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School
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General Physics1
Alternative Delivery Mode
Quarter 1 – Module 5: Newton’s second law, DOT product, and Work
First Edition, 2020

Republic Act 8293, section 176 states that: No copyright shall subsist in any work of
the Government of the Philippines. However, prior approval of the government agency or office
wherein the work is created shall be necessary for exploitation of such work for profit. Such
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Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names,
trademarks, etc.) included in this module are owned by their respective copyright holders.
Every effort has been exerted to locate and seek permission to use these materials from their
respective copyright owners. The publisher and authors do not represent nor claim ownership
over them.

Published by the Department of Education, Division of Palawan

School Division Superintendent:
Natividad P. Bayubay, CESO VI
Assistant Schools Division Superintendents:
Loida P. Olavario, Ph.D.
Felix M. Famaran

Development Team of the Module

Writer: Argie Paybay
Editor: Fe Kenneth Gadiano – Aban and Alvin P. Cajiles
Management Team: Aurelia B. Marquez
Rodgie S. Demalinao
Rosalyn Caabay – Gadiano

Printed in the Philippines by ________________________

Department of Education – MIMAROPA Region – Division of PALAWAN

Office Address: PEO Road, Barangay Bancao-Bancao, Puerto Princesa City

Telephone: (048) 433-6392
E-mail Address: palawan@deped.gov.ph
Website: www.depedpalawan.com

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 Senior High School
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iii
Introductory Message
For the facilitator:

Welcome to the General Physics 1 Alternative Delivery Mode (ADM) Module

on Newton’s Second Law, Dot Product and Work!

This module was collaboratively designed, developed and reviewed by

educators both from public and private institutions to assist you, the teacher or
facilitator in helping the learners meet the standards set by the K to 12 Curriculum
while overcoming their personal, social, and economic constraints in schooling.

This learning resource hopes to engage the learners into guided and
independent learning activities at their own pace and time. Furthermore, this also
aims to help learners acquire the needed 21st century skills while taking into
consideration their needs and circumstances.

In addition to the material in the main text, you will also see this box in the
body of the module:

Notes to the Teacher

In view to the new normal world we are facing, this module
was created to answer that education should not stop for
our learners.
This General Physics 1 Module 5 for the Quarter 1 is
all about Newton’s Second Law, Dot product, and Work.
With this we are trying to allow our learners to work
independently in discovering through simple and enjoyable
activities/ experimentation that are aligned to the
competencies that they should learn.

As a facilitator, you are expected to orient the learners on how to use this
module. You also need to keep track of the learners' progress while allowing them to
manage their own learning. Furthermore, you are expected to encourage and assist
the learners as they do the tasks included in the module.

iv
For the learner:

Welcome to the General Physics 1 Alternative Delivery Mode (ADM) Module

on Newton’s Second Law, DOT Product and Work!

This module was designed to provide you with fun and meaningful
opportunities for guided and independent learning at your own pace and time. You
will be enabled to process the contents of the learning resource while being an active
learner.

This module has the following parts and corresponding icons:

What I Need to Know This will give you an idea of the skills or
competencies you are expected to learn in the
module.

This part includes an activity that aims to

What I Know
check what you already know about the
lesson to take. If you get all the answers
correct (100%), you may decide to skip this
module.

What’s In This is a brief drill or review to help you link

the current lesson with the previous one.
In this portion, the new lesson will be
What’s New
introduced to you in various ways; a story, a
song, a poem, a problem opener, an activity
or a situation.
This section provides a brief discussion of the
What is It
lesson. This aims to help you discover and
understand new concepts and skills.
This comprises activities for independent
What’s More
practice to solidify your understanding and
skills of the topic. You may check the
answers to the exercises using the Answer
Key at the end of the module.
This includes questions or blank
What I Have Learned
sentence/paragraph to be filled in to process
what you learned from the lesson.
This section provides an activity which will
What I Can Do
help you transfer your new knowledge or skill
into real life situations or concerns.

Assessment This is a task which aims to evaluate your

level of mastery in achieving the learning
competency.

v
 Additional Activities In this portion, another activity will be given
to you to enrich your knowledge or skill of the
lesson learned.

Answer Key This contains answers to all activities in the

module.

At the end of this module you will also find:

References This is a list of all sources used in developing

this module.

The following are some reminders in using this module:

1. Use the module with care. Do not put unnecessary mark/s on any part of the
module. Use a separate sheet of paper in answering the exercises.
2. Don’t forget to answer What I Know before moving on to the other activities
included in the module.
3. Read the instruction carefully before doing each task.
4. Observe honesty and integrity in doing the tasks and checking your answers.
5. Finish the task at hand before proceeding to the next.
6. Return this module to your teacher/facilitator once you are through with it.
If you encounter any difficulty in answering the tasks in this module, do not
hesitate to consult your teacher or facilitator. Always bear in mind that you are not
alone.

We hope that through this material, you will experience meaningful learning
and gain deep understanding of the relevant competencies. You can do it!

vi
 What I Need to Know

This module was designed and written with you in mind. It is here to help you
master the nature of Newton’s second law, DOT product, and Work. The scope of this
module permits it to be used in many different learning situations. The language
used recognizes the diverse vocabulary level of students. The lessons are arranged
to follow the standard sequence of the course. But the order in which you read them
can be changed to correspond with the textbook you are now using.

The module is divided into three lessons, namely:

• Lesson 1 – Newton’s Second Law
• Lesson 2 – Dot or Scalar Product
• Lesson 3 – Work done by a force acting on a system

After going through this module, you are expected to:

1. Apply Newton’s 2nd law and kinematics to obtain quantitative and qualitative
conclusions about the velocity and acceleration of one or more bodies, and the
contact and noncontact forces acting on one or more bodies.
2. Solve problems using Newton’s law of motion in contexts such as but not
limited to, ropes and pulleys, the design of mobile sculptures, transport of
loads on conveyor belts, force needed to move stalled vehicles, determination
of safe driving speeds on banked curved roads.
3. Calculate the dot or scalar product of vectors.
4. Determine the work done by a force acting on a system.

1
 What I Know

Directions: Choose the letter of the best answer. Write the chosen letter on a
separate sheet of paper.

1. What is produced when an unbalanced force acts on an object or body?

a. acceleration b. inertia c. interaction d. momentum

2. A highly motivated farmer pulls a 60-kg sack of rice across a smooth floor, with
negligible friction. If the sack of rice accelerates horizontally at 0.50m/s 2, how
much horizontal force does the man exert in pulling it?
a. 0.0083N b. 30N c. 59.50N d. 120N

3. Which of the following is a written equation for dot product?

a. B = AB cos b. A = A.B cos c. A.B = AB cos d. AB cos = A.B

4. What object will have lesser acceleration if same amount of force is being
applied?
Object a Object b

20kg. 45kg.

a. object a b. object b c. both a and b d. either a or b

5. If a certain force accelerates a 4.35kg mass by 3.5m/s 2. What would be the

acceleration of a 1.27kg mass when the same amount of force acts on it?
a. 9.97m/s2 b. 11.99m/s2 c. 13.96m/s2 d. 19.34m/s2

6. The following describes work. EXCEPT for__________?

a. scalar c. has a unit of J or N.m
b. vector d. product of force and distance

7. This variable is directly proportional to unbalanced force and inversely

proportional to mass.
a. acceleration c. interaction
b. inertia d. momentum

8. A bird is shot by a 5.0g bullet travelling at 205m/s, occasionally strikes its wing
and comes to rest in 501µseconds. Calculate the force imparted by the wing to
the bullet?
a. 2045N b. - 2045N c. 4.09 x 105N d. - 4.09 x 105N

2
9. What variable can be associated with the illustration below?
a. direction
b. magnitude B
c. both magnitude and direction
d. either magnitude or direction
)
A

10. There are two masses in a system, m 1= 8kg and m2= 10kg and are connected by
an inextensible light cord over a frictionless pulley. What happens to the
acceleration of the system? Refer your answer to the diagram below.
a. m1 will have lesser acceleration
b. m2 will experience bigger acceleration
c. both masses will experience equal acceleration
d. both masses will experience unequal acceleration

m1
11. Which of the following is an example of dot product?
a. acceleration c. work m2
b. momentum d. torque

12. Calculate the dot product when A=2i-3j+5k; B=i+2j+8k.

a. 32 b. 36 c. 48 d. 52

13. What happens if an object experiences no net force at all?

a. there is an acceleration
b. the value of mass becomes zero
c. the product of mass and acceleration is zero
d. the object is expected to either move slower or faster

14. How much work is done when 45N is applied by a porter to his load pushing it
through a distance of 5m?
a. 40J b. 150J c. 225J d. 245J

15. A volleyball with a mass of 0.325kg is hit above the net with a net force of 20N.
Find the acceleration of the ball.
a. 0.016m/s2 b. 6.5m/s2 c. 19.68m/s2 d. 61.54m/s2

You have completed the test! Now, you are ready to open the initial pages
which will bring you enlightenment. Take a deep breath and have fun as you begin
your journey.

3
 Lesson
Newton’s Second Law:
1 Acceleration
Perhaps you have already had the knowledge and understanding about force
and this is not new to you. The push or pull force being acted or exerted to a certain
body or an object will basically serve as your foundation in understanding the new
lesson. In this lesson, the concept as well as the theory about Newton’s second law
which is acceleration will be discussed.

What’s In

Activity 1: Think of an answer!

A certain empty bottle at rest caught your attention while walking at

Talacanen beach, you were being rude for a moment and kicked the bottle towards
the shore. The bottle flew for a couple of seconds before landing on the shore and
suddenly the waves repeated what you’ve done by pushing the bottle slowly back to
the beach. This scenario made you realized that though the waves moved the bottle
it was different when it comes to distance and the time taken by the bottle before it
completely stopped.

1. What do you think is/are the reason/s?

__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________

2. Does applied force has something to do with it?

__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________
__________________________________________________________________________________

4
 What’s New

Activity 2: Newton’s Second Law!

The second law states that, “when a net force acts on a body or an object, it
will be accelerated in the same direction with the force”.

Have you ever wondered how directly related is acceleration to the net force
and inversely proportional to the mass of the body or an object?

Think about this illustration!

2N 20.0g
Calculate for the acceleration of a
ball?

________________________________
________________________________
________________________________

What is It

There are a lot of ways to write the equation/kinematic equation of Newton’s

second law. Depending on what is/are the unknown variable/s and what is/are
given, the general equation can be derived

Mathematically, Newton’s second law can be written this as

F = ma

where F is the force applied to an object

m is the mass of the object
a is acceleration
The unit for force is N (newton). 1 N = 1 kg m/s2. Remember that the force in this
equation is a net force.

Example: How much force is exerted to an object whose mass is10 kg and
acceleration is 2.5 m/s2?
Given: m=10 kg, a= 2.5 m/s2, F=?
Solution: F=ma; F=(10 kg)( 2.5 m/s2); F= 25N

5
 The first part of the second law tells us that the greater the unbalanced force
acted upon a body is, the greater the acceleration of the body will be.
𝐹1 𝑎1
=
𝐹2 𝑎2
So, if we push a motorcycle with a certain force F 1 and then push it twice as
hard as the first push for the second time; we will have an acceleration that is twice
as the first time it is pushed.
The second part of Newton’s second law tells us that the greater the mass of
a body or of an object the lesser the acceleration is.

m1a1=m2a2
The second law can be stated in another way. Just recall that acceleration
𝑣𝑓 − 𝑣𝑜
a=
𝑡
𝑣𝑓−𝑣𝑜
If F = ma, then F = 𝑚
𝑡

𝑚𝑣𝑓−𝑚𝑣𝑜
F=
𝑡

Ft = 𝑚𝑣𝑓 − 𝑚𝑣𝑜
In some cases, when displacement is related to acceleration and time; then,
the equation will be

d = ½ at2
When we know the acceleration in a system and if distance and initial velocity
are given, the kinematic equation below can be used:

vf2 – vo2 = 2ad; vf2 = vo2 + 2ad

vf2 = √2ad

Tension is also a force that may act on a certain mass (kg).

Mathematically, tension T = ma

6
 What’s More

Activity 3: Concept application and Problem Solving!

Learning the Skill: Application of concept and problem solving
1. Use your knowledge about the concept of Newton’s second law. Understand
each statement to correctly answer them.
2. Apply the different equation about Newton’s second law or derive the
equations if needed in answering the problems. Find the given and calculate
for what is being asked in every problem.
Fill in the blank using your understanding of Newton’s second law.
1. A man pushes a box to slide it up the plane and exerted 10N on the first push
then 5N on the next push. Then, the first push would give us _________
acceleration and the second push will result to ________ acceleration.

2. Refer to the illustration below. The box on the right will slide _______________
than the box on the left. The box on the right experiences ________ as much
force is directed to the right.

m1 m2

F F2
1

3. If the same amount of force is applied to two objects with different masses,
then, the less massive object is expected to move ________ when experiencing
a net force.

4. Stalled vehicle requires _______ push to accelerate from rest.

5. If block1 whose mass is 12kg and block2 whose mass is 15kg are joined by
an inextensible light cord over a frictionless pulley. As shown below, block 1
will move _______ and block 2 will move _________.

Block 1
Block 2
) 30°

7
 What I Have Learned

Activity 4: Complete each statement!

Directions: Try to recall what you have just learned and complete the following
statements. Fill in the blanks with the term that will best complete the
thought.
1. The massive the object the ______________________________ to accelerate.
2. Acceleration acts in the same direction as the ________________________.
3. The acceleration of an object is _____________________________ to the mass of an
object.
4. When an object does not accelerate, it does not mean that __________________.
5. If mass is (kg) and acceleration is (m/s 2). Then, force is _____________________.
6. The direction of acceleration tells us that _________________________________.
7. A 20N force is exerted to a ball whose mass is 0.75kg, this has an acceleration
of ___________?
8. A motorcycle and a tricycle are both moving at 35km/h. Which of the two vehicles
is harder to stop? Why? __________________________________________.
9. A force of 7N gives an object an acceleration of 3.5m/s 2. What force is needed to
give the same object an acceleration of 2m/s2? _________________.
10. An acceleration of 25m/s2 is produced when an object whose mass is 2000g is
exerted with a force of _________.

What I Can Do

Directions: Analyze the following situations and fill in the blank/s with necessary
word/s that will complete the whole statement. Make sure to choose the
best word that will make the whole statement correct.
1. A 50kg sack of rice is _______ to push compared to a 25kg sack of rice.
2. The forces that may act on a stalled/stationary vehicle are ______ and _______.
3. When an object experiences no net force, it means that the product of mass and
acceleration is ______.
4. The ground exerts ________ force to the force given by a stationary vehicle.
5. If a truck weighs massive than the tricycle, then the truck requires ________ force
to stop.

8
 Assessment

Directions: Read and answer the following problems carefully. Each item has four
(4) suggested possible answers. Choose the letter of the best answer and
write it on a separate sheet of paper.

1. If there are two masses involved in the system. One is lighter than the other
and their acceleration is equal to both masses.
What happens to the tension in the system?
a. m1 will have lesser tension
b. m2 will experience bigger tension
c. both masses will experience equal tension
d. both masses will experience unequal tension

2. If applied force is constant, the mass and

acceleration produced are.
a. equal c. directly proportional
b. not related d. inversely proportional

3. A coin is thrown upward from a roof at the same time as the identical coin is
dropped from there. Which of the following statements is TRUE? The coins
_______.
a. will collide midair
b. reach the ground at the same time
c. have the same velocity when they reach the ground
d. have the same acceleration when they reach the ground

4. How should the two blocks move? Refer to the illustration.

a. block 1 will move upward
b. block 2 will move upward
c. both blocks will move upward
d. both blocks will move downward

5. Two masses m1= 9kg and m2=11kg and are connected to an inextensible light
cord over a frictionless pulley. Calculate the acceleration of the system.
0.98 m/s2 b. 2.04 m/s2 c. 10.2 m/s2 d. 19.6 m/s2

6. If a bullet of 4.75g travelled and hit the wall at 125 m/s then came to rest in
2.75s, how much force did the wall exert to the bullet causing it to stop?
-0.22 N b. -13.06 N c. -45.45 N d. -49.78 N

7. A certain force gives a 5 kg object an acceleration of 2 m/s 2. If the same force is

applied to a 20 kg object, what is the acceleration produced?
a. 0.5 m/s2 b. 2.0 m/s2 c. 4.9 m/s2 d. 8.0 m/s2

9
8. How heavy is a piece of log if it is pushed by a force of 45N and accelerates at
0.76m/s2?
12.4 kg b. 34.2 kg c. 44.24 kg d. 59.21 kg

9. Refer to item number 5. Calculate for the tension T in the cord.

8.9 N b. 10.78 N c. 11.22N d. 19.6N

10. A 30N force is exerted to a sack of 25kg rice and is pushed and piled up in a
corner. What is the acceleration of the sack?
a. 0.83 m/s2 b. 1.2 m/s2 c. 5.0 m/s2 d. 7.5 m/s2

11. If 25N is exerted to accelerate an object to 2.5 m/s2, how heavy is an object?
a. 0.1kg b. 10kg c. 25kg d. 62.5kg

12. A person’s mass is 55kg and steps into an elevator to move to another floor.
What is the person’s weight as the elevator accelerates downward at a rate of
1.75 m/s ?
2

a. 96.3N b. 442.8N c. 539.0N d. 808.5N

13. A 75N is applied to a 57.5kg object located on a frictionless area. What is the
acceleration of an object?
a. 17.5m/s2 b. 7.7m/s2 c. 1.3m/s2 d. 0.8m/s2

14. In which case is the car accelerating?

a. a car changing lanes at a constant speed.
b. a car slowing down while making a left turn.
c. a car speeding up in an effort to beat the red light.
d. all of the above.

15. A bird is shot by a 5.0g bullet travelling at 205m/s, occasionally strikes its
wing and comes to rest in 501µseconds. Calculate the force imparted by the
wing to the bullet?
a. 2045N b. - 2045N c. 4.09 x 105N d. - 4.09 x 105N

Additional Activities

Activity 6: True or False

Directions: Analyze the following situations. Before each number, write True if the
statement is correct and write False if the statement is wrong.

_____ 1. Massive objects require lesser amount of force to accelerate.

_____ 2. The mass of an object is inversely proportional to acceleration.

10
_____ 3. If we push a motorcycle with a certain force F1 and then we push it twice as
hard as the first push for the second time, it will give us an acceleration twice
as the first time it is pushed.
_____ 4. An airplane that is massive than a car requires greater amount of force to
take-off.
_____ 5. When an object does not accelerate, it means that no force is acting on it.

Congratulations for a job well done! You are now ready to take-off to another
level just make sure to have fun while learning. Do not force yourself so new tons of
things will be easy to learn.

Lesson

2 Dot Product
The dot product is derived from two vector quantities and it is also called as
the scalar product. In this lesson even though two vectors are being multiplied the
product is still a scalar quantity. A certain point and its projection along with another
point in relation to a certain cosine of the angle is what this lesson will talk about.

What’s In

Activity 1: Crossword Puzzle!

Directions: Look for words related to dot product in the crossword puzzle below.
Write your answer on a separate sheet of paper.

A N G L E M E I F E U M
I O A D D I T I O N N U
M L V O E N I N L O I L
S O I E V D O T R T V T
S V V I C T I R O I E I
C E B I C T E O G O R P
A D D I N E O N I N S L
L O N G E G E R O O A I
A C C E L E R A T I L C
R Q U A N T I T Y L A A
C O S I N E A R U I N T
A W E S O M E L O V E I
A D V E N T U R E S R O
K A G S A W A R U I N N

11
 What’s New

Activity 2: Complete the paragraph!

Vectors do not only focus on operations such as __________ and ____________.
In this lesson multiplication of ____________ is the essential focus. Two vectors
quantity are the main points in this lesson as well as the _________ angle or the
smaller ___________ in between them. We will also study one way to multiply vectors
which is _____________________.

What is It

Mathematically, the dot product or scalar product of two vector is

A º B = AB cos
where A and B are the magnitudes of two vectors and cos is the smaller angle
between the direction of the two vectors, simply put, two vectors with an angle in
between them.

Example: Calculate the dot product of two vectors A and B, where A = 10 N and B
=12 N separated by an angle of 45°.

Given: A = 10N, B = 12N, = cos45°

Solution: A.B = A.B cos ;

A.B=(10N)(12)(cos45°)
A.B= 84.85N

Dot product may be viewed as the product of A and the projection of B along
A. The common example of a dot product is work for it is a dot product of both force
and displacement and work is usually a scalar quantity. Dot product can also be
calculated through component method. In equation:
A.B = A1B1+A2B2+A3B3

12
 What’s More

Activity 3: Calculate the dot or scalar product of vectors!

Read and answer the following problem set carefully. Write your answer on a
separate sheet of paper.
1. There are two vectors, A and B and they are perpendicular to each other as
shown below. Find the dot product of the two vectors.

A=3.4N

90°

2. Find the dot product of two vectors given; F=12N, G=10N, and cos30.
B= 3.7N

3. Calculate for the dot product of the two vectors S and T.

S = (6i + 4j + 2k)m; T = (-3i + 4j + 5k)m
4. Find A.B given that A = 3m and B = 5m at an angle of 45°.
5. Calculate the dot product of C = (-2, -6)N and D = (-2, 3)N.

What I Have Learned

Activity 4: Fill in the blank!

1. If A.B and B.A are the same then dot product shows _____________ property.
2. Dot product is a _________ quantity and product of two __________.
3. The angle in between the two vectors is the ________ angle.
4. The range of the angle in between two vectors is _____ to ______.
5. Basically, the common example of dot or scalar product is _________.

13
 What I Can Do

Activity 5: Complete each statement!

1. The angle in which a certain satellite is placed is _________ to the signal received.
2. Not all the houses have the same roof design. That is why solar panels are placed
or positioned depending on the angle of tilt of the roof for ____________ electrical
power.
3. Street lights are also positioned in a certain angle to have _________ light for the
street.
4. When you are standing on a moving truck, the common thing you do with your
base is that you create _________ angle in order to stay still.
5. In choosing the path to walk on, we usually take the ________ route.

Assessment

Multiple Choice

Directions: Read and answer the following problems carefully. Each item has four
(4) suggested possible answers. Choose the letter of the best answer and
write it on a separate sheet of paper.

1. Find the dot product of vectors X and Y if they have the same value which is 12N
at 35°.
a. 9.83 N b. 34.24N c. 117.96 N d. 410.82 N

2. Calculate the dot product of two vectors.

Q = (2i + 6j + 4k) m; R = (-3i + 3j + 5k) m
a. 32m b. 42m c. -32m d. -42m

3. What is the dot product of two vectors perpendicular to each other with an angle
of 40° in between them? Vector 1 is 5 m as vector 2 is 7 m.
a. 5.36m b. 26.81m c. 39.85m d. 56.0m

4. Two men walk at different paths. The first man walks north at 12 m and the second
man takes northeast at 15 m. What is the dot product of the two men?
a. 40.10m b. 44.02m c. 127.28m d. 521.0m

5. An arrow is accidentally shot to a certain post and creates an angle of 65° in

between them. If an arrow is 0.75 m long and the post is 5 m, what is the dot
product of these two vectors?
a. 1.58m b. 2.82m c. 6.67m d. 11.30m

14
6. Calculate the dot product when A=2i-3j+5k; B=i+2k+8k.
a. 32 b. 36 c. 48 d. 52

7. What variable can be associated with the illustration below?

a. direction
b. Magnitude B
c. both magnitude and direction
d. either magnitude or direction
)
A
8. Which of the following is an example of dot product?
a. energy b. force c. torque d. work

9. If two non-zero vectors have a dot product equal to 0, then the two vectors are
said to be ____________ to each other.
a. Perpendicular c. parallel (pointing in the same direction)
b. cannot be determined d. parallel (pointing in the opposite direction)

10. Which of the following is a written equation for dot product?

a. B = AB cos c. A.B = AB cos
b. A = A.B cos d. AB cos = A.B

11. Determine if the dot product is parallel or not using the given below.
Q = (4i -6j + 4k) m; R = (-3i + j + 2k)
a. parallel c. not parallel
b. perpendicular d. can’t be determined

12. Calculate the dot product of C = (-2, -6)N and D = (-2, 3)N.
a. 13N b. 14N c. -13N d. -14N

13. Find the dot product of two vectors given; F=12N, G=10N, and cos30.
a. 113.36N b. 109.08N c. 103.92N d. 101.75N
14. The angle formed is orthogonal if the value of dot product is________?
a. 0N b. 1N c. 2N d. 3N

15. What does dot product means?

a. addition c. multiplication
b. division d. subtraction

15
 Additional Activities

Activity 6: Remember your paths and maybe your past!

Directions: Think of the routes/paths that you take. It could be a nearest store, a
school, and etc. Design your own short cut route that you think you
could take and that would lead to your chosen destination. Do this on a
short bond paper or a clean sheet of paper. You may use markers,
pencils, and other drawing materials present at home.

Lesson
3 Work
Now let’s talk about work!
This lesson is not new to you at all. Work can be described by anyone
depending on their perception about it. On a daily basis, moving the chair to a
distance, lifting objects to a certain height, and even the normal walking involve
work.

What’s In

Activity 1: Think, just think!

You are asked to pull a cabinet and so you exerted so much effort in order to
pull the cabinet. Sadly, no matter how much you pulled it, you were not able to move
it.
Think of any reason/s why does the cabinet didn’t move at all?

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16
 What’s New

Activity 2: Work it!

Remember how you brought your cellphone to a certain height or distance
just to take selfies and nice glam shot? That’s actually a work already. In that case
you have exerted an upward force to overcome the force of gravity. And this basically
results to a vertical displacement of your cellphone parallel to the applied force.

What it means for a force to do work on a body?

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How to calculate the amount of work done?

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What is It

In lesson 1 of this module we talked about Newton’s second law which

basically involved the force. If a force acted on an object and brought it to a certain
height or distance, work is there. The involvement of force and displacement led to
the existence of work in a particular system. Moreover, work is also related to energy
for it is transferred if you do something to an object.
When a force succeeds in moving the body in which it acts then work exists.
Work is a scalar quantity and therefore there is no direction associated to it. The SI
unit of work is joule (J) or simply a combination of Newton meter (N.m) which is a
force of one Newton in a moving body through a distance of one meter.

W = Fd

where W – work (J)

F – force (N)
d – distance (m)

Example: A consumer exerted 30 N on a push cart through a distance of 5 m. How

much work is done by the consumer?

Given: F = 30 N, d = 5 m, W = ?

17
Solution: W = F d
W = (30N)(5m)
W = 150 J

Work can also be written in this equation if height is involved:

W = mgh
where m = mass of an object (kg)
g = acceleration due to gravity (9.8m/s 2)
h = height (m)

If the force (F) and displacement (d) are not parallel to each other, it says that only
the component of the force parallel to the displacement does the work.

Given the angle in between the directions of (F) and (d), then the magnitude of the
component of (F) parallel to (d) is F cos . If the calculated work is negative, this
means that force is acting in a direction opposite to the motion. This case is written
in equation as
W = Fd (cos )
If the force exerted is at the right angle to the displacement, then no work is done.
Right angle is 90° and cos90° is equal to zero, therefore work is equal to zero.
Example: A dog whose weight is 30 N runs up the stairs where the vertical distance
between the floors is 3.5 m. Calculate for the work done by gravity.

Given: F = 30 N, d =3.5 m, cos180°

Solution: W = Fd cos
W = (30N)(3.5m)(cos180°)
W = - 105 J

What’s More

Problem Solving!
Directions: Read and answer the following problem set carefully. Write your
answer on a separate sheet of paper.
1. A 3.7 kg rock on the shore is moved to a distance of 6.4 m caused by a 5.5 N
strong wave. What is the work done by a force to an object?

2. Calculate for the work done by a force on the system. Refer your answer to the
illustration below.
F = 10N

35° )
d = 5N
18
3. A 2.3 kg box is lifted to a 1.3 m high shelf. What is the work done?

4. If a load weighing 160 N is attached to the end of the rope and dragged across a
horizontal floor; how much work is done if the rope makes a horizontal angle of
35° and the load is dragged over a distance of 15 m?

5. The porter exerted a 225 N force to push a crate to a distance of 7.2 m. Calculate
for the work done.

What I Have Learned

Activity 4: Fill in the blank!

Directions: Complete the following statements by filling in the blanks with the
correct terms you have just studied.

1. Work is done if the acting force is _____________ to the displacement.

2. Cos0° is equal to 1 then cos90° is ______ it means that __________ is done.

3. If there is a movement to the direction of the ________ then there is work done.

4. There is no work done if displacement is ______________ to the applied force.

5. Work is also a transfer of _________ that causes movement or sets an object in

motion.

What I Can Do

Activity 5: True or False!

Directions: Read and analyze the following statements. Write True if the statement
describes correctly about work and False if not. Write your answer on
the separate sheet of paper.
1. Work is done if you are able to push your table at a distance.
2. An object moving at a constant speed on a frictionless floor
3. Children crossing the streets
4. A box is being pushed across a horizontal ground at a constant speed.
5. Stowed vehicle on a parking lo

19
 Assessment

Multiple Choice

Directions: Read and answer the following problems carefully. Each item has four
(4) suggested possible answers. Choose the letter of the best answer and
write it on a separate sheet of paper.

1. Calculate for the work if 125N force is applied by a porter to his load pushing it
through a distance of 15m?
1587J b. 1785J c. 1875J d. 1885J

2. When an object is at rest, there is no work done because____________________.

a. the floor is frictionless.
b. there is no force acting on the object.
c. there is no friction exerted on the object.
d. there is no distance moved or taken by an object.

3. A box is pushed horizontally on a smooth surface by a 10 N force to 6 m distance.

What is the work done on the surface of the box by the 10 N force?
0.6 J b. 16 J c. 1.7 J d. 60 J

4. Which of the following situations has no work done?

a. a saw cuts a piece of wood.
b. a space ship cruises in space.
c. a boy drags his bag along the floor.
d. a ball falls from a tree due to gravity.

5. A dog whose weight is 30N runs down the stairs with 3.5m vertical distance
between the floors. Calculate for the work done by gravity.
105J b. -105J c. 125J d. -125J

6. A 40kg crate is being pushed along the floor with a force of 15N and covered a
distance of 7.5m. calculate for the work done?
73.5N b. 80.0N c. 112.5N d. 300.0N

7. If 10J of energy is transferred to certain toy car which mass is 25g and being
exerted with 3N to moved. How far does it go?
2.5m b. 3.3m c. 8.3m d. 11.7m

8. Olivia exerts a force of 200N on a boulder but she wasn’t able to move it. How
much work is done?
0J b. 1J c. 4.9J d. 5.1

9. The following describes work EXCEPT

a. scalar c. has a unit of J or N.m
b. vector d. product of force and distanc

20
10. Across a horizontal surface a box weighing180N is dragged and is moved using
a rope attached to the front end. If the rope makes an angle of 30° with the
horizontal surface, how much work is done by the man in dragging the box over
a distance of 8m?
720.0J b. 831.4J c. 1247.1J d. 1140.0J

11. An old lady applied a force of 50N to move her bag to a distance of 10m. How
much work is done?
a. 5J b. 50J c. 500J d. 5000J

12. A certain 200.0g coin at rest falls freely from a height of 3m. Determine the work
done by the gravity?
a. 2.6J b. 5.9J c. 12.8J d. 19.6J

13. If a 10N acted on a box along a horizontal surface to a distance of 1m. Calculate
for the work done?
a. 1J b. 10J c. 15J d. 21J

14. How much work is done when 45N is applied by a porter to his load pushing it
through a distance of 5m?
a. 40J b. 150J c. 225J d. 245J

15. If a load weighing 160 N is attached to the end of the rope and dragged across a
horizontal floor; how much work is done if the rope makes a horizontal angle of
35° and the load is dragged over a distance of 15 m?
a. 1965.96J b. 1970.70J c. 1979.35J d. 1992.72J

Additional Activities

Directions: Think about the things that you have done today, all the chores or
movements where work is related. In a clean sheet of paper, list down
at least ten (10) work you have done today.
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21
 Answer Key

What I Know Lesson 2 Lesson 3

1. A What's More What's More
2. B 1. 0 1. 35.2 J
3. C 2. 103.92 N
3. 8 m
2. 40.96 J
4. B
4. 10.61 m 3. 29.30 J
5. B
6. B 5. -14 N 4. 1965.96 J
7. A 5. 1,620 J
8. D
9. B
10. C Lesson 2 Lesson 3
11. C Assessment Assessment
12. B 1. C 1. C
13. C 2. A 2. D
14. C 3. B 3. D
15. D 4. C 4. B
5. A 5. A
Lesson 1 6. B 6. C
What's More 7. B 7. B
1. Twice, twice 8. D 8. A
2. Much faster, twice 9. A 9. B
3. Faster 10. C 10. C
4. Bigger 11. C 11. C
5. Zero / 0 12. D 12. B
13. C 13. B
14. A 14. C
Lesson 1
15. C 15. A
Assessment
1. C
2. A
3. D
4. A
5. A
6. A
7. A
8. D
9. B
10. B
11. B
12. B
13. C
14. C
15. D

22
References

Angelina A. Silverio. Exploring Life Through Science Physics (2013). Phoenix

Publishing House, Inc
Hugh D. Young, Carnegie Melon University., Roger A. Freedman, University
of California, Santa Barbara., A Lewis Ford, Texas A&M University.
Sean and Zemansky’s University Physics With Modern Physics 13th
Edition
Carmelita U. Cruz, Ph.,D. Contemporary Physics (2000). Instructional
Coverage System Publishing, Inc

23
For inquiries or feedback, please write or call:

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Curriculum Implementation Division Office

2nd Floor Deped Palawan Building
Telephone no. (048) 433-3292

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Telephone no. (048) 434-0099