air can be neglected, Galileo is right; all bodies at a particular location fall
with the same downward acceleration, regardless of their size or weight. (The
constant acceleration of a free falling body is called the acceleration due to
gravity.) If the distance of the fall is small compared to the radius of the earth, the
acceleration is constant.
In the following discussion we use an idealized model in which we neglect the
effects of the air, the earth’s rotation,, and the decrease of acceleration with
increasing altitude.
a.
b.
c.
d.
e.
y = (vi + vf)t/2
vf = vi + gt
y = vit + ½ gt2
y = vft - ½ gt2
vf2 = vi2 + 2gy
Where:
y = vertical displacement of the free falling body
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g = constant acceleration due to gravity
vf = final velocity
vi = initial velocity
t = time elapsed
Note: for convenience, the following concept should be applied:
 The value of acceleration due to gravity “g” is always negative (g = -9.81 m/s2; g
= -32.2 ft/s2; g = - 981 cm/s2)
 Velocities directed upward are positive, while velocities directed downward are
negative.
 If the vertical displacement “y” is above the reference point (starting point) of
the free falling body, it is positive, otherwise it is negative.
 The velocity at the highest point of a free falling body thrown vertically upward is
zero.
EXAMPLE PROBLEM ON FREE-FALL:
You throw a ball vertically upward from the roof of a tall building. The ball leaves
your hand at a point even with the roof railing with an upward speed of 15 m/s
,the ball is then in free fall. On its way back down, it just misses the railing. Find (a)
the ball’s position and velocity 1.00 s and 4.00 s after leaving your hand; (b) the
ball’s velocity when it is 5.00 m above the railing; (c) the maximum height
reached; (d) the ball’s acceleration when it is at its maximum height.
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Solution:
Assume that starting (y = 0 ) is the point where the object was released.
Given: vi = 15 m/s upward --- initial velocity of the ball
a) Find the ball’s position and velocity 1.00 s and 4.00 s after leaving your hand
At t = 0:
Position:
y = vit + ½ gt2
y = (15m/s)(1s) + 1/2 (-9.81m/s2)(1s)2
y = 10.095 m
Velocity:
vf = vi + gt
vf = 15m/s +(-9.81m/s2)(1s)
vf = 5.19 m/s
1 second after release, the ball is 10.095 m above starting point with a velocity of
5.19 m/s and directed upward
At t = 4s:
Position:
y = vit + ½ gt2
y = (15m/s)(4s) + 1/2 (-9.81m/s2)(4s)2
y = -18.48 m
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Velocity:
vf = vi + gt
vf = 15m/s +(-9.81m/s2)(4s)
vf = -24.24 m/s
4 second after release, the ball is 18.48 m below starting point with a velocity of
24.24 m/s m/s and directed downward.
b) ) the ball’s velocity when it is 5.00 m above the railing
vf2 = vi2 + 2gy
vf2 = (15 m/s)2 + 2(-9.81m/s2)(5m)
vf = ±11.26 m/s
We get two values of because the ball passes through the point y = 5m,twice, once
on the way up (so is positive) and once on the way down (so is negative)
c) the maximum height reached
vf2 = vi2 + 2gy at the maximum height final velocity vf =0
0 = (15m/s)2 + 2(-9.81m/s2)y
y = 11.47m
d) the ball’s acceleration when it is at its maximum height.
At any point along the path of the projectile acceleration is always constant,
therefore a = 9.81 m/s2 directed downward.
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2.1 Motion in Two-Dimensions
PROJECTILE MOTION
Projectile motion is an important special case of two dimensional motion.
A projectile is any body that is given an initial velocity and then follows a path
determined entirely by the effects of gravitational acceleration and air resistance.
The path followed by a projectile is called its trajectory.
To analyze this common type of motion, we’ll start with an idealized model,
representing the projectile as a single particle with an acceleration (due to gravity)
that is constant in both magnitude and direction. We’ll neglect the effects of air
resistance and the curvature and rotation of the earth.
Projectile motion is always confined to a vertical plane determined by the
direction of the initial velocity (velocity of projection). This is because the
acceleration due to gravity is purely vertical; gravity can’t move the projectile
sideways. Thus projectile motion is two-dimensional. We will call the plane of motion
the xy-coordinate plane, with the x-axis horizontal and the y-axis vertically upward.
The x-component of acceleration is zero, and the y-component is constant and
equal to –g (By definition, g is always positive; with our choice of coordinate
directions, ay is negative.)
In projectile motion
a = g = 9.80 m/s2 directed downward
Thus,
ax = 0
and
ay = - 9.80 m/s2
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EQUATIONS USED IN ANALYZING PROJECTILE MOTION
1.
HORIZONTAL COMPONENT
x = vxt
2.
VERTICAL COMPONENT
𝑣𝑖𝑦 +𝑣𝑓𝑦
a. y = (
b.
c.
d.
e.
2
)t
vfy = viy + gt
y = viyt + ½ gt2
y = vfyt - ½ gt2
vfy2 = viy2 + 2gy
EXAMPLE PROBLEM ON PROJECTILE MOTION
1. An object is launched at a velocity of 20 m/s in a direction making an
angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground)
of the object?
c) What is the horizontal range (maximum x above ground) of the object?
a) What is the maximum height reached by the object?
The formulas for the components Vx and Vy of the velocity and
components x and y of the displacement are given by:
Vx = V0 cos(θ)
Vy = V0 sin(θ) - g t
x = V0 cos(θ) t
y = V0 sin(θ) t - (1/2) g t2
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51
In the problem V0 = 20 m/s, θ = 25° and g = - 9.8 m/s2.
The height of the projectile is given by the component y, and it reaches its
maximum value when the component VBY is equal to zero. That is when the
projectile changes from moving upward to moving downward.(see figure above)
VBY = V0 sin(θ) + g t
0 = (20m/s)sin25 +(-9.81m/s2)t
Solving for t = 0.86s (time to reach maximum height)
Find the maximum height by substituting t by 0.86 seconds in the
formula for y
𝒗𝟎𝒚 +𝒗𝒃𝒚
y=(
𝟐
)t
y = (20sin25 +0)(0.86s)/2
y = 3.635m
b) What is the total flight time (between launch and touching the
ground) of the object?
The time of flight (t) is the interval of time between when projectile is launched until
the projectile touches the ground and is located at y = 0.
Thus
y = voyt + ½ gt2
0 = (20sin25)(t) + ½(-9.81m/s2)t2
Solving for t
t = 1.72 s
c) What is the horizontal range (maximum x above ground) of the
object?
x = voxt
Where t is the time from launch to point where projectile nearly
reaches the ground. In this case it is the time solved in letter (b)
t =.172 s
Therefore X = (20cos25)(1.72s)
X = 31.176m
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ASSIGNMENT 2: DUE ON JUNE 24
1. A subway train starts from rest at a station and accelerates at a rate of 1.6 m/s2
for 14 s. It then runs at constant speed for 70 s after which it slows down at a
rate of 3.5 m/s2 until it stops at the next station. Find the total distance covered.
2. A freight train moving at an initial speed of 40 m/s puts on its breaks, producing
a deceleration of 0.5 m/s2. (a) How long will it take the train to travel the next
100 m? (b) At what speed will it be traveling the end of this 100 m?
3. How fast must a ball be thrown vertically upward to reach a height of 12 m
from the point where it was thrown? How long will it take for the ball to go back
to its original position?
4. A champagne bottle is held upright 1.2 m above the floor as the wire around
its cork is removed. The cork then pops out, rises vertically and falls to the floor
1.4 s later. (a) What height above the bottle did the cork reach? (b) What
was the cork’s initial velocity? (c) What is its velocity jus before it strikes the
ground?
5. A boy throws a stone from the top of a building 46.0 m above ground. The
stone is thrown at an angle of 33.0° below the horizontal and strikes the ground
55.6 m away from the building, find the following: (a) Time of flight. (b) Initial
speed. (c) The magnitude and the direction of the velocity of the stone just
before it strikes the ground.
6. A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s
and at an angle of 36.9o above the horizontal. You can ignore air resistance.
(a) At what two times is the baseball at a height of 10 m above the point at
which it left the bat? (b) Calculate the horizontal and vertical components of
the baseball’s velocity at each of the two times calculated in part (a). (c)
What are the magnitude and direction of the baseball’s velocity when it
returns to the level at which it left the bat?
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MODULE 3:
NEWTON’S LAW OF MOTION
References
In Uniformly Accelerated Rectilinear Motion (UARM), we concentrated on the
analysis of the motion of a body irrespective of the factors that influenced the
motion. This is called kinematics. We now proceed to the study of the relation
between the motion of a body and the forces acting on it. It is here where we will
find out why a body remains at rest, why it accelerates, why it moves along a curve,
etc. This area of study is a branch of dynamic physics called kinetics.
Newton’s Laws of Motion:
The principles of dynamics are based on a set of laws formulated by Sir Isaac Newton.
(1642-1727)
The 1st Law (Law of Inertia)
Any body will remain at rest or in motion along a straight line with constant velocity
unless acted upon by a net external force.
 When the resultant of the forces acting on a body is zero, the acceleration of the
body is also zero.
Inertia – it is a property of a body that tends to preserve the state of rest of the body
when it is at rest or to maintain the motion of a body when it is in motion. The mass of
the body is a measure of its inertia.

The 2nd Law (Law of Acceleration)
When the vector sum of the forces acting on a body is not equal to zero, the body
will move with an acceleration that is: (a) in the same direction as the vector sum,
(b) directly proportional to the vector sum, (c) inversely proportional to the mass
of the body.
The 2nd Law therefore is just an extension of the 1st Law. The 1st Law tells us the state
of motion of the body if the vector sum is zero while the 2nd Law tells us the state of
motion if the vector sum is not equal to zero.

The 3rd Law (Law of Action and Reaction)


To every action there is an equal and opposite reaction.
Whenever one body exerts a force a second body, the second body exerts a
force back on the first that is equal in magnitude and opposite in direction.
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3.1 Basic Concepts
Before we go into further discussion of the laws and into their applications,
there are several related concepts and quantities that we should first be familiar with
in addition to the concepts on motion which were already considered in the previous
chapter.
1. Force – it is hard to define force with a single statement. It is better understood by
considering the different aspects about it as a physical quantity:
a. Force is exerted either as a push or a pull.
b. Force is a vector quantity.
c. In many systems, force is exerted by one body to another through contact
(contact force) but force can be exerted without the bodies in contact (noncontact force).
d. Any force acting on a body can be replaced by its components.
e. Any number of forces acting on a body can be replaced by a single force
which is the vector sum or the resultant of the forces.
2. Mass (m) – amount of matter, which a certain body, contains. It is a scalar
quantity that remains constant wherever it is.
3. Weight (W) – is the force exerted on a body by the earth due to gravity. It is always
considered as acting at the center of gravity of the body and, as a vector
quantity, it is directed downward.
Relation between Mass and Weight:
From Newton’s 2nd Law of motion:
W = mg
Mks
cgs
English
m
kg
g
Slug
W
N
dyne
lb
g
m/s2
cm/s2
ft/s2
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Example problem for Mass and Weight
1. At the surface of Mars the acceleration due to gravity is g = 3.72 m/s 2. A
watermelon weighs 52 N at the surface of the earth. (a) What is its mass on the
earth’s surface? (b) What are its mass and weight on the surface of Mars?
Given:
gm = 3.72 m/s2 = gravitational acceleration at the surface of moon
We = 52 N = weight of watermelon at the earth’s surface
a) To determine the mass of on the earth’s surface
me = We/g
me = 52N/9.81m/s2 = 5.3kg
b) To determine the weight of Watermelon on Mars (Wm)
Wm= mmgm
Note that the mass of a body is constant wherever it is located thus m m =me
Wm = 5.3kg(3.72m/s2) = 19.716N
4. Frictional force (f) and Normal force ()
Whenever two surfaces (bodies) in contact move or tend to move past each
other, the two surfaces will always exert two forces on each other. One is always
perpendicular to the surfaces in contact and it gives its name Normal Force while
the other is always parallel to the surface in contact and always opposite to the
direction of motion or impending motion. This is called Frictional Force or simply
friction.
Two general types of Frictional Force:
a. Static friction (fs) – frictional force generated when one surface starts to slide
across a surface (acts when motion is just impending, no actual motion yet).
b. Kinetic friction (fk) – Frictional force acting when the body is in motion
 For the same two surfaces in contact, the sliding friction is always less than the static
friction.
 The ratio of frictional force to the normal force is always constant. This constant
is called the coefficient of friction (µ). The value of this constant is always between 0
and 1 except in idealized cases wherein the surfaces are assumed to be smooth
wherein µ = 0.
Relation between Frictional Force and Normal Force:
f = 
Where  is the coefficient of friction.
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5. Tension (T)
A pulling force exerted on an object by a rope, cord, etc.
6. Free Body Diagram (FBD)
The construction of free body diagrams is an integral part of the analysis of
systems using Newton’s Law of motion. The FBD is a diagram of a body or object in
question isolated from the other parts of the system and showing all the forces
acting on it.
Example problem on Friction and Normal Force:
A box of weighing 20 N rests on a horizontal surface. The coefficient of static
friction between the box and the surface is 0.4 and the coefficient of sliding friction
is 0.2.
a. How large is the frictional force exerted on the box?
W = 15N
floor
N
Normal force (always perpendicular
To surfaces in contact)
Since the box is at rest, frictional force is static friction, thus
fs = sN
Since the box is at rest there are no unbalanced force along the vertical axis
therefore N = W = 15N
fs = (0.4)(15N) = 6N = static frictional force
** note that if an external horizontal force is greater than the static frictional
force the box will start to move along the direction of the horizontal force
applied.
b. How great will the friction force be if a horizontal force of P = 5 N is exerted on the
box?
W=15N
P=5N
W
N
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**Since the horizontal force P of 5N is less than the static frictional force of 6N,
the will remain at REST. Therefore, frictional force is static friction, fs = 6N
c. What is the minimum force that will start the box in motion?
A minimum force of 6N will start the box in motion (impending motion)
d. What is the minimum force that will keep the box in motion once it has been
started?
The minimum force that will keep the box in motion once it has been started is
equal to kinetic friction, thus:
Fk = kN = (0.2)(15N) =3N
e. If the horizontal force P is 10 N, how great is the frictional force?
Since the horizontal force P (15N) is greater than static friction (6N), the block
will be in motion, therefore the fictional force is kinetic friction fk =3N
3.2 Newton’s 1st Law and Equilibrium of a Particle
A very common application of Newton’s First Law is in the analysis of a particle in
equilibrium. A particle is an idealized model for anybody where the forces acting
can be considered to be acting at the same point.
A particle in equilibrium is a body that is either at rest or is moving along a straight
line at constant speed. Thus, as stated by Newton’s 1st Law, if a particle remains at
rest or is moving along a straight line, there is no unbalanced force acting on it. The
resultant of the forces acting on the body is equal to zero.
Therefore for anybody that is at rest or moving at constant velocity along a straight
line, the equation R = 0 can be applied to the forces acting on it, where R is the
resultant or vector sum of all forces. Using components, the equation can be broken
down into two:
1. Rx = 0 or the algebraic sum of all components along the x-axis is equal to zero.
2. Ry = 0 or the algebraic sum of all components along the y-axis is equal to zero.
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Example
1. A man is pushing a piano with mass 160 kg at constant velocity up a ramp that is
inclined at 36.90 above the horizontal. Coefficient of kinetic friction k = 0.2. If the
force applied by the man is parallel to the incline, calculate the magnitude of this
force.
k =0.2
F
36.90
F
Wx
FBD OF SYSTEM
fk
36.90
N
53.1
0
WY
W = 160kg(9.81m/s2)=1569.6N
STEP1: Make sure forces are along x or y axis. If not resolve forces into components.
In this case the weight W.
Wx = Wcos 53.10 = (1569.6N)cos 53.1 =942.4196 N
Wy = W sin 53.10 = (1569.6N)sin 53.1 = 1255.185 N
STEP2: Apply NEWTON’S FIRST LAW of MOTION
RY = 0
N – WY =0
N = WY =1255.185N
RX = 0
F – fK – WX = 0
F = fK – WX = 0.2(1255.185N) + 942.4196N = 1,193.4566N
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2. crates connected by a rope lie on a horizontal surface. Create A has a mass of100
kg, and crate B a mass of 150 kg. The coefficient of kinetic friction between the
crates and the surface is 0.4. The crates are pulled to the right at constant velocity
by a horizontal force P. determine:
a. the magnitude of the force P
b. the tension in the rope connecting the blocks.
A
FBD of BLOCK A:
fk (friction)
P
B
FBD of BLOCK B:
Y
T (Tension)
fk (friction)
Y
P
X
NA (Normal Force)
WA = 100kg(9.81m/s2) = 981N
STEP 1: Make sure forces are along x
or y axis. If not resolve forces into
components. In this case all forces lie
along the x or y axis.
STEP 2: Apply Newton’s First Law of
motion.
Ry = 0
NA –W = 0
NA = W = 981N
Rx =0
T –fk =0
T = fk =0.4(981N) =392.4 N
X
T (Tension)
NA (Normal Force)
WA = 150kg(9.81m/s2) = 1471.5N
Ry = 0
NA –W = 0
NA = W = 1471.5 N
Rx =0
P - T –fk =0
T =T - fk =392.4 - 0.4(1471.5N) =981 N
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FORMATIVE PROBLEM:
3. Determine the value of force F so that the block will move to the right at constant
velocity Coefficient of kinetic friction = 0.20 (Ans.. F = 130.544N)
Given:
F
300
W = 500N
Newton’s 2nd Law of Motion:
When a body accelerates, it means that the vector sum of the forces acting on it is
not equal to zero. It is not in equilibrium and Newton’s 1st Law does not apply, instead
it’s Newton’s 2nd Law.
In more direct terms Newton’s 2nd Law states that “The vector sum of the forces
acting on a moving body is equal to the product of its mass and its acceleration.”
In equation form:
R = ma
Where: R = the resultant or vector sum of the forces applied on the body
m = mass of the body
a = acceleration of the body
mks
cgs
English
R
N
dyne
Lb
m
kg
g
Slug
a
m/s2
cm/s2
ft/s2
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61
Using components, the equation can be broken down and applied in two forms:
1. Rx = max or the algebraic sum of all x components of forces is equal to the
product of the mass and the x-component of the acceleration.
2. Ry = may or the algebraic sum of all y components of forces is equal to the
product of the mass and the y-component of the acceleration.
Example problem.
1. A worker applies a constant horizontal force with magnitude 20 N to a box
with mass 40 kg resting on a level floor with negligible friction. What is the
acceleration of the box?
2. You walk into an elevator, step onto a scale, and push the “up” button. You
also recall that your normal weight is 625 N. Start answering each of the
following questions by drawing a free body diagram. (a) If the elevator has
an acceleration of magnitude of 2.5 m/s2, what does the scale read? (b) If
you start holding a 3.85-kg package by a light vertical string, what will be the
tension in this string once the elevator begins accelerating?
a) To determine your mass:
m = W/g = 625N/9.81=63.71kg
Ry = may
N – W = ma
N – 625N = 63.71kg (2.5m/s2)
N = 784.27N scale reading
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b) To Determine the tension in string:
Ry = may
T – W = ma
T -3.85kg (9.81m/s2) = 3.85kg (2.5m/s2)
T = 47.4N
3. A block of mass 6 kg resting on a horizontal surface is connected by a cord
passing over a light, frictionless pulley to a hanging block of mass 4 kg. The
coefficient of kinetic friction between the block and the horizontal surface is
0.5. After the blocks are released find: (a) the acceleration of each block (b)
The tension on the cord
6 kg
4 kg
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FBD of BLOCK A:
FBD of BLOCK B:
Y
ax = a
Y
T (Tension)
fk (friction)
x
aY = a
T
NA (Normal
Force)
x
Y
WA = 6kg(9.81m/s2) = 58.86N
STEP 1: In this case all forces lie
along the x or y axis.
STEP 2: Apply Newton’s 2ND Law of
motion.
RY = (mA)(aY) =0; (since ay = 0)
NA – WA =0
NA = WA =58.86N
**Forces having the same direction as
acceleration are POSITIVE; Forces whose
direction are opposite the direction of
acceleration are NEGATIVE.
RX = mA(ax)
T – fK = mAa
WB = 4kg(9.81m/s2) = 39.24N
Ry = (mB)aY
** note that since body Block A and B are
connected by a single cord, the acceleration
of block A is Equal to the acceleration of block
B. Thus ax = ay = a
WB – T = mBa
39.24N – T = (4kg)a
T = 39.24N – 4kg(a) -- equation 2
Solving equations 1 and 2
simultaneously:
a = 0.981 m/s2
T = 35.316 N
T – 0.5(58.86N) = 6kg(a)
T = (6kg)(a) + 29.43N – equation 1.
ASSIGNMENT #3 due on July 1, 2020
1. The object weighs 50 N and is supported by a cord. Find the tension on the
cord.
2. A bag of sugar weighs 5.00 lb on Earth. What would it weigh in Newtons on the
Moon, where the free-fall acceleration is one-sixth that on Earth? Repeat for
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3.
4.
5.
6.
7.
8.
Jupiter, where g is 2.64 times that on Earth. Find the mass of the bag of sugar
in kilograms at each of the three locations.
A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal
floor at constant speed. What is the coefficient of friction between floor and
box?
A 40 lb block is on a horizontal surface. A force of 70 lb is applied to the block
parallel to the surface. The coefficient of sliding friction is 0.2. What is the
acceleration of the block?
A 600 N man stands on a bathroom scale in an elevator. As the elevator
starts moving, the scale reads 800 N. (a) Find the magnitude and direction of
the acceleration. (b) What is the acceleration if the scale reads 450 N? (c) If
the scale reads zero, should the man worry? Explain.
A block of mass 6 kg resting on a horizontal surface is connected by a cord
passing over a light, frictionless pulley to a hanging block of mass 4 kg. The
coefficient of kinetic friction between the block and the horizontal surface is
0.5. After the blocks are released find: (a) the acceleration of each block (b)
The tension on the cord
A 15 lb block slides down a plane inclined at 300 to the horizontal. Find the
acceleration of the block: (a) if the plane is frictionless (b) if the coefficient of
kinetic friction is 0.4.
A 150-N bird feeder is supported by three cables as shown in the figure below.
Find the tension in each cable
.
9. Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light
string that passes over a frictionless pulley as in the figure above. The 5.00-kg
crate lies on a smooth incline of angle 40.0°. Find (a) the acceleration of the
5.00-kg crate and (b) the tension in the string.
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MODULE 4:
WORK, ENERGY AND POWER
WORK: The physicist's definition of work is based on these observations. Consider a
body that undergoes a displacement of magnitude s along a straight line. (For now,
we'll assume that anybody we discuss can be treated as a particle so that we can
ignore any rotation or changes in shape of the body.) While the body moves, a
constant force F acts on it in the same direction as the displacement s (Fig. 6.2).
We define the work W done by this constant force under these circumstances as the
product of the force magnitude F and the displacement magnitude s:
𝑾 = 𝑭𝒔 (constant force in direction of straight − line displacement)
The SI unit of work is the Joule (abbreviated J, pronounced “jewel," and named in
honor of the 19th-century English physicist James Prescott Joule). The British unit of
work is the foot-pound (ft-lb).
𝟏 𝑱𝒐𝒖𝒍𝒆 = (𝟏 𝑵𝒆𝒘𝒕𝒐𝒏)(𝟏 𝒎𝒆𝒕𝒆𝒓) = 𝟏 𝑵·𝒎
𝟏 𝑱 = 𝟎. 𝟕𝟑𝟕𝟔 𝒇𝒕 − 𝒍𝒃
𝟏 𝒇𝒕 − 𝒍𝒃 = 𝟏. 𝟑𝟓𝟔 𝑱
Think of a person pushing a stalled car. If he pushes the car through a displacement
s with a constant force F in the direction of motion, the amount of work he does on
the car is given by W = Fs. But what if the person pushes at an angle Φ with the car's
displacement (Fig. 6.3)?
In this case only the parallel component Fll is effective in moving the car, so we define
the work as the product of this force component and the magnitude of the
displacement. Hence W = Flls = (FcosΦ)s, or
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𝑾 = 𝑭𝒔𝒄𝒐𝒔𝜱 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒, 𝑠𝑡𝑟𝑎𝑖𝑔ℎ𝑡 − 𝑙𝑖𝑛𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡)
ENERGY: Energy can be defined as the capacity for doing work. The simplest case
of mechanical work is when an object is standing still and we force it to move.
Types of Mechanical Energy.
1. Kinetic Energy = Energy of Motion
2. Potential Energy = Stored Energy
Like work, the kinetic energy of a particle is a scalar quantity; it depends on only the
particle's mass and speed, not its direction of motion. A car has the same kinetic
energy when going north at 10 mls as when going east at 10 mls. Kinetic energy can
never be negative, and it is zero only when the particle is at rest.
𝟏
𝒎𝒗𝟐 (𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦)
𝟐
where m is the mass of the object, and v is the velocity of the object.
𝑲𝑬 =
Potential energy (also a scalar quantity) is the ability of a system to do work due to
its position or internal structure. Gravitational potential energy is energy of position.
An object's gravitational potential energy with respect to a reference level is
𝐏𝐄 = 𝐦𝐠𝐡 = (𝐖𝐞𝐢𝐠𝐡𝐭)𝐡(definition of potential energy)
Where m is the mass of the object, g=9.81 m/s2, and h is the vertical distance
above the reference level.
The total mechanical energy (TME) is given by the equation: 𝐓𝐌𝐄 = 𝐏𝐄 + 𝐊𝐄
If no energy is lost or gained from the movement of a body, the final TME of the
body is equal to its initial TME. The unit of energy is the same with work.
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POWER: Power is the time rate at which work is done. Like work and energy, power
is a scalar quantity. When a quantity of work W is done during a time interval t, the
average work done per unit time or average power Pave is defined to be
𝑾
𝑷 =
𝒕
The SI unit of power is the watt (W), named for the English inventor James Watt.
𝟏 𝑾 = 𝟏 𝑱/𝒔
𝟏 𝒌𝑾 = 𝟏𝟎𝟎𝟎 𝑾
𝟏 𝑴𝑾 = 𝟏𝟎𝟔 𝑾
In the British system, the unit of power is the foot-pound per second (ft-lb/s). A larger
unit called the horsepower (hp) is also used.
𝟏 𝒉𝒑 = 𝟓𝟓𝟎 𝒇𝒕 − 𝒍𝒃/𝒔
𝟏 𝒉𝒑 = 𝟕𝟒𝟔 𝑾 = 𝟎. 𝟕𝟒𝟔 𝒌𝑾
1. CJ is out with her friends. Misfortune occurs and CJ and her friends find
themselves getting a workout. They apply a cumulative force of 1080 N to push
the car 218 m to the nearest fuel station. Determine the work done on the car.
s = 218m
F = 1080N
𝑾 = 𝑭𝒔𝒄𝒐𝒔𝜱
W = (1080N)(218m)cos 0 = 235,440 N-m or J =235.44 kJ
*connote that angle Φ = 0 because the angle between force direction of F
and direction of s is 0.
2. A weight lifter lifts a 350-N set of weights from ground level to a position over
his head, a vertical distance of 2.00 m. How much work does the weight lifter
do, assuming he moves the weights at constant speed?
𝑾 = 𝜟𝑷𝑬 + 𝜟𝑲𝑬
Since velocity is uniform ΔKE = 0
W = (weight)(Δz)
W = (350N)(2m) = 700 J
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3. During the Powerhouse lab, Jerome runs up the stairs, elevating his 102 kg body
a vertical distance of 2.29 m in a time of 1.32 s at a constant speed.
a. Determine the work done by Jerome in climbing the stair case.
b. Determine the power generated by Jerome.
𝑾 = 𝜟𝑷𝑬 + 𝜟𝑲𝑬
Since velocity is uniform ΔKE = 0
W = (102kg)(9.81m/s2)(2.29m) = 2291.42J
𝑾
𝑷 =
𝜟𝒕
𝟐𝟐𝟗𝟏.𝟒𝟐𝑱
𝑷 = 𝟏.𝟑𝟐𝒔 = 1735.92 J/s or Watts
4. A new conveyor system at the local packaging plan will utilize a motorpowered mechanical arm to exert an average force of 890 N to push large
crates a distance of 12 m in 22 s. Determine the power output required of such
a motor.
W = FscosΦ
Assuming that Force has the same direction as displacement s, Φ = 0
W = (890N)(12m)cos0 = 10680 J
P = 10860J/22s =485.45 watts
5. A 78-kg skydiver has a speed of 62 m/s at an altitude of 870 m above the
ground.
a. Determine the kinetic energy possessed by the skydiver.
b. Determine the potential energy possessed by the skydiver.
c. Determine the total mechanical energy possessed by the skydiver.
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
KE = ½(78kg)(62m/s)2 = 149,916 J OR 149.916 kJ
𝐏𝐄 = 𝐦𝐠𝐡
2
PE = (78kg)(9.81m/s )(870m) = 665,706.6 J or 665.7066 kJ
TME = PE + KE
TME = 665.7066 KJ +149.916 KJ
Assignment #4: due on July 1, 2020
1. Olive Udadi is at the park with her father. The 26-kg Olive is on a swing following
the path as shown. Olive has a speed of 0 m/s at position A and is a height of
3.0-m above the ground. At position B, Olive is 1.2 m above the ground. At
position C (2.2 m above the ground), Olive projects from the seat and travels
as a projectile along the path shown. At point F, Olive is a mere picometer
above the ground. Assume negligible air resistance throughout the motion.
Use this information to fill in the table. (TME=total mechanical energy)
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Position Height PE
(m)
(J)
A
3
B
1.2
C
2.2
F
0
KE
(J)
TME
(J)
Speed
(m/s)
0
2. Suzie (m=56 kg) is skiing at Bluebird Mountain. She is moving at 16 m/s across
the crest of a ski hill located 34 m above ground level at the end of the run.
Determine:
a. Suzie's kinetic energy.
b. Suzie's potential energy relative to the height of the ground at the end
of the run.
c. Suzie's total mechanical energy at the crest of the hill.
d. If no energy is lost or gained between the top of the hill and her initial
arrival at the end of the run, then what will be Suzie's total mechanical
energy at the end of the run?
e. Suzie's speed as she arrives at the end of the run and prior to braking to
a stop.
3. Ima (m=56.2 kg) is traveling at a speed of 12.8 m/s at the top of a 19.5-m high
roller coaster loop.
a. Determine Ima's kinetic energy at the top of the loop.
b. Determine Ima's potential energy at the top of the loop.
c. Assuming negligible losses of energy due to friction and air resistance,
determine Ima's total mechanical energy at the bottom of the loop (h=0
m).
d. Determine Ima's speed at the bottom of the loop.
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71
MODULE 5:
IMPULSE AND MOMENTUM
IMPULSE (J)
Impulse of a force is defined as the product of a force and the time during
which it acts.
If we let F be the force and t be the time during which it acts, then
Impulse of the force
J = F(t); (N-s; Newton-seconds)
 Impulse (J) is a vector quantity. Its direction is the same as the direction of the
force.
 If the force varies with time, then
t
J = ∫t 2 Fdt
1
Where F is given as a function of time, t.
Example 1) A force of 100 N to the right is applied to a body for 5 seconds. What
is the impulse of the force?
J
F
J = Ft = (100N)(5s) = 500 N-s in the same direction as F
MOMENTUM (p)
Momentum is the product of the mass (m) of a body and its velocity (v).
p = m(v); (kg-m/sec)

Momentum is a vector quantity. Its direction is the same as the direction of
the velocity
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Example: A 10 kg block is initially moving to the right at 20 m/s. Determine the
change in momentum if : a ) finally the body is moving at 10 m/s to the left and b)
the body is finally moving at 5 m/s downward.
V1 = 20 m/s
Solution: a) initial condition:
p1
10 kg
p1 = mv1 = 10kg (20m/s) = 200 kg-m/s
condition:
final
V2 = 10 m/s
10 kg
p2
p2 = mv2 = 10kg (100m/s) = 100 kg-m/s
p = p2 – p1 = p2 + (–p1) = (-100 kg/m-s) + (– 200 kg-m/s) = - 300 kg-m/s
** FOLLOW THE RULES OF VECTOR ADDITION AND SUBTRACTION.
b) Initial condition:
V1 = 20 m/s
p1 =10kg(20m/s) = 200 kg-m/s
10 kg
Final condition:
10 kg
V1 = 5 m/s
p2
p2 = mv2 = 10kg (5m/s) = 50 kg-m/s
p = p2 – p1 = p2 + (-p1) (vector operation )
p = p2 – p1
p2 = 50 kg-m/s

- p1 = 200 kg-m/s
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Using triangle method of adding vectors:
p = √𝒑𝟐𝟐 + 𝒑𝟐𝟏 = √𝟓𝟎𝟐 + 𝟏𝟎𝟎𝟐 = 111.8 kg-m/s
Tan  = 200/50
 = 75.96o
The IMPULSE – MOMENTUM relation
As individual concepts, impulse and momentum would have few practical
applications. If we want to apply them in the analysis of physical phenomena, we
have to look at the relation between them.
Let a body of mass m be initially moving with a velocity v1. A force resultant
force F is then applied for a time t. The body will accelerate and attain a final
velocity v2 after time t.
“Impulse applied to a body or system is equal to the change in the momentum of
the body or system. “
J = p2 - p1
EXAMPLE PROBLEM ON IMPULSE-MOMENTUM RELATION:
1. You throw a ball with a mass of 0.40 kg against a brick wall. It hits the wall
moving horizontally to the left at 30 m/s and rebounds horizontally to the right
at 20m/s (a) Find the impulse of the net force on the ball during its collision
with the wall. (b) If the ball is in contact with the wall for 0.010 s, find the
average horizontal force that the wall exerts on the ball during the impact.
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2. A soccer ball has a mass of 0.40 kg. Initially it is moving to the left at but then
it is kicked. After the kick it is moving at 45° upward and to the right with
speed of 30 m/s). Find the impulse of the net force and the average net
force, assuming a collision time t = 0.01 seconds.
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The LAW of CONSERVATION of MOMENTUM
From the Impulse-Momentum equation, it can be seen that if no force F is
applied to a body or system, the final momentum is equal to the initial momentum
or the total momentum of the body or system is conserved. This is the Law of
Conservation of Momentum.
“The total momentum of the body or system is conserved as long as no external
force is applied to the body or system.”
Computation of the Velocity of Recoil
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A marksman holds a rifle of mass 3 kg loosely, so it can recoil freely. He fires a bullet
of mass 5 grams horizontally with a velocity relative to the ground of vBX = 300 m/s.
What is the recoil velocity of the rifle? What are the final momentum and kinetic
energy of the bullet and rifle?
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COLLLISION OF BODIES
Collision is the forceful contact (impacts) between two bodies or among
several bodies. Collision can be classified as elastic when the kinetic energy of the
colliding bodies is conserved before and after collision or inelastic when the total
energy after collision is less than the total energy before collision. Super elastic
collision is when total kinetic energy after collision is greater than the total kinetic
energy before collision.
In all types of collision, the law of conservation of momentum applies, that is
“The total momentum before collision = The total momentum after collision”
Collision is quite complicated to analyze especially when the bodies
disintegrate in several parts after collision. However, if we assume that the colliding
bodies remain intact after collision, then the analysis can be highly simplified. We
will only be concerned with the changes in the velocities of the colliding bodies.
In many cases, the law of conservation of momentum is not enough to
analyze and predict the outcome after collision. That’s why if necessary, the
following additional relations can be used:
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1. For elastic collisions:
“The total kinetic energy before collision = The total kinetic energy after
collision”
2. For all types of collisions
The coefficient of restitution (e) is equal to the negative ratio of the
relative velocity after collision to the relative velocity before collision.
𝑒= −
𝑢1 − 𝑢2
𝑢2 − 𝑢1
=
𝑣1 − 𝑣2
𝑣1 − 𝑣2
Where: v1 = velocity of body 1 before collision
V2 = velocity of body 2 before collision
u1 = velocity of body 1 after collision
u2 = velocity of body 2 after collision
If the collision is elastic, e = 1. If inelastic, e has value less than 1.
If applicable and necessary, it is also possible to break down the law of
conservation of momentum into component, that is:
“The total momentum along the x-axis before collision = total momentum
along the x-axis after collision.” and
“The total momentum along the y-axis before collision = total momentum
along the y-axis after collision.”
This is usually being done in cases where the collision is oblique, that is, the bodies
do not move along the same line before and after collision.
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81
Assignment #4: due on July 9, 2020
1. A 20 kg body is initially moving at 30 m/s to the right along a frictionless
horizontal surface when a force F = 80 N is applied to it for 10 s. Determine its
final velocity if a) F is directed to the right, b) F is directed to the left, and c) F
is applied at an angle of 60o N of W.
2. A 0.25 kg baseball is travelling horizontally to the right at 40 m/s when it was
hit by a bat and it bounced off at 60 m/s 60o above the horizontal to the left.
How much impulse was applied to the ball by the baseball bat?
3. A 2.5 kg rifle fires a 50 gram bullet with a velocity of 300 m/s. Determine a)
the velocity of recoil of the rifle and b) the kinetic energy that will be
absorbed by the one firing the rifle.
4. A 5 kg block moving along a frictionless surface at 2 m/s to the right collides
with a 10 kg block moving to the left at 5 m/s. Determine their velocities after
collision if a) they stick together after collision;
b) the collision is elastic;
and c) the collision is inelastic wherein the coefficient of restitution is 0.8. How
much kinetic energy is lost in this particular collision?
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MODULE 6:
ROTATIONAL MOTION
Rotational motion plays an important role in nature, and here we investigate the
behavior of rigid bodies when they rotate. A rigid body is one that does not deform
as it moves. The equations involved here are similar to those that describe linear
translational motion.
What is a radian?
Consider a planar object rotating about an axis perpendicular to its plane.
We describe the position of a point on the object by the coordinates r and θ,
where θ is measured with respect to the
x-axis, as in Figure 1. When the object turns through an angle θ, the point moves a
𝒔
distance s along the arc. We define the angle θ in radians as 𝜽 = 𝒓 or 𝒔 = 𝒓𝜽 . 1
radian is an angle subtended at the center of the circle by an arc of length equal
to the radius of the circle.
Figure 1
You can see that if θ is doubled, the arc length s will also be
doubled. Since θ is the ratio of two lengths, it is a dimensionless
quantity. The circumference of a circle is 𝒔 = 𝟐𝝅𝒓 so θ for a full
circle is 2π. Thus 2π rad = 360o. It is easy to convert radians to
degrees or degrees to radians using a ratio
𝜽 (𝒓𝒂𝒅𝒊𝒂𝒏𝒔)
𝟐𝝅
=
𝜽 (𝒅𝒆𝒈𝒓𝒆𝒆𝒔)
𝟑𝟔𝟎𝒐
What is RPM?
RPM means revolutions per minute (rev/min). One revolution is equal to 2π
rad or 360o. Sometimes the angular velocity of a rotating body is expressed in RPM.
EXAMPLES:
1. What angle in radians is subtended by an arc 3 m in length, on the
circumference of a circle whose radius is 2 m?
𝒔
3𝑚
SOLUTION: 𝜽 = 𝒓 = 2𝑚 = 𝟏. 𝟓 𝒓𝒂𝒅𝒊𝒂𝒏
2. What angle in radians is subtended by an arc of length 78.54 cm on the
circumference of a circle of diameter 100 cm? What is the angle in degrees?
𝒔
78.54 𝑐𝑚
SOLUTION: 𝜽 = 𝒓 = 100
= 𝟏. 𝟓𝟕𝟎𝟖 𝒓𝒂𝒅𝒊𝒂𝒏
(
2
) 𝑐𝑚
𝜃 = 1.5708 𝑟𝑎𝑑𝑖𝑎𝑛 ∗
360𝑜
= 𝟗𝟎𝒐
2𝜋 𝑟𝑎𝑑
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3. The angle between the two radii of a circle of radius 2 m is 0.60 rad. What
length of arc is intercepted on the circumference of the circle by the two
radii?
SOLUTION: 𝑠 = 𝑟𝜃 = 2𝑚 ∗ 0.60 = 𝟏. 𝟐 𝒎
4. What is the angular velocity (in rad/s) of the crankshaft of an automobile
engine that is rotating at 4800 RPM?
𝑟𝑒𝑣
1 𝑚𝑖𝑛
2𝜋 𝑟𝑎𝑑
𝒓𝒂𝒅
SOLUTION: 𝜔 = 4800 𝑚𝑖𝑛 ∗ 60 𝑠 ∗ 1 𝑟𝑒𝑣 = 𝟓𝟎𝟐. 𝟔𝟓𝟒𝟖 𝒔
Average angular velocity
Average angular velocity 𝜔𝑎𝑣𝑒 of the body in the time interval ∆𝑡 = 𝑡2 − 𝑡1 is
the ratio of the angular displacement ∆𝜃 = 𝜃2 − 𝜃1 to ∆𝑡.
𝝎𝒂𝒗𝒆 =
∆𝜽
∆𝒕
=
𝜽𝟐 − 𝜽𝟏
𝒕𝟐 − 𝒕𝟏
Instantaneous angular velocity 𝜔
The angular velocity of a rotating body may be constant, may be increasing
or decreasing. At one particular moment, the angular velocity of the body is called
the instantaneous angular velocity. It is the limit of the average angular velocity as
∆𝑡 approaches zero, i.e., the first derivative of 𝜃 with respect to time.
EXAMPLE:
A merry-go-round is being pushed by a child. The angle the merry-go-round has
𝒓𝒂𝒅
turned through varies with time according to the equation 𝜽𝒕 = (𝟐 𝒔 ) 𝒕 +
(𝟎. 𝟎𝟓
𝒓𝒂𝒅
𝒔𝟑
) 𝒕𝟑 .
a) Calculate the angular velocity of the merry-go-round as a function of time.
b) What is the initial value of the angular velocity?
c) Calculate the instantaneous value of the angular velocity at t = 5 s and the
average angular velocity for the time interval t = 0 to t = 5 s.
SOLUTION:
a) The angular velocity as a function of time is the first derivative of θ with
respect to time.
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𝝎(𝒕) =
𝒅𝜽
𝒓𝒂𝒅
𝒓𝒂𝒅
=𝟐
+ (𝟎. 𝟏𝟓 𝟑 ) 𝒕𝟐
𝒅𝒕
𝒔
𝒔
b) The initial value of the angular velocity is taken when time t = 0.
𝑟𝑎𝑑
𝑟𝑎𝑑
𝒓𝒂𝒅
𝜔𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝜔𝑖 = 2
+ (0.15 3 ) 02 = 𝟐
𝑠
𝑠
𝒔
c) When the time t = 5 s, the instantaneous angular velocity is
𝑟𝑎𝑑
𝑟𝑎𝑑
𝒓𝒂𝒅
𝜔5 𝑠𝑒𝑐 = 2
+ (0.15 3 ) (5 𝑠)2 = 𝟓. 𝟕𝟓
𝑠
𝑠
𝒔
For the average angular velocity for the time interval from t = 0 to t = 5 s, first
calculate the value of θ at time t = 0, then at time t = 5 s.
𝑟𝑎𝑑
𝑟𝑎𝑑
When t = 0; 𝜃1 = (2 𝑠 ) (0) + (0.05 𝑠3 ) (0)3 = 0
When t = 5 s;
𝜔𝑎𝑣𝑒 =
𝜃2 = (2
∆𝜃
∆𝑡
=
𝜃2 − 𝜃1
𝑡2 − 𝑡1
=
𝑟𝑎𝑑
𝑠
) (5 𝑠) + (0.05
16.25 𝑟𝑎𝑑−0
5 𝑠−0
𝑟𝑎𝑑
= 𝟑. 𝟐𝟓
𝑠3
) (5 𝑠)3 = 0 = 16.25 𝑟𝑎𝑑
𝒓𝒂𝒅
𝒔
Angular acceleration α
If the angular velocity of a rotating body changes, there is an angular
acceleration. It is understood that the angular velocity of the body either increases
or decreases.
Average angular velocity 𝜶𝒂𝒗𝒆 =
Instantaneous angular
∆𝝎
∆𝒕
=
𝝎𝟐 − 𝝎𝟏
𝒕𝟐 − 𝒕𝟏
acceleration
EXAMPLE:
𝒓𝒂𝒅
A rigid object rotates with an angular velocity that is given by 𝝎(𝒕) = 𝟒 𝒔 −
(𝟎. 𝟖
𝒓𝒂𝒅
𝒔𝟑
) 𝒕𝟐
a) Calculate the angular acceleration as a function of time.
b) Calculate the instantaneous angular acceleration at t = 2 s and the average
angular acceleration for the time interval t = 0 to t= 2 s.
SOLUTION:
a) The angular acceleration as a function of time is the first derivative 𝜔 of with
respect to time.
𝒅𝝎
𝒓𝒂𝒅
𝜶=
= − (𝟏. 𝟔 𝟑 ) 𝒕
𝒅𝒕
𝒔
b) When the time t = 2 s, the instantaneous angular acceleration is
𝑟𝑎𝑑
𝛼2 𝑠𝑒𝑐 = − (1.6 3 ) (2 𝑠) = −𝟑. 𝟐 𝒓𝒂𝒅/𝒔𝟐
𝑠
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For the average angular acceleration for the time interval from t = 0 to t = 2 s,
first calculate the value of 𝜔 at time t = 0, then at time t = 2 s.
𝑟𝑎𝑑
𝑟𝑎𝑑
When t = 0; 𝜔1 = 4 𝑠 − (0.8 𝑠3 ) 02 = 0
When t = 2 s;
𝛼𝑎𝑣𝑒
𝜔2 = 4
𝑟𝑎𝑑
𝑠
− (0.8
𝑟𝑎𝑑
𝑠3
) (2 𝑠)2 = 0.8 𝑟𝑎𝑑/𝑠
𝑟𝑎𝑑
0.8 𝑠 − 0
∆𝜔
𝜔2 − 𝜔1
𝒓𝒂𝒅
=
=
=
= 𝟎. 𝟒 𝟐
∆𝑡
𝑡2 − 𝑡1
2𝑠−0
𝒔
MOTION WITH CONSTANT ANGULAR ACCELERATION
Whenever a body rotates with constant angular acceleration, its change of
angular velocity is constant for equal time interval. For example, a body which is
initially at rest, rotates with constant acceleration of 4 rad/s 2. When time is 0, the
angular velocity of the body is 0, after 1 sec its angular velocity is 4 rad/s, after 2 sec
(from starting time), its angular velocity is 8 rad/s, after 3 sec ( from starting time), its
angular velocity is 12 rad/s, etc.
Relation of Linear and Angular Quantities:
Motion with constant linear
acceleration
(a = constant)
𝑣𝑖 + 𝑣𝑓
𝑥=
𝑡
2
Motion with constant angular
acceleration
(α = constant)
𝜔𝑖 + 𝜔𝑓
𝜃=
𝑡
2
𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
1
𝑥 = 𝑣𝑖 𝑡 + 𝑎𝑡 2
2
1
𝜃 = 𝜔𝑖 𝑡 + 𝛼𝑡 2
2
1
𝑥 = 𝑣𝑓 𝑡 − 𝑎𝑡 2
2
1
𝜃 = 𝜔𝑓 𝑡 − 𝛼𝑡 2
2
𝑣𝑓2
=
𝑣𝑖2
𝜔𝑓2 = 𝜔𝑖2 + 2𝛼𝜃
+ 2𝑎𝑥
EXAMPLES:
1. An electric motor is turned off, and its angular velocity decreases uniformly
from 1000 RPM to 400 RPM in 5 sec.
a) Find the angular acceleration of the motor.
b) Find the number of revolutions the motor made in the 5-s interval.
c) After the 5-s interval, how many more seconds are required by the motor
to come to rest?
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SOLUTION:
𝑟𝑒𝑣 1 𝑚𝑖𝑛 2𝜋 𝑟𝑎𝑑
𝑟𝑎𝑑
∗
∗
= 104.7198
𝑚𝑖𝑛 60 𝑠
𝑟𝑒𝑣
𝑠
𝑟𝑒𝑣 1 𝑚𝑖𝑛 2𝜋 𝑟𝑎𝑑
𝑟𝑎𝑑
𝜔1 = 400
∗
∗
= 41.8879
𝑚𝑖𝑛 60 𝑠
𝑟𝑒𝑣
𝑠
𝜔𝑖 = 1000
a) 𝜔1 = 𝜔𝑖 + 𝛼𝑡
𝑟𝑎𝑑
𝑟𝑎𝑑
41.8879
= 104.7198
+ 𝛼 (5 𝑠)
𝑠
𝑠
𝜶 = −𝟏𝟐. 𝟓𝟔𝟔𝟒
b) 𝜃 =
𝜔𝑖 +𝜔1
2
𝑡=
𝑟𝑎𝑑
𝑟𝑎𝑑
+41.8879
)
𝑠
𝑠
(104.7198
2
𝜃 = 366.5192 𝑟𝑎𝑑 ∗
𝒓𝒂𝒅
𝒔𝟐
∗5𝑠
1 𝑟𝑒𝑣
= 𝟓𝟖. 𝟑𝟑𝟑𝟑 𝒓𝒆𝒗
2𝜋 𝑟𝑎𝑑
c) 𝜔𝑓 = 𝜔1 + 𝛼𝑡
0 = 41.8879
𝒕 = 𝟑. 𝟑𝟑𝟑𝟑 𝒔
𝑟𝑎𝑑
𝑟𝑎𝑑
+ (−12.5664 2 ) 𝑡
𝑠
𝑠
2. The angular velocity of a bicycle wheel is 4 rad/s at t = 0, and its angular
acceleration is constant and equal to 2 rad/s2. A spoke OP on the wheel is
horizontal at t =0.
a) What angle does this spoke make with the horizontal at time t = 3 s?
b) What is the wheel’s angular velocity at this time?
SOLUTION:
a) The angle θ is given as a function of time by the equation
1
𝜃 = 𝜃𝑖 + 𝜔𝑖 𝑡 + 𝛼𝑡 2
2
Since the spoke initially positioned horizontally, 𝜃𝑖 = 0
𝑟𝑎𝑑
1 𝑟𝑎𝑑
𝜃 = 0 + (4
) (3 𝑠) + (2 2 ) (3 𝑠)2 = 21 𝑟𝑎𝑑
𝑠
2
𝑠
1 𝑟𝑒𝑣
𝜃 = 21 𝑟𝑎𝑑 ∗
= 3.3423 𝑟𝑒𝑣
2 𝜋 𝑟𝑎𝑑
 The body turns through three complete revolutions plus an additional
360𝑜
0.3423 rev
( 0.3423 𝑟𝑒𝑣 ∗ 1 𝑟𝑒𝑣 = 123.228𝑜 ). The line OP thus turns
through 123.228o and makes an angle of 56.772o(180o-123.228o) with
the horizontal.
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b) At time t = 3 s
𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡 = 4
𝑟𝑎𝑑
𝑟𝑎𝑑
𝒓𝒂𝒅
+ (2 2 ) (3 𝑠) = 𝟏𝟎
𝑠
𝑠
𝒔
TORQUE:
The ability of a force to rotate a body about some axis is measured by a
quantity called torque  (Greek letter ‘tau’).
a) The torque due to a force of F has a magnitude of
  Fd
i)  is torque (N·m).
ii) F is the applied force (N).
iii) d is the lever arm (also called the moment arm) distance (m).
b) The lever arm is the  distance from the axis of rotation to a line drawn
along the direction of the force. Note that
d  r sin 
Where r is the magnitude of the displacement from the axis to the point of
the applied force F and Φ is the angle between the direction of r and the
direction of F.
c) As a result, we can rewrite the torque equation as
  Fr sin 
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d) Torque is actually a vector which points in a direction perpendicular to the
plane defined by the F and r vectors.
e) The net torque is found by summing all torques (i.e., multiple forces acting
on a rotating object).
N
 net   i   1   2  ...   N  F1d1  F2 d 2  ...  FN d N
i 1
i)  is positive if the rotation is counter clockwise (CCW).
ii)  is negative if the rotation is clockwise (CW).
Assignment #6. Due July 9, 2020
1. (a) What angle in radians is subtended by an arc 1.50 m long on the
circumference of a circle of radius 2.50 m? What is this angle in degrees? (b)
An arc 14.0 cm long on the circumference ofa circle subtends an angle of
128o. What is the radius of the circle?
(c) The angle between two radii of a circle with radius 1.50 m is 0.700 rad.
What length of arc is intercepted on the circumference of the circle by the
two radii?
2. A turntable rotates with a constant 2.25 rad/s2 angular acceleration. After
4.00 s it has rotated through an angle of 60.0 rad. What was the angular
velocity of the wheel at the beginning of the 4.00-s interval?
3. A circular saw blade 0.200 m in diameter starts from rest. In 6.00 s it
accelerates with constant angular acceleration to an angular velocity of 140
rad/s. Find the angular acceleration and the angle through which the blade
has turned.
4. At t=0 a grinding wheel has an angular velocity of 24 rad/s. It has a constant
angular acceleration of 30 rad/s2 until a circuit breaker trips at t = 2s. From then
on, it turns through 432 rad as it coasts to a stop at constant angular
acceleration. (a) Through what total angle did the wheel turn between t = 0
and the time it stopped? (b) At what time did it stop? (c) What was its
acceleration as it slowed down?
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MODULE 7:
OSCILLATION
PERIODIC MOTION
A marble rolling back and forth on its springs, and a pendulum bob keeping time in
a clock are all examples of periodic motion or oscillation, a motion that repeats itself
in a definite cycle. Periodic motion occurs whenever a body has a stable equilibrium
position and a restoring force that acts when it is displaced from equilibrium.
Basic Concepts:
1. Equilibrium Position (e.p.) – the position of the body when the forces acting on
it have zero resultant; the position of the body when it is at rest.
2. Restoring Force (F) – the net force, acting on a body, directed back toward
the equilibrium position. It is called restoring force because it acts to restore
equilibrium. (newtons)
3. Displacement (x) – the distance of the body from the e.p. at any instant.
(meter)
4. Amplitude (A) – is the m maximum magnitude of displacement from
equilibrium – that is, the maximum value of x. (meter)
5. Period (T) – is the time for one cycle (one complete roundtrip). It is always
positive. (seconds)
6. Frequency (f) – is the number of cycles in a unit of time. (hertz)
A particular kind of periodic motion is known as simple harmonic motion. When the
restoring force is directly proportional to the displacement from equilibrium, the
oscillation is called simple harmonic motion (SHM). When an object is disturbed from
equilibrium, its motion is probably simple harmonic motion. Here are some examples
of periodic motion that approximate simple harmonic motion:
Harmonic oscillator – a body that undergoes simple harmonic motion.
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Equations for SHM:
Our prototype for SHM is a mass attached to a spring.
Using Hooke's law, we have that
Fx
F = kx
F = -kx
 equation 1
By Newton’s 2nd Law
F = ma
 equation 2
Equate 1 and 2 yields
a=-
𝒌
𝒎
x
acceleration of a body in SHM
where: k = force constant of spring, N/m
m = mass of body, kg
x = displacement, m
SHM and Conservation of Energy
From conservation of energy, we then find that the energy at any point is
Total Energy = Kinetic energy + Potential Energy
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91
E=K+U
E = ½mv2 + ½kx2
 constant
Consider the body at the extreme position or at xmax = A, v = 0, then
E = U = ½kA2
Because E is constant : ½kA2 = ½mv2 + ½kx2
A
Then, v 
 x2
2
 mk
instantaneous velocity
of the vibrating body
But Period
m
k
T  2
and Frequency
f 
1
1

T 2
k
m
It is usually more convenient to work with the angular frequency, . This is defined to
be   2 f so

Thus, v  
A
2
k
m
 x2

and
a = - 2x
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Displacement, Velocity, and Acceleration in SHM
The reference circle compares the circular motion of an object with its horizontal
projection.
The displacement as a function
of time for SHM with phase
angle  is
X(t) = Acos(t + φ)
vx (t)= -Asin(t + φ)
ax (t)= -2Acos(t + φ)
φ = phase angle in SHM
φ= arctan(- vox/ xo)
Amplitude, A:
𝒗𝟐
𝒐𝒙
A =√𝒙𝟐𝒐 + (
𝝎𝟐
)
SHM occurs whenever :
i.
ii.
iii.
there is a restoring force proportional to the displacement from
equilibrium: F ∝ −x
the potential energy is proportional to the square of the displacement: PE
∝ x2
the period T or frequency f = 1 / T is independent of the amplitude of the
motion. iv. the position x, the velocity v, and the acceleration a are all
sinusoidal in time.
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THE SIMPLE PENDULUM
SIMPLE PENDULUM – an idealized model consisting of a point mass suspended
by a massless, unstretchable string.
The path of the point mass is not a straight line but the arc of a circle with
radius L equal to the length of the string.
-
Laws of the Simple Pendulum
1. The amplitude does not affect the period of a simple pendulum.
2. The angle of swing does not affect the period.
3. The mass of the body does not affect the period.
4. The period is directly proportional to the square root of its length.
5. The period is inversely proportional to the square root of the acceleration.
A pendulum executes SHM, if the amplitude is not too large.
Equation for a Simple Pendulum
𝑔
 = √𝐿
𝐿
T = 2√𝑔
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Assignment: due July 9, 2020
1. A simple harmonic oscillator takes 12.0 s to undergo five complete vibrations.
Find (a) the period of its motion, (b) the frequency in hertz, and (c) the angular
frequency in radians per second.
2. A 7.00 kg object is hung from the bottom end of a vertical spring fastened to
an overhead beam. The object is set into vertical oscillations having a period
of 2.60 s. Find the force constant of the spring.
3. At an outdoor market, a bunch of bananas attached to the bottom of a
vertical spring of force constant 16.0 N/m is set into oscillatory motion with an
amplitude of 20.0 cm. It is observed that the maximum speed of the bunch of
bananas is 40.0 cm/s. What is the weight of the bananas in newtons?
4. A block-spring system oscillates with an amplitude of 3.50 cm. The spring
constant is 250 N/m and the mass of the block is 0.500 kg. Determine (a) the
mechanical energy of the system, (b) the maximum speed of the block, and
(c) the maximum acceleration.
5. ) A 326-g object is attached to a spring and executes simple harmonic motion
with a period of 0.250 s. If the total energy of the system is 5.83 J, find (a) the
maximum speed of the object, (b) the force constant of the spring, and (c)
the amplitude of the motion.
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MODULE 8:
FLUIDS
OSC
We begin our study with fluid statics, the study of fluids at rest in equilibrium situations.
Like other equilibrium situations, it is based on Newton’s first and third laws. We will
explore the key concepts of density, pressure, and buoyancy. We can analyze many
important situations using simple idealized models and familiar principles
such as Newton’s laws and conservation of energy. Even so, we will barely scratch
the surface of this broad and interesting topic ILLATION
8.1 PROPERTIES OF FLUID:
1. DENSITY (ρ)
An important property of any material is its density, defined as its mass per unit
volume. A homogeneous material such as ice or iron has the same density
throughout. We use ρ (the Greek letter rho) for density. If a mass m of homogeneous
material has volume V, the density is:
𝛒=
𝐦
𝐕
Two objects made of the same material have the same density even though they
may have different masses and different volumes. That’s because the ratio of
mass to volume is the same for both objects.
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2. SPECIFIC GRAVITY (SG)
The specific gravity of a material is the ratio of its density to the density of
water at it is a pure number without units. It is also called the “relative
density” of a material.
𝐒𝐆𝐅 =
𝛒𝐅
𝛒𝐇𝟐𝟎
WHERE: SGF and ρF are the specific gravity and density of the fluid respectively and
ρH2O = 1000 kg/m3
3. SPECIFIC WEIGHT (δ)
The specific weight is defined as its weight (W) per unit volume.
𝛅=
𝐖 𝐦𝐠
=
= 𝛒𝐠
𝐕
𝐕
Example Problem:
Saponification is a process wherein soap is added to a certain type of oil to produce
grease. One such grease is said to have 75.7 % by volume oil and 24.3% by volume
soap, wherein the oil and soap have densities 760 kgm/m3 and 6,250 kgm/m3,
respectively. This kind of grease is sold by packs shaped like a box with dimensions 20
cm x 40 cm x 10 cm. Calculate: a) the mass of oil per pack (kg), b) the mass of soap per
pack (kg),c) the weight of each pack (N), d) the specific weight of grease (N/m3), e)
the specific gravity of grease
Solution:
The total volume per pack VT = 0.2m x 0.4m x 0.1m = 0.008 m3
To solve for mass of oil per pack:
mo = ρo(Vo)
mo =(760 kg/m3)[(0.757)(0.008 m3) = 4.603 kg
To solve for mass of soap per pack:
ms= ρs(Vs)
3
ms =(6250 kg/m )[(0.243)(0.008 m3) = 12.15 kg
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To solve for the weight of each pack:
WT = (mT)(g)
WT = (4.603 +12.15)kg (9.81 m/s2) = 164.347 N
To solve for the specific weight of grease:
δT = WT/VT = 164.347 N / .008m3 = 20,543.37 N/m3
To solve for the specific gravity of grease:
SGT = ρT /ρH2O
Where ρT = mT /VT = (4.603 +12.15)kg / 0.008 m3 =2094.1 kg/m3
SGT = (2094.125 kg/m3) / (1000 kg/m3) = 2.0941
8.2 BUOYANCY
Buoyancy is a familiar phenomenon: A body immersed in water seems to
weigh less than when it is in air. When the body is less dense than the fluid, it floats.
The human body usually floats in water, and a helium-filled balloon floats in air.
Example Problem:
A 15.0-kg solid gold statue is raised from the sea bottom. What is the tension in the
hoisting cable (assumed massless) when the statue is (a) at rest and completely
underwater and (b) at rest and completely out of the water?
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Assignment # 8: due on July 16, 2020
1. A spherical tank 2 m in diameter contains steam. If the mass of steam is 0.8
kg:
a. Determine the density of steam.
b. What is its specific weight if g = 9.81 m/s2
2. An 11-m3 rigid tank of air is separated by a thin membrane into side A with a
volume of 6 m3 and side B with an initial 2.4 kg/m3. The membrane is broken
and the resulting density of the mixture is 1.82 kg/m3. Find the initial density of
air inside A in kg/m3.
3. On a part-time job, you are asked to bring a cylindrical iron rod of length 85.8
cm and diameter 2.85 cm from a storage room to a machinist. Calculate
the weight of the rod.
4. A cube 5.0 cm on each side is made of a metal alloy. After you drill a
cylindrical hole 2.0 cm in diameter all the way through and perpendicular to
one face, you find that the cube weighs 7.50 N. (a) What is the density of this
metal? (b) What did the cube weigh before you drilled the hole in it?
5. A 950-kg cylindrical can buoy floats vertically in salt water. The diameter of
the buoy is 0.900 m. Calculate the additional distance the buoy will sink when
a 70.0-kg man stands on top of it.
6. A slab of ice floats on a freshwater lake. What minimum volume must the slab
have for a 45.0-kg woman to be able to stand on it without getting her feet
wet?
7. An ore sample weighs 17.50 N in air. When the sample is suspended by a light
cord and totally immersed in water, the tension in the cord is 11.20 N. Find the
total volume and the density of the sample.
8. A hollow plastic sphere is held below the surface of a freshwater lake by a
cord anchored to the bottom of the lake. The sphere has a volume of 0.65
m3 and the tension in the cord is 900 N. (a) Calculate the buoyant force
exerted by the water on the sphere. (b) What is the mass of the sphere? (c)
The cord breaks and the sphere rises to the surface. When the sphere comes
to rest, what fraction of its volume will be submerged?
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MODULE 9:
HEAT TRANSFER
9-1 THERMODYNAMICS AND HEAT TRANSFER
- The science of thermodynamics deals with the amount of heat transfer as a
system undergoes a process from one equilibrium state to another, and makes
no reference to how long the process will take. But in engineering, we are often
interested in the rate of heat transfer, which is the topic of the science of heat
transfer.
- The basic requirement for heat transfer is the presence of a temperature
difference. There can be no net heat transfer between two mediums that are
at the same temperature.
9-2 TEMPERATURE
Temperature – measure of the hotness or coldness of a body. It is also defined as
the measure of the internal energy of a body.
Conversion of temperature reading to another temperature scale:
TF = 1.8(Tc) + 32
𝐓𝐅 − 𝟑𝟐
𝐓𝐂 =
𝟏. 𝟖
TK =TC + 273
TR = TF + 46O
WHERE:
TF = Temperature expressed in degree Fahrenheit (oF)
TC = Temperature expressed in degree Celsius (oC)
TK = Temperature expressed in Kelvin (K)
TR = Temperature expressed in Rankine (R)
NOTE: Tk and TR are called the absolute temperatures
Example:
1. A fluid system has a temperature of 26 oC. What is its temperature
expressed in Rankine.?
Solution:
TF = 1.8(Tc) + 32
TF = 1.8 (26) +32 = 78.8 oF
TR = TF + 46O
TR = 78.8 + 460 = 538.8 R
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2. A Celsius and Fahrenheit thermometer are used to measure the
temperature of a certain substance. If the temperature readings from
both thermometers are numerically equal, What is the temperature of the
substance.
Solution:
TF = 1.8(Tc) + 32
Since temperatures are numerically equal: TF =TC =T
T = 1.8T + 32
T = - 40
This means that – 40 oC = - 40 oF
9-3 MODES OF HEAT TRANSFER
1. Conduction is the transfer of energy from the more energetic particles of a
substance to the adjacent, less energetic ones as a result of interactions
between the particles.
-
-
-
The rate of heat conduction through a medium depends on the geometry of
the medium, its thickness, and the material of the medium, as well as the
temperature difference across the medium.
Experiments have shown that the rate of heat transfer 𝑸̇ through the wall is
doubled when the temperature difference ΔT across the wall or the area A
normal to the direction of heat transfer is doubled, but is halved when the wall
thickness L is doubled.
Thus we conclude that the rate of heat conduction through a plane layer is
proportional to the temperature difference across the layer and the heat
transfer area, but is inversely proportional to the thickness of the layer.
In the limiting case of thickness →0, the relation above reduces to the
differential form, which is called Fourier’s law of heat conduction.
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Where: the constant of proportionality k is the thermal conductivity of the
material, which is a measure of the ability of a material to conduct heat
expressed in (W/m-oC or BTU/hr-ft-oF).
CONDUCTION IN PLANE WALLS
Consider steady heat conduction through a large plane wall of thickness Δx and
area A, as shown in Fig. 1–21. The temperature difference across the wall is ΔT =
T2 - T1.
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CONDUCTION CYLINDERS OF LENGTH L
Example.
A temperature difference of 85◦C is impressed across a fiberglass layer of 13 cm
thickness. The thermal conductivity of the fiberglass is 0.035 W/m·◦C. Compute the
heat transferred through the material for a heat transfer surface area of 1 m 2.
T1 − T2
Δx
Q = (0.035 W/m-oC)(1m2)(85 oC)/(0.13m) = 0.387 W or watts
Q = kA
Steam at flows in a stainless steel pipe (k = 15 W/m · °C) whose inner and outer
diameters are 5 cm and 5.5 cm, respectively. Determine the rate of heat loss from
the steam for 1 m length of the pipe if the temperature drop across the thickness of
the pipe is 10 oC.
T1 − T2
Q = 2πLk
𝑟
ln(𝑟2 )
1
o
o
Q = 2π(1m)(15 W/m- C)(10 C) / ln(2.75cm/2.5cm)
Q = 9888.532 W or watt
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2. Convection is the mode of heat transfer between a solid surface and the
adjacent liquid or gas that is in motion, and it involves the combined effects
of conduction and fluid motion.
The rate of convection heat transfer is observed to be proportional to the
temperature difference, and is conveniently expressed by Newton’s law of
cooling as:
Where:
h - the convection heat transfer coefficient in W/m 2 · °C or Btu/h · ft2 · °F
As - the surface area through which convection heat transfer takes place
Ts is the surface temperature
T∞ - the temperature of the fluid sufficiently far from the surface.
Note that at the surface, the fluid temperature equals the surface
temperature of the solid.
Example:
A 5-cm-external-diameter, 10-m-long hot water pipe at 80°C is losing heat to the
surrounding air at 5°C by natural convection with a heat transfer coefficient of 25
W/m2 · °C. Determine the rate of heat loss from the pipe by natural convection, in
Watts
Q = hAs(Ts - T∞)
2
o
Q = (25 W/m - C)[ π(0.05m)(10m)] [(80 -5)oC] = 2945.24 W or watts
Note that the heat transfer surface area of a cylinder with length L is:
As = π(diameter)(length)
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3. Radiation is the energy emitted by matter in the form of electromagnetic
waves (or photons) as a result of the changes in the electronic configurations
of the atoms or molecules.
The rate of radiation that can be emitted from a surface is given by the
Stefan-Boltzmann law as:
When a surface of emissivity Ɛ and surface area As at an absolute temperature Ts is
completely enclosed by a much larger surface at absolute temperature Tsurr
separated by a gas (such as air) that does not intervene with radiation, the net rate
of radiation heat transfer between these two surfaces is given by:
Where:
Ɛ – Emissivity of the surface ( a measure of how close a surface approximates a
black body) 0 ≤ Ɛ ≤ 1
Ơ - Stefan-Boltzmann constant (5.67 x 10 -8 W/m2-K4 and 0.1714 x 10-8 BTU/hr-ft2 –R4)
As – surface area of radiating body
Ts and Tsurr – Absolute temperature of radiating body (Temperatures in K or R)
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Example:
Consider a person whose exposed surface area is 1.7 m2, emissivity is 0.7, and
surface temperature is 32°C. Determine the rate of heat loss from that person by
radiation in a large room having walls at a temperature of (a) 300 K and (b) 280 K.
Solution:
Ts = 32 oC + 273 = 305 K
a) Q = ЄσAs(Ts4 –Tsurr4)
Q = 0.7 (5.67 x 10 -8 W/m2-K4)(1.7 m2)[(3054 - 3004)K4]
Q = 37.4 W
b) Q = 0.7 (5.67 x 10 -8 W/m2-K4)(1.7 m2)[(3054 - 2804)K4]
Q = 169.2 W
Assignment #9: due July 16, 2020
1. The inner and outer surfaces of a 5-m by 6-m brick wall of thickness 30 cm
and thermal conductivity 0.69 W/m ·°C are maintained at temperatures of
20°C and 5°C, respectively. Determine the rate of heat transfer through the
wall, in Watts.
2. The inner and outer surfaces of a 0.5-cm-thick 2-m by 2-m window glass in
winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass
is 0.78 W/m · °C, determine the rate of heat loss, in Watt, through the glass.
What would your answer be if the glass were 1cm thick?
3. Consider a person standing in a room maintained at 20°C at all times. The
inner surfaces of the walls, floors, and ceiling of the house are observed to be
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at an average temperature of 12°C in winter and 23°C in summer. Determine
the rates of radiation heat transfer between this person and the surrounding
surfaces in both summer and winter if the exposed surface area, emissivity,
and the average outer surface temperature of the person are 1.6 m 2, 0.95,
and 32°C, respectively.
4. Hot air at 80°C is blown over a 2-m by 4-m flat surface at 30°C. If the average
convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of
heat transfer from the air to the plate, in kW.
5. Two surfaces of a 2-cm-thick plate are maintained at 0°C and 80°C,
respectively. If it is determined that heat is transferred through the plate at a
rate of 500 W/m2, determine its thermal conductivity.
6. For heat transfer purposes, a standing man can be modeled as a 30-cmdiameter, 170-cm-long vertical cylinder with both the top and bottom
surfaces insulated and with the side surface at an average temperature of
34°C. For a convection heat transfer coefficient of 15 W/m2 · °C, determine
the rate of heat loss from this man by convection in an environment at 0°C.
7. Hot air at 80°C is blown over a 2-m by 4-m flat surface at 30°C. If the average
convection heat transfer coefficient is 55 W/m2 · °C, determine the rate of
heat transfer from the air to the plate, in kW.
8. Convert 125 oC to a) oF, b) K, c) R
9. The increase in temperature of water as it is being heated is 80 oC. Determine
this change in temperature expressed in a) oF and b) oR
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MODULE 10: HEAT TRANSFER
ENGAGE
How is heat transferred?
How do you explain the heat transfer processes in cooking using an electric stove?
Cite simple situations where you can observe heat transfer.
What do you think happens to objects when they transfer heat?
EXPLORE
Read Module 10: Heat Transfer (pp 86 – 89)
EXPLAIN
HEAT TRANSFER
- The movement of energy from a warmer object to a cooler object.
HEAT
- Form of energy transferred from one body to another due to difference in
temperature
- usually measured by the temperature (Celsius, C; Fahrenheit, F; Kelvin, K)
FACTORS that affect Energy gained or lost by an object during heat transfer:
o Mass
o specific heat
o change in temperature
*3 Processes by which heat is being transferred from one body to another:
1. Conduction
2. Convection
3. Radiation
86
CONDUCTION
 The transfer of thermal energy between neighboring molecules in a substance
due to a temperature gradient.
 Temperature gradient – is the rate of change of temperature with
distance.
 It is a process by which heat energy is transferred from particle to particle by
collisions or direct interactions.
** For example, if you hold one end of a long metal bar and insert the other end
into a flame, you will find that the temperature of the metal in your hand soon
increases. The energy reaches your hand by means of conduction.
INSULATORS VS CONDUCTORS:
INSULATORS – these are materials that do not transfer heat well
e.gAir, wood, wool
CONDUCTORS – materials that transfer heat well
e.gSilver, fiberglass, tile
EQUATIONS FOR CONDUCTION:
where:
= heat conducted (cal or J)
= area (cm2 or m2)
= temperature difference (°C) =
= time (sec)
= thickness (cm or m)
= coefficient of thermal conductivity (
⋅
⋅
⋅ ⋅
)
= power (watts)
CONVECTION
 It is the transfer of heat by the motion of a volume of hot fluid from one place to
another.
 Heat transfer by the actual movement of the heated material itself.
** 2 Process:
Force Convection – if the heated substance is forced to move by a fan or pump.
Natural Convection - the change in density that takes place when a fluid is heated.
*example: Airflow at a beach
87
EQUATIONS for CONVECTION:
where:
= heat transfer coefficient (
)
= area (m2)
= temperature difference between the surface and the bulk of the fluid away
from the surface (K)
= rate of convective heat flow (watts)
Value if h applies to natural convection of air at atmospheric pressure:
1. Horizontal plate with air passing the top surface
2. Horizontal plate with air passing the bottom surface
3. Vertical plate
4. Horizontal plate or vertical pipe of diameter D
(
)
RADIATION
 It takes place by means of electromagnetic waves which require no material
medium for their passage.
 It consists of electromagnetic waves which transmit energy from a source to an
absorber.
EQUATION for RADIATION
where:
= Stefan Boltzmann constant
=
= emissivity
e = 0, ideal reflector
88
e = 1, ideal absorber
= surface temperature (K)
= area (m2)
= power (watts)
Figure 10.1 In a fireplace, heat transfer occurs by all three
methods: conduction, convection, and radiation. Radiation is
responsible for most of the heat transferred into the room. Heat
transfer also occurs through conduction into the room, but at a
much slower rate. Heat transfer by convection also occurs
through cold air entering the room around windows and hot air
leaving the room by rising up the chimney.
ELABORATE
EXAMPLE SITUATIONS:
CONDUCTION:
1. when you hold one end of an iron rod while the other end is in direct contact with a
flame. Heat will transfer through the metal by conduction.
2. The knives and forks use a wooden handle to break with the conduction of heat.
3. When you pour hot coffee to the cup containing it. Hot liquids transfer the heat to the
container containing them, causing the latter to warm up a bit.
4. Compresses (hot water bags) used to relax muscles. The heat is transferred from the
compress to the skin and from there to the muscles.
5. An ice cube melts when placed on apersons hand. The heat from the skin is transferred
to the ice cube causing it to melt.
CONVECTION:
1. The heat transfer of a stove.
2. Hot air balloons, which are held in the air by hot air. If it cools, the balloon immediately
begins to fall.
3. When the water vapor fogs the glass of a bath, by the hot temperature of the water
when bathing.
RADIATION
1. The transmission of electromagnetic waves through the microwave oven.
2. The light emitted by an incandescent lamp.
3. The emission of gamma rays by a nucleus.
89
COMPUTATION EXAMPLES:
CONDUCTION:
1. The wall of shed in which ice is stored consists of an outer layer of wood 2cm and an
inner layer of rock wool 3cm thick. Find the heat conducted through 50m 2 of the wall in
1hr. When the outer wood surface is at 20°C and the inner rock wool surface is at 5°C.
Also find the temperature of the wood-rock wool interface.
CONVECTION:
1. A Bathroom is heated by a floor-to-ceiling steam pipe that is 10cm in diameter. The
ceiling height is L=3.0m and the temperature of the bulk of air in the room is 22°C. If the
pipe surface is at 90°C, what is the rate of convective heat transfer?
RADIATION:
1. A copper ball 2cm in radius is heated in a furnace to 400°C. If the emissivity is 0.3, at
what rate does it radiate energy?
2. The sun‘s surface has a temperature of about 5800K, and the radius of the sun is about
7x108m. Calculate the total energy radiated by the sun each day, assuming the
emissivity is e=1. By way of comparison, the total energy consumed worldwide each
year by humans is about 1021J.
EVALUATE
1. Identify whether the given situations are examples of conduction, convection or
radiation.
a. Solar ultraviolet radiation, precisely the process that determines the Earth‘s
temperature.
b. Ironing of clothes
c. When you walk barefoot on the hot street, and it burns your toes.
d. The heat transferred by hand or hair dryer
e. When teaspoons get hot when placed in hot coffee
f. The heat transfer generated by the human body when a person is barefoot.
g. The heat emitted by a radiator.
h. A thermometer works because the liquid in it contracts when heated.
2. Identify the word/s defined in each item
a. the measure of the amount of energy in matter
b. The movement of energy from a warmer object to a cooler object
c. Materials that do not transfer energy (heat) easily
d. The transfer of heat through empty space
90
3. Answer the following questions as either TRUE or FALSE
a.
b.
c.
d.
e.
the Sun directly heat the air in our atmosphere
higher temperature means faster moving molecules
Air a great conductor of heat
Heat always comes from cooler temperature to warmer temperature
The transfer of heat When the warmth of the sun heats rocks is an example of
radiation
f. A thermometer works because the liquid in its contracts when heated which is
an example of conduction
g. temperature is a measure of total kinetic energy
PRACTICE PROBLEMS:
1. The thermal conductivity of ice is 5.2 x 10-4 kcal/m-s-oC. At what rate is heat lost by the
water in a 6m by 10m outdoor swimming pool covered by a layer of ice 1cm thick if the
water is at a temperature of 0oC and the surrounding air is at a temperature of
-10oC?
2.
Forced air flows over a heat exchanger in a room heater, with a convective heat
transfer coefficient, h = 150 BTU/hr-ft2-oF. The surface temperature of the heat exchanger is
held at 200oF while the air in the room is maintained at 72 oF. Find the surface area of the
heat exchanger if 20,000 BTU/hr is delivered by the heater.
3.
An incandescent lamp filament, with a surface area of 100mm 2, operates at a
temperature of 2300oC. Assume that the filament acts like a blackbody(e = 1). (a) What is
the rate of radiation from the filament? (b) If the walls of the room in which the lamp
operates are at 27oC, What is the rate at which the filament absorbs radiation? (c) At what
rate does electrical energy have to be supplied to the filament to keep its temperature
constant? (Ignore conduction and convection losses from the filament.)
91
MODULE 11: ELECTROSTATICS
ENGAGE
What happens when two atoms like charges are near each other?(i.e. both are positively
charged)
An electron is negatively charged particle of an atom. What happens when an atom loses
an electron?
What happens to the force between two atoms when they are placed too far from each
other?
When a rubber comb is run through the hair and is then placed near small pieces of paper,
the pieces of paper are attracted to the rubber comb. Explain why?
EXPLORE
Read Module 11: Electrostatics (pp 92 - 95)
EXPLAIN
Electrostatics
- Branch of physics that deals with the study of charges at rest
- "Electrostatic" pertains to electric charges at rest or to fields or phenomena
produced by stationary charge(s)
- Static electricity is the accumulation of electrical charges on the surface of a
material; may result in sparks, shocks or materials clinging together.
Electric Charges
- one of the basic properties of the elementary particles of matter giving rise to all
electric and magnetic forces and interactions.
There are two types of electric charges: positive and negative
these charges are commonly known as protons (+) and electrons (–)
92
Atomic Structure:
Particle
Electron (e)
Proton (p)
Neutron (n)
Charge
-1.6 x 10-19 C or
-4.8 x 10-10esu (or statC)
+1.6 x 10-19 C or
+4.8 x 10-10esu (or statC)
0
Mass (kg)
9.1094 x 10-31
1.6726 x 10-27
1.6749 x 10-27
*1C = 3 x 109statC
ELECTRON THEORY:
1. A neutral body is one that has exactly as many electrons as there are protons.
2. When a neutral body gains electron from an outside source, it acquires a negative
charge.
Hence, a negatively charged body has more electrons than protons.
3. When neutral body loses some of its own electrons, it acquires a positive charge.
Hence, a positively charged body has fewer electrons than protons.
LAWS OF ELECTROSTATICS:
1. INTERACTION OF CHARGES:
a. Like charges repel
b. opposite charges attract , and each other
c. Charged particle and neutral attract:
Example: a charged comb (ran through the hair) attracts bits of uncharged
(or neutral) bits of paper
+
-
-
-
-
+
Figure 11.1 Direction of interaction of Charges
93
2. LAW OF CONSERVATION OF CHARGE
- The algebraic sum of the electric charge in any closed system remains constant
- charge is always conserved
- When all objects involved are considered prior to and after a given process, the
total amount of charge among the objects is the same before the process starts as
it is after the process ends.
ELECTRIFICATION (CHARGING OF BODY):
- The process of gaining or losing electron (transfer of electrons from one material to
the other)
The Triboelectric Series
- a list of materialsshowing which have a greater tendency to become positive (+)
and which have a greater tendency to become negative (−)
- atoms of different materials hold on to their electrons with different strengths
the materials would acquire the same amount of charge (but opposite type
or sign)
- the greater the separation in the series, the greater the exchange of charges
Methods of Electrification/ Charging:
1. Friction
- results in a transfer of electrons between two objects that are rubbed
together
2. Conduction
- charging a neutral body by bringing into contact with a charged body
- (also known as charging by contact) involves the contact of a charged
object to a neutral
3. Induction
- The charging body is brought near the neutral body, thus inducing it to have
an opposite charge as that of the charging body
- object is charged without actually touching the object to any other charged
object
**OTHER PROCESSES:
Polarization is the process of separating opposite charges within an object. The
positive charge becomes separated from the negative charge. By inducing the
movement of electrons within an object, one side of the object is left with an excess
of positive charge and the other side of the object is left with an excess of negative
charge. Charge becomes separated into opposites.
94
Grounding
: the process of removing the excess charge on an object by means of the transfer
of electrons between it and another object of substantial size. When a charged
object is grounded, the excess charge is balanced by the transfer of electrons
between the charged object and a ground.
Ground: an object that serves as a seemingly infinite reservoir of electrons; the
ground is capable of transferring electrons to or receiving electrons from a charged
object in order to neutralize that object. (ex. ground or earth)
3. COULOMB’S LAW OF ELECTROSTATICS (CHARLES AUGUSTIN DE COULOMB)
- The electric force F between two electric charges is directly proportional to the
product of the charges and inversely proportional to the square of the distance
between them.
ELECTRIC FORCE
- The attractive or repulsive interaction between any two charged objects; an
action-at-a-distance force
- hold atoms together; makes possible the existence of material things, human
activities, and properties and attributes (tasting, smelling, thinking, etc)
COULOMB’S LAW
|
||
|
Where:
F = force between the charges; N or dynes
q1& q2 = magnitude of the charges; C or esu (or statC)
r = distance between them; m or cm
k = Coulomb‘s constant (depending on
Where:
95
ELABORATE
ELECTRIFICATION: (Amounts of charge given are fictitious and given for purposes of
illustration only)
CHARGING BY FRICTION:
1. What type of charge would each material acquire for each of the following pairs
rubbed together?
Assume they are all initially uncharged or neutral.
a) acetate and wool
b) paper and polyester
2. Which pair would acquire higher amount of individual charge?
96
CHARGING BY CONDUCTION:
1.
Initially, object A is positively charged
with +8μC; object B is neutral
2.
A is brought in contact with B. Since A is
positively charged, electrons in B are
attracted towards A.
Then the A and B are pulled apart.
3.
B loses electrons; so, it becomes positively charged. A gains those electrons lost by B; so,
A becomes less positively charged.
As shown in no. 1 and 3, by conservation of charge:
4. The total charge in the system of charges is the same before and after the charging
process.
CHARGING BY INDUCTION
1. Initially, object A is negatively
charged with -15μC; objects B and
C are neutral
3.
2.
A is brought close to but not in contact with
B. Since A is negatively charged, electrons in
B are repelled and move as far away as
possible from A. These electrons transfer to C.
Then the A and B are pulled apart.
B loses electrons; so, it becomes positively charged C gains those electrons lost by B; so,
C becomes negatively charged. A retains its original charge
As shown in no. 1 and 3, by conservation of charge:
4. The total charge in the system of charges is the same before and after the charging
process.
97
POLARIZATION:
1. Initially, object A is negatively charged
with -10μC, and B is neutral.
2. Since A is negatively charged, electrons
on the left edge of Bare repelled and
move as far away as possible from A.
There are no other objects these
electrons can go to. So, they
accumulate at the right edge of B.
3. B neither loses nor gains electrons; the electrons only move to a different part of the
object. So, B remains in its initial state which is being neutral.
4.
GROUNDING
1. Identify the following particles as being
charged or uncharged. If charged,
indicate whether they are charged
positively or negatively.
(n = neutron, p = proton, e = electron)
98
2.
b)
3.12 x 1014 protons and 4.5488 x 1016
a) 8.25749 x 1017 protons and 5.26 x 1014
electrons; the charge on this object is ____ C.
electrons; the charge on this object is ____
+
C.
+
(Amounts of charge given are fictitious and given for purposes of illustration only)
a. Neutralizing a positively
charged object
Object A is brought in contact with
ground, which has abundance of
electrons. Electrons from ground
are attracted to A because this
object is positively charged (means
that it has ―electron vacancies‖).
When enough electrons have
transferred into A (or filled up the
―electron vacancies‖), A becomes
neutral.
b. Neutralizing a negatively charged object
Object A being negatively
charged repel electrons at
the surface of the ground.
These electrons move as far
away as possible from A.
The surface of the ground
now becomes positively
charged. In turn, electrons
from
object
A
are
attracted to the ground.
When enough electrons
have transferred out of A, A
becomes neutral.
99
COULOMB’S LAW:
Given:
located at (+0.2 m,0)
located at (+0.6 m,0)
Determine the magnitude and direction of
the force on each charge.
Draw the charges along a line.
and
have opposite types of charge;
so, they are attracted to each other.
is pulled by
is pulled by
The force vector of each charge is
directed toward the other charge:
100
Given:
located at (0,+0.3 m)
located at (0, -0.2 m)
Determine the magnitude and direction of
the force on each charge.
EVALUATE
1. Three charged particles are aligned along the x axis as shown. Find the electric force
between the three charges.
2. Three point charges are located on a circular arc as shown. aFind the electric force
that would be exerted on a 5nC point charge placed at P.
3. Two particles, with charges of 20nC and -20nC, are placed at the points with
coordinates (0, 4cm) and (0, -4cm) respectively. Find the electric charge on a 10nC
located at the origin.
101
MODULE 12: ELECTRICITY
ENGAGE
In your opinion, how do electricity work, does it move from positive to negative or
otherwise?
Why is it important to follow the electric specifications on chargers and appliances?
A small amount of current is generated by putting your tongue to a 9V battery, however
the amount of current cannot be felt by holding 9V battery using your dry hands. What is
the concept behind the difference between the amount of current generated?
EXPLORE
Read Module 12: Electricity (pp142–147)
EXPLAIN
CURRENT, RESISTANCE AND ELECTROMOTIVE FORCE
I. ELECTRIC CURRENT
 charges in motion
DC (Direct Current)
: is a constant flow between two points having a different electrical potential and
thecharge flow is one way (ex. cell or battery)
AC (Alternating Current)
:the charge flows alternately backwards then forwards in a circuit many times
everysecond (ex. power plant, generator, mains supply)
Direction of Current
CONVENTIONAL FLOW
ELECTRON FLOW
`
In a metallic conductor, the moving
charges are electrons — but the
current still points in the direction
positive charges would flow.
A conventional current is treated as a
flow of positive charges (would move
from the positive battery terminal
and toward the negative terminal).
102
The motion of positive charge carriers in one direction has the same effect as the
actual motion of negative charge carriers in the opposite direction. So for historical
reasons, however, we use the following convention:
A current arrow is drawn in the direction in which positive
charge carriers would move, even if the actual charge carriers
are negative and move in the opposite direction.

Current as the rate of flow of electric charge through a conductor connected
between two points of different potentials
where:
I = current, Ampere (amp or A)
t = time, s
Q = total charge that flows, coulomb
* Note that for a metallic conductor, charges that flow are electrons
Q= Nq
where:
N = number of electrons that flow during a time t
q = charge of an electron, |qe| = 1.6x10-19 C
 Current in Relation to the Drift Velocity of the Charges
When an electric field ⃗ is established in a conductor, the charge carriers (assumed
positive) acquire a drift speed
in the direction of ⃗ ; the current is related to the drift
speed by
where:
A = cross sectional area of conductor (m2)
n = free electron(conduction electron) density or concentration of particles
= the number of carriers per unit volume (n = N/V)
 (nq) = carrier charge density
 Current Density in a Conductor, J
 It is the current per unit cross-sectional area.


103
II. RESISTIVITY AND RESISTANCE
 Resistivity,
 The quantitative measure of a material‘s opposition to the flow of current. It is an
intrinsic property of the material (metal element) which depends only (if
temperature is constant) on the chemical composition of the material and
temperature, not its shape or size.
 Good conductors have small resistivity; good insulators have large resistivity.
 Resistance, R
 It is the obstruction or opposition offered by the material (conductor) in the flow
of current through it. It is the extrinsic property of the material which depends
upon the amount of material present (shape and size).
Ohm’s Law: For many materials (including most metals), the ratio of the current density to
the electric field is a constant that is independent of the electric field producing the
current.
Materials and devices that obey Ohm‘s Law are said to be ohmic
Resistance: the ratio of the potential
difference across a conductor to the
current in the conductor
Resistance of a block of material along
the length :
Resistance and Temperature
Over a limited temperature range:
III. ELECTROMOTIVE FORCE, emf or ℰ
the maximum potential difference between two electrodes of a galvanic or voltaic cell
energy per unit electric charge that is imparted by an energy source, such as an
electric generator or a battery.
Sources of Electrical Energy: burning oil or coal, hydroelectric plant, geothermal, wind,
solar, nuclear, natural gas, hydrogen, biofuel, biomass, fruits and vegetables with
moderate to high levels of acidity, cell
104
IV. OHM’S LAW
 Georg Simon Ohm (1787 – 1854), a German physicist, is credited with finding the
relationship between current and voltage for a resistor.
 The voltage across a resistor is directly proportional to the current, i flowing through
the resistor.
 Ohm further defined the constant of proportionality for a resistor to be the
resistance, R.
 Thus, the resistance, R is the ability of an element to resist the flow of electric current;
measured in ohms (Ω).
 The power dissipated by a resistor is given by:
( )
V. SERIES RESISTORS AND VOLTAGE DIVISION
 What is Series?
- Two or more elements are in series if they exclusively share a single node and
consequently carry the same current.
- The equivalent resistance of any number of resistors connected in a series is the sum
of the individual resistances.
∑
-
For the given figure:
-
The current in a series circuit is the same for each
of the elements or resistors.
-
The total voltage in a series circuit is equivalent to
the summation of all resistance voltages or
voltage drops in the circuit
105
 To determine the voltage across each resistor;
Ohm‘s Law
,
Substitute
-
Note that source voltage is divided among the resistors in direct proportion to their
resistances
This is regarded as the principle of voltage division
In general, if a voltage supply has resistors in series with the source voltage , the
th resistor
can be expressed as:
VI. Parallel Resistors and Current Division
 What is Parallel?
- Two or more elements are in parallel if they are connected to the same two nods
and consequently have the same voltages across them.
- The equivalent resistance o 2 parallel resistors is equal to the product of their
resistance divided by their sum.
-
-
-
Generally, the equivalent resistance of
circuit with resistors in parallel is:
a
The current in a parallel circuit is equivalent to the summation of all branch currents
in the circuit
The resistance voltages are each equal to the source voltage of the parallel circuit.
106
 To determine the current through each resistor
Ohm‘s Law
Substitute
-
Note that the total current is shared by the resistors in inverse proportion to their
resistances.
This is regarded as the principle of current division
Generally
ELABORATE
CURRENT, RESISTANCE and EMF
1. Suppose you wish to fabricate a uniform wire from 1.00 g of copper. If the wire is to have a
resistance of 0.500 Ω and all the copper is to be used, what must be a) the length and b) the
diameter of this wire? Density of copper
g/cm3, and resistivity
Ω·m.
2. A 34.5 m length of copper wire at 20.0°C has a radius of 0.25 mm. If a potential difference of 9.00
V is applied across the length of the wire, determine the current in the wire. b) If the wire is
heated to 30.0°C while the 9.00 V potential difference is maintained, what is the resulting current
in the wire? Temperature coefficient of resistivity is
Resistivity is
.
3. A toaster has a heating element made of Nichrome wire (resistivity is 150x 10 -8Ω·m and
temperature coefficient of resistivity is 0.4x10 -3/°C). When the toaster is first connected to a 120 V
source (and the wire is at a temperature of 20.0°C), the initial current is 1.50 A. The current
decreases as the heating element warms up. When the toaster reaches its final operating
temperature, the current is 1.20 A. Find a) the power delivered to the toaster when it is at its
operating temperature b) the final temperature of the heating element
107
SERIES CONNECTION
1. Given 4 resistors connected in series, R1=2Ω, R2=1Ω, R3=4Ω and R4=8, connected to a 30 Vol
source, find for a) Total Resistance b) Total Current and c) Voltage drops across each resistor.
PARALLEL CONNECTION
2. If the resistors in Problem 4 are connected in parallel, find for the a) Total Resistance, b) Total
Voltage and c) Currents through each resistor.
SERIES-PARALLEL COMBINATION
Problem 1
By combining the resistors in the given
circuit, find
Problem 2
Find
for the circuit shown:
EVALUATE
1. A steady current of 2.5 A exists in a wire for 4.0 min. (a) How much total charge passed
by a given point in the circuit during those 4.0 min? (b) How many electrons would this
be?
2. A 200-km-long high-voltage transmission line 2 cm in diameter carries a steady current
of 1000 A. If the conductor is copper with a free charge density of 8.50 x 1028 electrons
per cubic meter, how many years does it take one electron to travel the full length of
the cable?
3. An 18-gauge copper wire (ρ = 1.72x10-8 .m) has a diameter of 1.02 mm and a crosssectional area of 8.20x10-7m2. It carries a current of 1.67 A. Find (a) the electric-field
magnitude in the wire; (b) the potential difference between two points in the wire 50m
apart; (c) the resistance of a 50m length of this wire.
108
4. A certain lightbulb has a tungsten filament with a resistance of 19 V when at 20°C and
140 V when hot. Assume the resistivity of tungsten varies linearly with temperature even
over the large temperature range involved here. Find the temperature of the hot
filament. (αTungsten = 4.5 x 10-3/OC)
5. When an external resistance of 10 ohms is connected to a source, a current of 0.5A
flows, when the resistance is changed to 20 ohms, the current becomes 0.3 amperes.
What are the open circuit emf and the internal resistance of the source?
6. A battery has an emf of 15.0 V. The terminal voltage of the battery is 11.6 V when it is
delivering 20.0 W of power to an external load resistor R. (a) What is the value of R? (b)
What is the internal resistance of the battery?
7. Find the voltage needed so that a current of 10 A will
flow through the series circuit shown in the figure.
8. Find the voltage across each resistor in the circuit
of Problem 7. Show that the voltage drop equals
the applied voltage of 100V.
9. For
the given circuit, find (a) the total
resistance, (b) each branch current, and (c)
the total current.
10. Find I3 in the parallel circuit using current divider.
109
MODULE 13: MAGNETISM
ENGAGE
In your opinion, do you use electromagnetism in your daily lives? Can you cite example
where you apply or use them?
How do compass (navigation) work?
In your opinion, what is the importance of the earth‘s magnetic field?
EXPLORE
Read Module 13: Magnetism (pp110 - 115)
EXPLAIN
ELECTROMAGNETISM
Many applications of electricity are based on the fact that electric current passing
thru a conductor produces a magnetic field around the conductor or vice versa, i.e. when
a conductor moves thru a magnetic field, an electromotive force (emf) is generated and
hence current could flow if there is a complete loop. This is called electromagnetism. The
operation of such devices or machines like the electric motor, generator, the transformer,
the inductor coils, etc. are all based on this phenomenon.
A. The Magnetic Field
It is a region in space wherein a force acts on a charge moving through it.
Ex.
The space around a magnet
The space around a current-carrying conductor
110
One characteristic of a magnetic field is its direction which is usually indicated by a
set of lines drawn in the magnetic field called magnetic field lines (also called magnetic
lines of force or lines of induction).
Figure13.1 Different Examples of Magnetic field
Note: When the lines are curved, the magnetic field is non-uniform and the direction at
any point is taken by the tangent to a line passing through that point.
Figure13.2 Magnetic field taken at point P
B. Magnetic Flux (∅)
A set of magnetic field lines or a group of lines taken or computed as one
C. Magnetic Flux Density ( ) at a point
Magnetic flux passing per unit area of a surface placed at the point perpendicular
to the magnetic field.
111
Note: 1. If the surface area A is PERPENDICULAR to βthen,
2. If the surface area A is NOT PERPENDICULAR to βthen,
∅ = βA
∅ = βAcosθ
Magnetic flux density is a vector quantity and its direction is the same as the direction of
the magnetic field at a point.
SOURCES OF MAGNETIC FIELD
I.
MAGNETIC FIELD OF A MOVING CHARGE
When a charge moves through a portion of space, a
magnetic field is created around the charge.
To obtain the magnetic field produced by a charge, q,
moving at velocity v, at a point located at a
displacement r from charge, we need a mathematical
expression the field in terms of q, v and r.
where:
β = magnetic field at any point ―P‖ near the moving charge (Tesla or Gauss)
q = electric charge (Coulomb, statcoulomb)
v = speed of the electric charge (m/s, cm/s)
r = distance of the point from the charge
θ = angle between r and v
μ0= permeability of free space (Permeability is the measure of the ability of a
material to support the formation of a magnetic field within itself.)
= 4π x 10-7T.m/A
112
Direction: RIGHT HAND RULE
Thumb:
velocity, v
4 fingers: magnetic field, β
NOTE: If the point charge is negative, the directions of the field and field lines are the
opposite with that of positive charge.
II.
MAGNETIC FIELD at any POINT NEAR a STRAIGHT CONDUCTOR
a) Short Straight Conductor
The magnetic field due to the electric current in a straight wire is
such that the field lines are circles with the wire at the center.
Where:
βP = magnetic field at a point P away from the
conductor
I = current through the conductor
r = perpendicular distance between point P and conductor
θ1& θ2 = angles it make at the point with the conductor
USING RIGHT HAND RULE:
Thumb:
current, I
4 Fingers: magnetic filed,
113
b) Long
Straight Conductor
For infinite/long straight conductor, θ1& θ2 = 90O and sin(90O) = 1, so the equation
now becomes
III.
MAGNETIC FIELD of a CIRCULAR COIL or LOOP
The field depends on the distance x along the axis from the center of the loop to
the field point.
Where: βP = magnetic field at a point P along the axis
of the coil
N = number of turns
I = current in the coil
a = radius of the coil
r = distance from the point P to the radius of coil
Note: At the center of the coil (r=a),
IV.
MAGNETIC FIELD due to a SOLENOID
A long coil of wire consisting of many loops is called a solenoid and each loop
produces a magnetic field. Between any two wires, the fields due to each loop
tend to cancel. Toward the center of the solenoid, the fields add up to give a
field that can be fairly large and fairly uniform.
114
Let ―P‖ be any point along the axis of solenoid
Where:
field at
solenoid
βP = magnetic
point P along the axis of the
a
N = number of turns
I = current in the solenoid
L = length of the solenoid
ELABORATE
Magnetic Field:
1. The figure below is a perspective view of a flat surface with area 3cm 2 in a uniform
magnetic field. The magnetic flux through this surface is +0.9 mWb. Find the magnitude
of the magnetic field.
2. Determine the flux that passes through the area as shown on the figure below.
115
3. A pair of point charges, q1 = 5µC and q2 = -3µC are moving in a reference frame as
shown. At this instant, what are the magnitude and direction of the net magnetic field
produced at the origin? v1 = 6x105m/s and v2 = 8x105 m/s.
4. A coil consisting of 25 turns has a radius of 20cm and carries a current of 8 Amp.
Determine the magnetic field at (a) its center and (b) a point on its axis 10cm from its
center. Draw a diagram to indicate the current and the magnetic field. (The coil is
along the x-z plane whose current is moving from the z to the x-axis.)
EVALUATE
5. A +6µC point charge is moving at a constant speed of 8x10 6 m/s in the +y-direction,
relative to a reference frame. At the instant when the point charge is at the origin of this
reference frame, what is the magnetic-field it produces at point with coordinates
x=0.5m and y=0.5m?
6. Two infinitely long wires carrying 12A and 8A in
opposite directions are laid parallel in air 10 cm
apart. Find the resultant magnetic field at a point
a) midway between the wires.
b) 2cm from 8A and 12cm from 12A.
c) 8cm from 8A and 10cm from 12A.
7. In the figure, solve for the resultant magnetic
field at point ―O‖.
Coil 1: I 1 = 2A
N1 = 30 turns
a1 =
Coil 2: I 2 = 4A
N2 = 25 turns
a2 =
116
8. A solenoid is positioned with its axis on the y-axis as shown. The current I = 10 A and its
radius is 0.1m. It has 100 turns. Solve for the flux density at point ―O‖.
117
MODULE 14: OPTICS
ENGAGE
What is the difference between refraction and reflection?
What is the difference between transparent, translucent and opaque objects?
How is it that when you raise your right hand infront of a mirror your image raises its left
hand?
Why is the word Ambulance is spelled backward in front of the vehicle?
Why do a spoon placed on a glass with water looks like it is broken?
If you can see the face of a friend who is under water can your friend also see you? Justify
your answer.
EXPLORE
Read Module 14: Optics (pp117 - 123)
EXPLAIN
OPTICS — NATURE OF LIGHT
Optics – branch of physics that deals with the study of the behavior and properties of light
*Light – a transverse wave, and it is the only visible wave/ray among the electromagnetic
spectrum. It can travel through a vacuum and many other media. In a vacuum, the
speed of light is a constant, c = 3 x 108 m/s. Light will travel slower through different
media.
118
REFLECTION AND REFRACTION OF LIGHT
A light ray that encounters a change in media will: reflect, and/or refract (pass through),
and/or be absorbed.
Reflection of Light – the return or bouncing of light wave from a surface
A reflection can occur in an organized way (smooth surface) or in a random way
(rough surface).
* Reflection off of a smooth surface is called specular reflection.
* Reflection off of a rough surface is called diffuse reflection.
Law of Reflection: The angle of incidence with respect to the normal of the reflecting
surface θi, equals the angle of reflection θr:
The amount of energy that is reflected compared to the amount incident is called the
reflectivity of the surface. This also is called albedo. The reflectivity of a mirror is about 96%
(albedo = 0.96).
119
Refraction of Light
When light travels from one medium to another, part of the light can be transmitted
across the media surface and refracted.
a) Refraction means that the light beam bends.
b) b) This bending takes place because the light beam‘s velocity changes as it goes
from one medium to the next.



If the light goes from the medium of high velocity to the one of low velocity, it is bent
toward the normal to the surface
If the light goes the other way, it is bent away from the normal.
Light moving along the normal is not deflected.
The index of refraction of a transparent medium is the ratio between the speed of light in
free space or vacuum, c and its speed in the medium, v:
* The greater its index of refraction, the more a beam of light is deflected on entering a
medium.
Law of Refraction (Snell’s Law):
According to Snell‘s law, the angles of incidence θi and refraction θr are related by the
formula
where v1 and n1 are, respectively, the velocity of light and index of refraction of the first
medium and v2 and n2 are the corresponding quantities in the second medium. Snell‘s
law is often written
120
THIN LENSES
The most familiar and widely used optical device (after the plane mirror) is the lens. A
lens is an optical system with two refracting surfaces. The simplest lens has two spherical
surfaces close enough together that we can neglect the distance between them (the
thickness of the lens); we call this a thin lens.
There are 2 basic types of lenses:
a) Convex Lens (Converging lens):
i) Lens thicker at center than edges.
ii) Light rays are refracted towards the focal point, F, on the other side of the lens.
b) Concave Lens (Diverging lens):
i) Lens thinner at center than edges.
ii) Light rays are refracted in a direction away from the focal point, F, on the inner side
of the lens.
Thin Lens Equation:
Just as we had for mirrors, the thin lens equation is
Where:
Di = Image distance
Do= Object distance
f = Focal length
f = (+) for convex lens; ( – ) for concave lens
121
Graphical Methods for Lenses
We can determine the position and size of an image formed by a thin lens by using a
graphical method very similar to the one we used for spherical mirrors.
1. A ray parallel to the axis emerges from the lens in a direction that passes through the
second focal point of a converging lens, or appears to come from the second focal
point of a diverging lens.
2. A ray through the center of the lens is not appreciably deviated; at the center of the lens
the two surfaces are parallel, so this ray emerges at essentially the same angle at which it
enters and along essentially the same line.
3. A ray through (or proceeding toward) the first focal point emerges parallel to the axis.
MIRROR
A mirror is a surface that can reflect a beam of light in one direction instead of either
scattering it widely in many directions or absorbing it.
A. Plane Mirrors.
Images formed by plane (i.e., flat) mirrors have the following properties:
 The image is as far behind the mirror as the object is in front.
 The image is unmagnified, virtual, perverted and upright.
*Perverted means that the image is flipped about the vertical,
i.e. when you raise your right hand infront of a plane mirror, your image raises its left hand
i.e. when you have a mole on your left cheek, your image will have its mole on the right
122
B. Spherical Mirrors.
Spherical mirrors have the shape of a segment of a sphere.
 Concave mirror: Reflecting surface is on the ―inside‖ of the curved surface.
 Convex mirror: Reflecting surface is on the ―outside‖ of the curved surface.
Focal Length and Focal Point
The focal point of a mirror is the point where parallel rays converge after reflection from
a concave mirror (converging mirror), or the point from which they appear to diverge after
reflection from a convex mirror (diverging mirror). The distance from the focal point to the
vertex is called the focal length, denoted as f.
If the radius of curvature of the mirror is R, the focal length f is R/2. For a concave mirror,
f is positive, and for a convex mirror, f is negative.
Mirror Equation
When an object is a distance Do from a mirror of focal length f, the image is located a
distance Di from the mirror, where:
123
This equation holds for both concave and convex mirrors.
Graphical Methods for Mirrors
We can also determine the properties of the image by a simple graphical method.
1. A ray parallel to the axis, after reflection, passes through the focal point F of a concave
mirror or appears to come from the (virtual) focal point of a convex mirror.
2. A ray through (or proceeding toward) the focal point F is reflected parallel to the axis.
3. A ray along the radius through or away from the Radius of curvature R intersects the
surface normally and is reflected back along its original path.
4. A ray to the vertex V is reflected forming equal angles with the optic axis.
ELABORATE:
REFRACTION:
1. A layer of oil (n = 1.45) floats on water (n = 1.33). A ray of light shines onto the oil with an
incidence angle of 40o. Find the angle the ray makes in water.
THIN LENSES:
1. An object OO’, 4 cm high, is 20 cm in front of a thin convex lens of focal length + 12
cm. Determine the position and height of its image II’ (a) by computation and (b)
by construction (ray diagram)
2. An object OO‘, is 9 cm high, is 27 cm in front of a concave lens of focal length -18
cm. Determine the position and height of its image II‘ (a) by computation and (b)
by construction (ray diagram).
MIRRORS
1. An object OO’, 5 cm high, is 25 cm from a concave mirror of radius 80 cm.
Determine the position and the relative size of the image II‘ (a) by the use of the
mirror equation and (b) by construction.
124
2. An object OO’ 6 cm high is located 30 cm in front of a convex spherical mirror of
radius 40 cm. Determine the position and height of its image II’ by (a) use of the
mirror equation and (b) construction.
EVALUATE
1. the speed of light in an unknown medium is measured to be 2.76 x 10 8 m/s. (a) What is
the index of refraction of the medium?
2. Optical fibers are generally composed of silica, with an index of refraction around
1.44. (a) How fast does light travel in a silica fiber, and (b) How long will it take for that
light to travel from Baguio to Manila (distance between the two cities 128miles)
3. Light travels from fiber optic cable into diamond with an angle of refraction (θ r) of 36.1
°. if the refractive index of fiber optic cable is 1.6 and the refractive index of diamond is
2.4, determine (a) the angle of incidence(θ i); (b) critical angle; (c) the speed of light in
each material.
4. A light ray strikes a reflective plane surface at an angle of 58° with the surface. (a) Find
the angle of incidence.(b) Find the angle of reflection.(c) Find the angle made by the
reflected ray and the surface.(d) Find the angle made by the incident and reflected
rays.
5. If the reflective surface in problem no. 4 is placed at an angel 10° above the horizontal.
Calculate all required values.
6. A ray of light is reflected by two parallel mirrors (1) and (2) at points A and B. The ray
makes an angle of 30° with the a parallel line between the two mirrors. Calculate for (a)
the angle of reflection at the point of incidence A; (b) the angle of reflection at the
point of incidence B (c) the apporxinatenumber of reflections made by the two mirrors
if the distance between the two mirrors id d = 4 cm and the length L of the two mirror
system is 3 meters, (d) In a real system, at each reflection, there are losses of the light
energy travelling between the two mirrors. If L and d are fixed, what can be done to
decrease the number of reflections on the mirrors and hence the energy lost by
reflection?
30°
125
7. An 2cm object is placed 20 cm from a concave mirror. The focal length is 10 cm.
Determine (a) The image distance (b) the image size (c) magnification and (d)
describe the image formed
8. An image formed by a concave mirror is 3 times greater than the object. If the radius of
curvature 20 cm, determine the object distance in front of the mirror and describe the
image formed
9. The focal length of a convex mirror is 10 cm and the object distance is 15 cm.
Determine (a) the image distance (b) the magnification of image (c) the image size of
the object is 3 cm tall and (d) describe the image formed
10. A biker sees the image of a motorcycle behind it 1/6 times its original size when the
distance between the biker and motorcycle is 30 meters. Determine (a) the radius of
curvature of the rear-view mirror (b) the image distance (c)describe the image formed
126
A. Main Reference
Freedman, R.A. and Young, H.D. (2012). University Physics: with Modern Physics. San
Francisco, CA: Pearson Education, Inc.
Giancoli, D. C. (2008). Physics for scientists and engineers. Pearson Education
International.
Serway, R. A., & Vuille, C. (2014). College physics. Cengage Learning.
B.
Books
Bord, D.J. & Ostdiek, V.J. (2012). The World of Physics, Philippines: Cengage Learning
Asia Pte Ltd.
Cummings, K et. al. (2004). Understanding Physics. John Wiley & Sons, Inc.
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