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Week 1-3: Unit Learning Outcomes 1(ULO-1): At the end of the unit, you are expected to
a. Recall the measurement units, significant figures, and identify vector and scalar
quantities
b. Recall and apply the principles of motion and kinematics
c. Recall and apply the concept of force and systems of forces
Big Picture in Focus: ULO-1a. To recall fundamental principles of measurement units
and significant figures, and identify vector and scalar quantities
Metalanguage
In this section, the units of measurement, significant figures, vector, and scalar quantities relevant
to the study of physics will be reviewed to demonstrate ULOa. You need to understand the
following principles to demonstrate ULOa as a foundation knowledge in solving physics problems.
1. There are three fundamental physical quantities of mechanics – length, mass, and
time, which in the SI system have the units meter (m), kilogram (kg), and second
(s), respectively.
2. Dimensions can be treated as algebraic quantities. Dimensional analysis involves
the breaking down of units into simpler and basic quantities.
3. Measurements must consider significant figures. Notice that the rule for addition
and subtraction is different from that for multiplication and division. For addition
and subtraction, the important consideration is the number of decimal places, not
the number of significant figures.
4. Quantities may be classified as scalar or vector. Scalar describes the magnitude of a
physical phenomena without direction. Vector, on the other hand, is a quantity that
defines both the magnitude and direction.
5. Scalar always positive, vector can either be positive or negative.
Essential Knowledge
To perform the aforesaid big picture (unit learning outcomes) for the first three
(3) weeks of the course, you need to understand the basic rules of measurements and
quantities necessary that will be laid down in the succeeding pages. Like all other
sciences, physics is based on experimental observations and quantitative measurements.
Measurements are associated with a physical quantity, hence, the units and significant
figures are relevant. On the other hand, physical quantities may be classified as scalar or
vector.
Please note that you are not limited to exclusively refer to these resources.
Thus, you are expected to utilize other books, research articles, and other resources
that are available in the university’s library e.g. ebrary, search.proquest.com etc.
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Keywords
Length
Mass
time
SI unit
Unit prefixes
English Unit
Significant Figures
Scalar
Vector
Vector algebra
Unit vector
Units of Measurement
Measurements are done to describe natural phenomena where each measurement is
associated with a physical quantity. In mechanics, there are three fundamental quantities
such as length, mass, and time – the building blocks of all other quantities.
In measurement, a standard must be defined. In 1960, SI (Système International) was set as
a standard for fundamental quantities of science, also called the metric system. Its standard
units of measure are meter, kilogram, and seconds, known as mks.
Length
Length is defined as the distance between two points in space. The history of unit length
started in 1120 when the king of England decreed that the standard length in his country is
as long as the distance from the tip of his nose to the end of his outstretched arm, he called
this unit as yard. Similarly, a foot is equivalent to the length of the royal foot of King Louis
XIV. However, none of these became constant as it keeps on changing every time a new king
takes the throne. Later until 1799, the standard for length became meter (m) which is
equivalent to one ten-millionth of the distance from the equator of the North Pole alone one
particular longitudinal line passes through Paris. However, the length of a meter was
continuously redefined by expert bodies over history. Finally, in October 1983, the meter
was defined as the distance traveled by light in vacuum during a time of 1/299 792 458
second. Table 1.1 lists approximate values of some measured lengths relevant to physics.
Table 1.1 Approximate Values of Some Measured Lengths
One light-year
Mean orbit radius of the Earth about the Sun
Mean distance from the Earth to the Moon
Distance from the equator to the North Pole
Mean radius of the Earth
Diameter of a hydrogen atom
Diameter of an atomic nucleus
Diameter of a proton
Length (m)
9.46 x1015
1.50 x1011
3.84 x108
1.00 x 107
6.37 x106
≈10x -10
≈10x -14
≈10x -15
Mass
In 1887, the SI standard unit for mass was established and called kilogram (kg). One
kilogram is defined as the mass of a specific platinum–iridium alloy cylinder kept at the
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International Bureau of Weights and Measures at Sèvres, France. Table 1.2 shows some
approximate masses of various objects relevant to physics.
Table 1.2 Approximate Values of Some Measured Masses
Sun
Mass (kg)
1.99 x1030
Earth
Moon
5.98 x1024
7.36 x1022
Hydrogen atom
Electron
1.67 x10-27
9.11 x10-31
Time
The SI standard unit for time is second (sec) measured from the period of vibration of
radiation from the cesium-133 atom.
Other SI standard units are kelvin (K) (read as Kelvin and not degree Kelvin) for
temperature, ampere (A) for electric current, (Candela, cd) for luminous intensity, and
(mole) amount of substance.
SI Unit Prefixes
A unit prefix precedes a basic unit to indicate a multiple or fraction of the unit. It denotes
multipliers of the basic units based on various powers of ten. Table 1.3 is a list of the various
powers of ten and their abbreviations. These prefixes are helpful especially for values that are
too small or too big to quantify, for example, the mean radius of the Earth is 6,370,000 meters
that is equivalent to 6, 370 kilometers (km) or 6.37 x106 which is a much easier notation.
Basic knowledge on unit conversion includes unit analysis and significant figures which will
be discussed later part on this ULOa.
Table 1.3 Unit Prefix
Power
Prefix
Abbreviation
Power
Prefix
Abbreviation
3
-24
yocto
y
10
kilo
k
10
6
zepto
z
10
mega
M
10-21
9
-18
atto
a
10
giga
G
10
fempto
f
tera
T
10-15
1012
-12
pico
p
15
peta
P
10
10
-9
nano
n
18
exa
E
10
10
micro
𝜇
zetta
Z
10-6
1021
-3
milli
m
24
yotta
Y
10
10
centi
c
10-2
-1
deci
d
10
The United States Customary System is another unit system similar to Imperial Units or what
is ambiguously called the English Units. An old British system of measuring units including
inch, foot, yard, mile, ounce, pound, gallon, and so. This system of units comes from a lot of
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history, mostly defined and redefined based on the reigning kings and queens of the British
Empire. Its standard units of measure are foot (ft), slug, and seconds (sec) for the length,
mass, and time, respectively. Table 1.4 lists the unit equivalence of some basic English and SI
units.
Table 1.4. English and SI Units
English Unit
Mile
Foot
Inch
Pound
Ounce
Gallon
Celsius
SI Unit
Kilometer
Meter
Centimeter
Grams
Grams
Liter
Kelvin
Conversion
1 mile = 1.609 km
1 ft = 0.305 m
1 in = 2.54 cm
1 lb = 453.59 g
1 oz = 28.35 g
1 gal = 3.79 L
0 °𝐶 = 273 K
Unit Conversion
Unit conversion is a process involving multiplication or division by a numerical factor to
obtain the desired unit. In solving problems, you have to consider the uniformity of the units
before going through. For example, solving the volume of a cardboard box whose dimensions
are 2 ft by 2ft by 3 meters would not give you 12 ft3. Instead, you need to convert first the
meter unit to feet before performing the calculations. Units can be treated as algebraic
quantities that can cancel each other, see the following examples.
Example 1
Refer to table 1.4 and convert 5 cm to inches.
Step 1: Identify your given and desired unit
Step 2: Use the factor of conversion.
In this case, 1 in = 2.54 cm.
Step 3: Make the conversion in step 2 a ratio, always aim to cancel out the given unit by
division. Multiply the ratio to the given.
(5 𝑐𝑚) (
1 𝑖𝑛
) = 𝟏. 𝟗𝟔𝟖 𝒊𝒏
2.54 𝑐𝑚
Notice how the centimeters canceled out, which is your main objective.
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Example 2
Express the diameter of a hydrogen atom using prefixes.
From Table 1.1, the diameter of a hydrogen atom is ≈x10-10 meters, say 1.2x10 -10 m.
Step 1: Identify the nearest prefix to the given power of ten. Refer to Table 1.2
In this case, x10-10 is nearest to nano which is x10-9.
Step 2: Express x10-10 to nano by moving the decimal point up to 1 decimal place to the
left. (Note that shifting the decimal point to the left means increment to the power while
shifting to the right means decrement to the power, considering the sign.)
1.2 x10-10 m = 0.12 x10-9 m = 0.12 nanometer = 0.12 nm
Step 3: Express x10-10 to micro by moving the decimal point of step 2 up to 3 decimal
place to the left.
0.12 x10-9 m = 0.00012 x10-6 m = 0.00012 micrometer = 0.00012 𝜇𝑚
Step 4: Express x10-10 to mili by moving the decimal point of step 3 up to 3 decimal place
to the left.
0.00012 x10-6 m = 0.00000012 x10-3 m = 0.00000012 milimeter = 0.00000012 mm
Now you see that 1.2x10 -10 m = 0.12nm = 0.00012 𝝁𝒎 = 0.00000012 mm. Therefore, it
would be smarter to express too small or too large values using unit prefixes.
Example 3
A rectangular building lot has a width of 75.0 ft and a length of 125 ft. Determine the area
of this lot in square meters.
The area is solved by
𝑨 = 𝑳𝑾 = 𝟕𝟓. 𝟎 𝒇𝒕 𝒙 𝟏𝟐𝟓 𝒇𝒕 = 𝟗 𝟑𝟕𝟓 𝒇𝒕𝟐
Now 1 meter = 3.280 ft
1𝑚 2
(9375 𝑓𝑡 2 )(
) = 871.412 𝑚2
3.280 𝑓𝑡
Notice that the conversion factor is squared to obtain the square-meter area. (This is a
common mistake for students who forget the need to square the conversion factor.)
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Example 4
At Mc Arthur Highway, a car is traveling at a speed of 38 m/s. Is the driver over
speeding the limit of 75.0 mi/h?
Know that 1 mile = 1 609 meters. Take note that our desired unit is mile, therefore, make it
as the numerator.
(38
𝑚
1 𝑚𝑖
𝑚𝑖
)(
) = 2.36𝑥10−2
𝑠 1609 𝑚
𝑠
Now convert seconds to hours,
(2.36𝑥10−2
𝑚𝑖
60 𝑠
60 𝑚𝑖𝑛
𝑚𝑖
)(
)(
) = 84.96
𝑠
1 𝑚𝑖𝑛
1 ℎ𝑟
ℎ𝑟
Therefore, the driver is over speeding the limit.
Often problems include dimensional analysis which is a form of proportional reasoning for
physical quantities expressed in different units. Dimensions can be treated as algebraic
quantities, for example, quantities can be added or subtracted only if they have the same
dimensions. Furthermore, any relationship can be correct as long as the terms on both
sides of an equation have the same dimensions.
Example 5
Show that the expression 𝒗 = 𝒂𝒕, where 𝒗 represents speed, 𝒂 acceleration, and 𝒕 an
instant of time, is dimensionally correct.
Speed is defined as the distance traveled over a given time, in terms of dimension,
𝑣=
𝐿
𝑇
Where L is the distance traveled and T is the time.
On the other hand, acceleration is defined as the rate at which the object changes its
velocity, in terms of dimension,
𝐿
𝑎 =
𝑇2
Substitute to 𝒗 = 𝒂𝒕, we get,
𝑣 = 𝑎𝑡
𝐿
𝑇
=(
𝐿
) (𝑇)
𝑇2
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𝑳
𝑻
=
𝑳
𝑻
Therefore, 𝑣 = 𝑎𝑡 is dimensionally correct because we have the same dimensions on both
sides. If the expression were given as 𝑣 = 𝑎𝑡2, it would be dimensionally incorrect. Solve
and see!
Example 6
One of the fundamental laws of motion states that the acceleration of an object is directly
proportional to the resultant force on it and inversely proportional to its mass. If the
proportionality constant is defined to have no dimensions, determine the dimensions of
force.
𝐹
Given that 𝑎 ∝ , we have 𝐹 ∝ 𝑚𝑎. Therefore the units of force are those of 𝑚𝑎. From
𝑚
Example 1, acceleration is 𝑎 =
𝐿
𝑇2
which means
𝐿
𝒎
𝐹 = 𝑚𝑎 = (𝑀) ( 2) = 𝒌𝒈 ∙ 𝟐 = 𝟏 𝑵𝒆𝒘𝒕𝒐𝒏
𝑇
𝒔
Newton, 𝑁, is the SI standard unit of force.
Significant Figures
The significant figures of a number are digits that express something about the uncertainty.
For example, we measure the diameter of a disc using a meter stick. If the diameter is 12.0
cm and assuming that the accuracy of the number of measurements performed is ± 0.1𝑐𝑚,
we can claim that the diameter is somewhere between 11.9 cm to 12.1 cm.
In identifying the significant figures, remember the following:
•
•
•
•
Nonzero digits are significant, for example, 1.3 and 1.3 x10-4 both have two
significant figures.
Zeros may or may not be significant figures. Zeros that are used to position the
decimal point is not significant. Say, 0.003, 3.0, and 0.000133 have one, two and
three significant figures respectively.
Zeros between two significant digits are significant.
Trailing zeros or the zeros that come after other digits is rather confusing. For
example, an object has a mass of 1 300 g is considered to have two significant figures.
However, if we express 1 300 g into scientific notation such as 1.30 x103 g of 1.30 kg
then we have three significant figures. Consequently, 1.300 x103 g has four
significant figures.
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Calculations considering significant figures follow:
•
•
When multiplying several quantities, the number of significant figures in the final
answer is the same as the number of significant figures in the quantity having the
smallest number of significant figures. The same rule applies to division.
For example, in solving the area of a disc whose diameter is 12.0 cm, is
𝐴 = 𝜋𝑟2 = 𝜋(6.0 𝑐𝑚)2 = 113.0973355 𝑐𝑚2
However, the rule says that the number of significant figures of the product
must be the same as the smallest number of significant digits among the
factors. In this case, the radius has the least number of significant figures
which is two. Therefore the area should be expressed as,
𝐴 = 113 𝑐𝑚2 = 𝟏. 𝟏 𝑥102 𝑐𝑚2 (which has two significant figures)
When numbers are added or subtracted, the number of decimal places in the result
should equal the smallest number of decimal places of any term in the sum or
difference.
For example,
33.3 + 5.553 = 38.853
But since the sum must have the least number of significant figures, then
33.3 + 5.553 = 𝟑𝟖. 𝟖
Scalar and Vector Quantities
In physics, quantities may be either scalar or vector. A scalar is a quantity that describes a
physical phenomenon by magnitude only. Examples are mass, length, time, temperature,
distance, speed, and so on. Operations with scalars follow the same rules as in elementary
algebra.
On the other hand, a vector is a quantity having both the magnitude and direction. Examples
are displacement, velocity, acceleration, force, and so on. Analytically, a vector is
represented by a letter with an arrow over it, say
There are fundamental rules in
vector algebra which are as follows:
1. Two vectors are equal if and only if they have the same magnitude and direction.
2. A vector having direction opposite to that of vector A but having the same magnitude
is denoted by – A.
3. The sum or resultant of vectors A and B is a vector C formed by placing the initial
point of B on the terminal point of A and then joining the initial point of A to the
terminal point of B, shown in Fig. 1.
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Fig. 1
A unit vector is a vector having unit magnitude. Any vector can be represented by a unit
vector. An important set of unit vectors are those having the directions of the positive x, y,
and z axes of a three-dimensional rectangular coordinate system, and are denoted
respectively by i, j, and k shown in Fig. 2.
Fig. 2
Any vector A in a three-dimensional coordinate system, shown in Fig. 3, can be
represented by its component vectors. In this case, the component vectors of vector A are
A1i, A2j, and A3k in the x, y, and z directions respectively.
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Fig. 3
The sum or resultant of the vector components is vector A, written as
𝑨 = 𝑨𝟏𝒊 + 𝑨𝟐𝒋 + 𝑨𝟑𝒌
And the magnitude of vector A is
|𝑨| = √(𝑨𝟏)𝟐 + (𝑨𝟐)𝟐 + (𝑨𝟑)𝟐
It follows that for a two-dimensional coordinate system, the sum of resultant is written as,
𝑨 = 𝑨𝟏𝒊 + 𝑨𝟐𝒋
And the magnitude of vector A is
|𝑨| = √(𝑨𝟏)𝟐 + (𝑨𝟐)𝟐
(Note: other books write the component vectors as 𝑨𝒙, 𝑨𝒚 and 𝑨𝒛)
Example 1
A ball is thrown with an initial velocity of 70 feet per second., at an angle of 35° with the
horizontal. Find the vertical and horizontal components of the velocity.
Let the resultant vector A = 𝟕𝟎
𝒇𝒕
𝒔𝒆𝒄
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To solve for the x and y components of vector A, we find that the relationship between the
components and the resultant forms a right triangle. Therefore the component vectors are,
𝑨𝒙 = 𝟕𝟎 𝒄𝒐𝒔 𝟑𝟓 ° = 𝟓𝟕. 𝟑𝟒 𝒊
𝑨𝒚 = 𝟕𝟎 𝒔𝒊𝒏 𝟑𝟓 ° = 𝟒𝟎. 𝟏𝟓 𝒋
And the resultant in terms of the component vector is,
𝑨 = 𝟓𝟕. 𝟑𝟒 𝒊 + 𝟒𝟎. 𝟏𝟓 𝒋
𝒇𝒕
𝒔𝒆𝒄
To double-check, you may solve the magnitude which should be equal to 70 ft/sec.
|𝑨| = √(𝑨 )𝟐 + (𝑨 )
𝒙
𝟐
𝒚
|𝑨| = √(𝟓𝟕. 𝟑𝟒)𝟐 + (𝟒𝟎. 𝟏𝟓)𝟐
|𝑨| = 𝟔𝟗. 𝟗𝟗
𝒇𝒕
𝒔𝒆𝒄
And the direction is,
𝒕𝒂𝒏 𝜽 =
𝜽 = 𝒕𝒂𝒏−𝟏
𝜽 = 𝟑𝟓° North of East
21
𝑨𝒚
𝑨𝒙
𝟒𝟎. 𝟏𝟓
𝟓𝟕. 𝟑𝟒
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Example 2
Given 𝒓𝟏 = 𝟑𝒊 − 𝟐𝒋 + 𝒌, 𝒓𝟐 = 𝟐𝒊 − 𝟒𝒋 − 𝟑𝒌, 𝒓𝟑 = −𝒊 + 𝟐𝒋 + 𝟐𝒌, find the magnitudes of (a)
𝒓𝟑, (b) 𝒓𝟏 + 𝒓𝟐 + 𝒓𝟑, (c) 𝟐𝒓𝟏 − 𝟑𝒓𝟐 − 𝟓𝒓𝟑.
a. |𝒓𝟑| = |−𝒊 + 𝟐𝒋 + 𝟐𝒌| = √(−𝟏)𝟐 + (𝟐)𝟐 + (𝟐)𝟐 = 𝟑
b. |𝒓𝟏 + 𝒓𝟐 + 𝒓𝟑|
Adding vectors is only possible for coplanar vectors which mean 𝒊 + 𝒊, 𝒋 + 𝒋, and 𝒌 +
𝒌 only.
𝒓𝟏 + 𝒓𝟐 + 𝒓𝟑 = (𝟑𝒊 − 𝟐𝒋 + 𝒌) + (𝟐𝒊 − 𝟒𝒋 − 𝟑𝒌) + (−𝒊 + 𝟐𝒋 + 𝟐𝒌)
= 𝟒𝒊 − 𝟒𝒋 + 𝟎𝒌
Then the magnitude is
|𝒓𝟏 + 𝒓𝟐 + 𝒓𝟑| = √(𝟒)𝟐 + (−𝟒)𝟐 = 𝟒√𝟐
c. 𝟐𝒓𝟏 − 𝟑𝒓𝟐 − 𝟓𝒓𝟑 = 𝟐(𝟑𝒊 − 𝟐𝒋 + 𝒌) − 𝟑(𝟐𝒊 − 𝟒𝒋 − 𝟑𝒌) − 𝟓(−𝒊 + 𝟐𝒋 + 𝟐𝒌
= 𝟔𝒊 − 𝟒𝒋 + 𝟐𝒌 − 𝟔𝒊 + 𝟏𝟐𝒋 + 𝟗𝒌 + 𝟓𝒊 − 𝟏𝟎𝒋 − 𝟏𝟎𝒌
= 𝟓𝒊 − 𝟐𝒋 + 𝒌
Then the magnitude is
|𝟐𝒓𝟏 − 𝟑𝒓𝟐 − 𝟓𝒓𝟑| = |𝟏𝟓𝒊 − 𝟐𝒋 + 𝒌 | = √(𝟓)𝟐 + (−𝟐)𝟐 + (𝟏)𝟐 = √𝟑𝟎
Self-Help: You can also refer to the sources below to help you further understand the
lesson:
*Giancoli, D. C. (2016). Physics: Principles with Applications (14th ed.). Boston, USA:
Pearson.
*Katz, D. (2017). Physics for Scientist and Engineers: Foundations and Connections.
Australia: Cengage Learning.
*Young, H. D. (2016). Sears and Zemanky's University Physics with Modern Physics (14th
ed.). Harlow, England: Pearson
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Let’s Check
Activity 1. Practice Problems
Problem 1. Unit Conversion
A certain car has a fuel efficiency of 25.0 miles per gallon (mi/gal). Express this efficiency
in kilometers per liter (km/L).
Problem 2. Unit Conversion
A house is 50.0 ft long and 26 ft wide and has 8.0-fthigh ceilings. What is the volume of the
interior of the house in cubic meters and in cubic centimeters?
Problem 3. Significant Figures
How many significant figures are there in
a. 78.9±0.2
b. 3.788 x109
c. 2.46 x10 -6
d. 0.003 2
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Problem 4. Significant Figures
The speed of light is now defined to be 2.997 924 58 x 108 m/s. Express the speed of light
to
a. three significant figures
b. five significant figures, and
c. seven significant figures.
Problem 5. Scalar and Vector Quantities
State which of the following are scalars and which are vectors.
a. weight
b. calorie
c. specific heat
d. momentum
e. density
f. energy
g. volume
h. distance
i. speed
j. magnetic field intensity
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Let’s Analyze
Problem 1. Unit Conversion
A small turtle moves at a speed of 186 furlongs per fortnight. Find the speed of the turtle in
centimeters per second. Note that 1 furlong = 220 yards and 1 fortnight = 14 days.
Problem 2: Dimension Analysis
Kinetic energy, KE, has dimensions 𝑘𝑔
p, and mass m as
𝑚2
𝑠2
. It can also be written in terms of the momentum,
𝐾𝐸 =
𝑝2
2𝑚
a. Determine the unit for momentum using dimensional analysis.
𝑚
b. The unit of force is the newton, N, where 1𝑁 = 1 𝑘𝑔 ∙ . What are the units of momentum
𝑠2
p in terms of a newton and another fundamental SI unit?
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Problem 3: Dimension Analysis
Show by dimensional analysis that the equations are correct or incorrect.
1
1
2
a. 2 𝑚𝑣 2 = 2 𝑚𝑣𝑜 + √𝑚𝑔ℎ
b. 𝑣 = 𝑣𝑜 + 𝑎𝑡2
c. 𝑚𝑎 = 𝑣2
Problem 4: Significant Figures
A block of gold has length 5.62 cm, width 6.35 cm, and height 2.78 cm.
a. Calculate the base area and round up the answer to the appropriate number of significant
figures.
b. Now multiply the rounded result of part (a) by the height and again round, obtaining the
volume. Let this be 𝑉1.
c. Now obtain volume 2, 𝑉2, by first finding the width times the height, rounding up to
appropriate significant figures, and multiply by the length.
d. Explain why the answers don’t agree with the third significant figure.
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Problem 5: Scalar and Vector Quantities
If 𝑟1 = 2𝑖 − 𝑗 + 𝑘, 𝑟2 = 𝑖 + 3𝑗 − 2𝑘, 𝑟3 = −2𝑖 + 𝑗 − 3𝑘, and 𝑟4 = 3𝑖 + 2𝑗 + 5𝑘, find the
scalars 𝑎, 𝑏, 𝑐 such that 𝑟4 = 𝑎𝑟1 + 𝑏𝑟2 + 𝑐𝑟3.
Problem 6: Scalar and Vector Quantities
Find a unit vector parallel to the resultant of vectors 𝑟1 = 2𝑖 + 4𝑗 − 5𝑘, 𝑟2 = 𝑖 + 2𝑗 + 3𝑘.
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In a Nutshell
Activity 1.
Physics is based on experimental observations and quantitative measurements. The
fundamental laws used in developing theories are expressed in the language of
mathematics, the tool that provides a bridge between theory and experiment.
Discrepancies and uncertainties must be defined to achieve satisfactory data. Discuss the
relevance of the significant figures.
Activity 2.
Physical phenomena can be described by physical quantities. Thoroughly discuss the
difference between scalar and vector quantities and enumerate examples.
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Big Picture in Focus: ULO-1b. Recall and apply the principles of motion and
kinematics
Metalanguage
This section focuses on motion in one and two dimensions without considering
external influences that may cause or affect the motion. Demonstrating ULOb means
understanding motion including the concepts of displacement, velocity, and acceleration.
You will go through the following formulas as the fundamentals of rectilinear motion and
kinematics:
Displacement:
Average velocity:
∆𝒙 = 𝒙𝒇 − 𝒙𝒊
∆𝒙 𝒙𝒇 − 𝒙𝒊
𝒗
̅ 𝒂𝒗𝒆 =
=
∆𝒕
𝒕 −𝒕
𝒇
Particle at constant velocity:
Average acceleration:
𝒊
𝒙𝒇 = 𝒙𝒊 + 𝒗 ∆𝒕
𝒂𝒂𝒗𝒆 =
∆𝒗 𝒗𝒇 − 𝒗𝒊
=
∆𝒕 𝒕 − 𝒕
𝒇
Particle at constant
acceleration:
𝒊
𝒗𝒇 = 𝒗𝒊 + 𝒂 𝒕
𝟏
𝒙𝒇 = 𝒙𝒊 + (𝒗𝒊 + 𝒗𝒇)(𝒕)
𝟐
𝟏
𝒙𝒇 = 𝒙𝒊 + (𝒗𝒊)(𝒕) + 𝒂𝒕𝟐
𝟐
𝒗𝟐 = 𝒗𝟐 + 𝟐𝒂(𝒙𝒇 − 𝒙𝒊)
𝒇
𝒊
Essential Knowledge
.
The motion of an object represents a continuous change in an object’s position.
Kinematics describes motion without regard to its causes. Motion involves the change in
position from one place in space and time to another and can be either scalar or vector. The
object’s position, distance, and speed are scalar quantities, while displacement, velocity, and
acceleration are vectors. These terms and concepts are the key principles in the analysis of
motion.
Keywords
Distance
Displacement
speed
velocity
acceleration
Free-fall
29
trajectory
Component vectors
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Position, Distance, and Displacement
Describing motion requires a coordinate system and a specified origin. The position is the
location of the particle from a chosen reference point that we can consider to be the origin
of a coordinate system.
Consider the movement of a car shown in Fig. 4. The reference point is x=0 and the car’s
initial position is at 30 m to the right of the origin. Let us define to the right of the origin as
the positive distance and negative distance to the left. For every ten seconds, the car moves
from one point to another. The car moves back and forth along a straight line from point A
to F.
Fig. 4
Table 1.5. Position of the car at time t
Position
t (s)
x (m)
A
0
30
B
10
52
C
20
38
D
30
0
E
40
-37
F
50
-53
Table 1.5 shows a tabular representation of the recorded data of the car’s change of position
with respect to time. Distance is the length of a path followed by a particle. It is the total
amount of space that the particle covered during its motion. The data suggest that the car
has a total travel distance of 127 m. Since we only have the idea of the magnitude of motion
but not its direction, then distance is scalar and is always represented as a positive number.
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Displacement ∆𝑥 of a particle is defined as its change in position in some time interval. As
the particle moves from an initial position 𝑥𝑖 to a final position 𝑥𝑓, its displacement is given
by
∆𝒙 = 𝒙𝒇 − 𝒙𝒊
Displacement is a vector quantity and can be either positive or negative. It is a quantity that
describes how far out of place the particle is. In our example, the initial position is at 30 m
and the final position is at -53 m, if we solve for the displacement
∆𝑥 = (−53) − (30) = −𝟖𝟑 𝒎
Meaning, the car is displaced to 83 m from its initial position and the negative sign indicates
that the final position is to the left of the initial position. This is a clear example of a vector
quantity that has both the magnitude and direction of motion. If the object’s initial and final
position is the same, then it is not displaced at all (∆𝑥 = 0).
Speed and Velocity
We hear the terms speed and velocity often interchangeable. However, in physics, the two
are defined distinctly.
The average speed denoted as 𝑣, of an object over a given time interval is the length of the
path it travels divided by the total elapsed time:
𝒑𝒂𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝒗𝒂𝒗𝒆 =
=
𝒆𝒍𝒂𝒑𝒔𝒆𝒅 𝒕𝒊𝒎𝒆
𝒕𝒊𝒎𝒆
By dimension analysis, distance is in meters and time in seconds, thus giving the SI standard
unit for speed as meters per second (m/s). Average speed is a scalar quantity that describes
how fast is the motion regardless of any variation in speed over the given time interval.
On the other hand, the average velocity denoted as 𝑣̅, during a time interval ∆𝑡 is the
displacement ∆𝑥 divided by ∆𝑡:
∆𝒙 𝒙𝒇 − 𝒙𝒊
𝒗
̅ 𝒂𝒗𝒆 =
=
𝒕𝒇 − 𝒕𝒊
∆𝒕
The SI standard unity for velocity is also meters per second (m/s). Average velocity is a
vector quantity having both the magnitude and a direction, which can also be a positive or
negative value.
From the same example of Fig. 4, we compare the average speed and average velocity:
𝑑 127 𝑚
𝒎
𝑣𝑎𝑣𝑒 = = 50 𝑠 = 𝟐. 𝟓 𝒔
𝑡
∆𝑥 𝑥𝑓 − 𝑥𝑖 −53𝑚 − 30𝑚
𝒎
=
=
−𝟏.
𝟕
𝑣𝑎̅ 𝑣𝑒 = = 𝑡
50𝑠 − 0𝑠
𝒔
𝑓 − 𝑡𝑖
∆𝑡
See the difference between speed and velocity?
To further illustrate the difference between speed and velocity. A man walks from point P
to Q, given two scenarios of the same starting position and final position over the same time
interval. First, he walks directly straight down the path. Second, he walks following the
curved path. The first and second walks have the same average velocity since 𝑥𝑓, 𝑥𝑖, and
time intervals are the same for both scenarios. However, the average speed of scenario 2 is
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greater than in scenario 1 because the length of the path traveled by the curved route is
larger than the straightforward route.
Fig. 5
Instantaneous Velocity
Consider the car in Fig. 4, the data in Table 1.5 shows the car’s position at a given time,
however, it doesn’t take into account the details of what happens during an interval of time.
The car may speed up or slow down between any given points along the path, the data
doesn’t show that.
Instantaneous velocity describes the speed and direction of a particle in motion at an instant
of time. It is the actual velocity at specific instants of time. Mathematically, instantaneous
velocity 𝑣 is the limit of the average velocity as the time interval ∆𝑡 becomes infinitesimally
small:
𝒅𝒙
𝒗 = 𝐥𝐢𝐦 ∆𝒙 =
∆𝒕→𝟎 ∆𝒕
𝒅𝒕
The SI standard unit for instantaneous velocity is meters per second (m/s). Graphically, the
instantaneous velocity is the slope of the tangent line to the position 𝑥 at a given time. Notice
that the definition is the same as the derivative of a function, and therefore, the
instantaneous velocity is the first derivative of the displacement with respect to time. The
magnitude of the instantaneous velocity is called instantaneous speed.
Example 1:
A particle moves along the x-axis. Its position varies with time according to the expression
𝑥(𝑡) = −4𝑡 + 2𝑡2, where x is in meters, and t is in seconds.
a. Determine the displacement of the particle in the time intervals 𝑡 = 0 to 𝑡 = 1𝑠, and 𝑡 =
1 to 𝑡 = 3𝑠.
Since the given position is a function in terms of time, you need to substitute the given
time to the function to get its position at the said time.
𝑥(𝑡) = −4𝑡 + 2𝑡2
For t = 1 to t = 3s:
𝑥𝑖 = 𝑥(1) = −2 𝑚
𝑥𝑓 = 𝑥(3) = 6 𝑚
For t = 0 to t = 1s:
𝑥𝑖 = 𝑥(0) = 0 𝑚
𝑥𝑓 = 𝑥(1) = −2 𝑚
The displacement is,
∆𝑥 = 𝑥𝑓 − 𝑥𝑖 = (−2) − 0 = −𝟐 𝒎
∆𝑥 = 𝑥𝑓 − 𝑥𝑖 = (6) − (−2) = +𝟖 𝒎
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b. Calculate the average velocity during these two time intervals.
For t = 0 to t = 1s:
For t = 1 to t = 3s:
∆𝑥 = −2𝑚
∆𝑥 = +8 𝑚
𝑣𝑎𝑣𝑒 =
∆𝑥
∆𝑡
=
−2𝑚
1−0 𝑠
= −𝟐
𝒎
𝑣𝑎𝑣𝑒 =
𝒔
c. Find the instantaneous velocity at t = 2.5s.
𝑥(𝑡) = −4𝑡 + 2𝑡2
𝑑𝑥
= −4 + 4𝑡
𝑑𝑡𝑑𝑥
at t = 2.5s
𝑣 = = −4 + 4(2.5) = +𝟔
𝑑𝑡
∆𝑥
∆𝑡
=
8m
3−1 𝑠
= +𝟒
𝒎
𝒔
𝒎
𝒔
Particle Under Constant Velocity
If the velocity of a particle is constant, its instantaneous velocity at any instant during a time
interval is the same as the average velocity over the interval. That is,
𝑣 = 𝑣𝑎𝑣𝑒
∆𝑥
𝑣=
∆𝑡
𝑥𝑓 − 𝑥𝑖
𝑣=
∆𝑡
Therefore,
𝒙𝒇 = 𝒙𝒊 + 𝒗 ∆𝒕
(for constant velocity)
This equation is the position as a function of time for the particle under a constant velocity
model.
Example 2:
Ana determines the velocity of an experimental subject while he runs along a straight line
at a constant rate. She starts the stopwatch at the moment the runner passes a given point
and stops it after the runner has passed another point 20 m away. The time interval
indicated on the stopwatch is 4.0 s.
a. What is the runner’s velocity?
The runner starts from rest, hence, 𝑡𝑖 = 0 𝑠 and 𝑥𝑖 = 0 𝑚. So that the velocity is,
𝑣=
𝒎
∆𝑥 20 − 0 𝑚
=
= +𝟓. 𝟎
∆𝑡
4−0𝑠
𝒔
b. If the runner continues his motion after the stopwatch is stopped, what is his position
after 10 s have passed?
Note the key phrase of the problem which is ‘runs along at a constant rate’, meaning,
the particle is moving at a constant velocity. Therefore, using the velocity obtained in (a)
the final position is
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Acceleration
𝑚
(0
𝑥𝑓 = 𝑥𝑖 + 𝑣 ∆𝑡 = 𝑚) + (5 ) (10 − 0 𝑠) = 𝟓𝟎 𝒎
𝑠
Acceleration is defined as the change in velocity over a given time. That happens when you
step harder on the gas pedal or slow down as you turn to another direction or step on the
brakes to slow down.
The average acceleration 𝑎 during the time interval ∆𝑡 is the change in velocity ∆𝑣 divided
by ∆𝑡.
∆𝒗 𝒗𝒇 − 𝒗𝒊
𝒂𝒂𝒗𝒆 = =
∆𝒕 𝒕𝒇 − 𝒕𝒊
If an object is moving at a constant velocity (𝑣𝑓 = 𝑣𝑖), then the acceleration is zero. By
dimension analysis to obtain the SI standard unit for acceleration, we have
𝑚𝑒𝑡𝑒𝑟
𝒎
𝑎 = 𝑠𝑒𝑐𝑜𝑛𝑑 =
𝑠𝑒𝑐𝑜𝑛𝑑 𝒔𝟐
Acceleration is a vector quantity, however, negative acceleration doesn’t necessarily mean
an object is slowing down. For the case of motion in a straight line, the direction of the
velocity of an object and the direction of its acceleration are related as follows: When the
object’s velocity and acceleration are in the same direction, the speed of the object increases
with time. When the object’s velocity and acceleration are in opposite directions, the speed
of the object decreases with time. Positive and negative accelerations specify directions
relative to chosen axes, not “speeding up” or “slowing down.” The terms speeding up or
slowing down refer to an increase and a decrease in speed, respectively.
Instantaneous Acceleration
Like the instantaneous velocity, instantaneous acceleration is defined at a certain instant of
time. The instantaneous acceleration 𝑎 is the limit of the average acceleration as the time
interval ∆𝑡 approaches to zero:
𝒂 = 𝐥𝐢𝐦 ∆𝒗
∆𝒕→𝟎 ∆𝒕
The standard SI unit for instantaneous acceleration is also 𝑚. Since instantaneous
2
𝑠
acceleration is the change in velocity over a given time, we can say that
𝒂=
𝒅𝒗
𝒅𝒕
=
𝒅𝟐 𝒙
𝒅𝒕𝟐
That is, in one-dimensional motion, the acceleration equals the second derivative of x with
respect to time.
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Example 3
The velocity of a particle moving along the x-axis varies according to the expression 𝑣 =
40 − 5𝑡2 m/s.
a. Find the average acceleration in the time interval t = 0 to t = 2.0 s.
Since the given velocity is a function in terms of time, you need to substitute the given
time to the function to get its velocity at the said time.
𝑣(𝑡) = 40 − 5𝑡2
For t = 0 to t = 2 s:
The average acceleration is,
𝑚
𝑚
𝑚
20 − 40
𝑣𝑓 − 𝑣
𝑣 = 𝑣(0) = 40
𝑠 = −𝟏𝟎 𝒎/𝒔𝟐
𝑖
𝑎=
𝑖
𝑠
𝑠
=
𝑚
𝑡𝑓 − 𝑡𝑖
2𝑠−0𝑠
𝑣𝑓 = 𝑣(2) = 20
𝑠
b. Determine the acceleration at t = 2.0s.
𝑎=
at t = 2.0s
𝑑𝑣
𝑑
(40 − 5𝑡2) = −10𝑡2
𝑑𝑡 𝑑𝑡
𝒎
𝑎 = −(2.0)2 = −𝟐𝟎
=
𝒔
Because the velocity of the particle is positive and the acceleration is negative at this
instant, the particle is slowing down.
Particle Under Constant Acceleration
When the acceleration is constant, the average acceleration is over any time interval is
numerically equal to the instantaneous acceleration at any instant within the interval.
𝑎 = 𝑎𝑎𝑣𝑒
∆𝑣
𝑎=
𝑎=
Therefore,
∆𝑡
𝑣𝑓 − 𝑣𝑖
𝑡
𝒗𝒇 = 𝒗𝒊 + 𝒂 𝒕 (for constant acceleration)
This equation enables us to determine an object’s velocity at any time t if we know the
object’s initial velocity and its (constant) acceleration. Also, the average velocity at constant
acceleration can be described as
𝒗𝒂𝒗𝒆 =
𝒗𝒊+𝒗𝒇
𝟐
(for constant acceleration)
Now we can derive an equation that describes the position as a function of velocity and time
for the particle under constant acceleration model by equating the two average velocity
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formulas. This formula helps solve the object’s position without knowing the magnitude of
its acceleration. (As long as acceleration is constant!)
𝑣
𝑎𝑣𝑒
=
∆𝑥
and 𝑣
𝑎𝑣𝑒
∆𝑡
=
𝑣𝑖+𝑣𝑓
2
∆𝑥 𝑣𝑖 + 𝑣𝑓
=
2
∆𝑡
𝑥𝑓 − 𝑥𝑖
1
= (𝑣𝑖 + 𝑣𝑓)
𝑡−0
2
𝟏
𝒙𝒇 = 𝒙𝒊 + (𝒗𝒊 + 𝒗𝒇)(𝒕)
𝟐
(for constant acceleration)
Another formula can be also derived to obtain an equation that describes the position as a
function of time for the particle under constant acceleration model. This equation provides
the final position of the particle at time t in terms of the initial position, the initial velocity,
and the constant acceleration.
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 + 𝑣𝑓)(𝑡)
2
Since
𝑣𝑓 = 𝑣𝑖 + 𝑎 𝑡
1
𝑥𝑓 = 𝑥𝑖 + [𝑣𝑖 + (𝑣𝑖 + 𝑎 𝑡)](𝑡)
2
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 )(𝑡) + 2 𝑎𝑡 2
(for constant acceleration)
Finally, we can obtain an expression for the final velocity that does not contain time as a
variable. This equation provides the final velocity in terms of the initial velocity, the
constant acceleration, and the position of the particle.
We get an equation of 𝑡 from
𝑣𝑓 = 𝑣𝑖 + 𝑎 𝑡
𝑣𝑓 − 𝑣𝑖
𝑎
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 + 𝑣𝑓 )(𝑡)
2
𝑣𝑓 − 𝑣𝑖
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 + 𝑣𝑓 )(
)
2
𝑎
𝑣𝑓 2 − 𝑣𝑖 2
𝑥𝑓 = 𝑥𝑖 +
2𝑎
2
2
𝑣𝑓 = 𝑣𝑖 + 2𝑎(𝑥𝑓 − 𝑥𝑖 )
(for constant acceleration)
𝑡=
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Remember that these equations of kinematics cannot be used in a situation in which the
acceleration varies with time. They can be used only when the acceleration is constant. The
equations may be used to solve any constant acceleration problem but the choice of which
equation you use in a given situation depends on what you know beforehand. One formula
is better or easier than the other depending on the given.
If you notice, these three rectilinear equations are the key to the other formulas. (Tip:
Understand the derivation of the formulas so you don’t need to memorize them all.)
𝑣𝑓 = 𝑣𝑖 + 𝑎 𝑡
𝑣𝑎𝑣𝑒 =
𝑣𝑎𝑣𝑒 =
𝑣𝑖 + 𝑣𝑓
2
𝑥𝑓 − 𝑥𝑖
𝑡
Example 1
A jet lands on an aircraft carrier at a speed of 140 mi/h (≈ 63 m/s).
a. What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable
that snags the jet and brings it to a stop?
First, examine the given. You are given the jet’s initial speed that is 63 m/s, constant
acceleration, time of 2.0 s, and final speed that is 0 m/s (because it stopped). Out of
the given, your goal is to solve the magnitude of its acceleration. Therefore, you use
the formula that has speed, acceleration, and time:
𝑎=
𝑎=
𝑣𝑓 − 𝑣𝑖
𝑡
𝑚
𝑚
0 𝑠 − 63 𝑠
= −𝟑𝟐 𝒎
2.0 𝑠
𝒔𝟐
b. If the jet touches down at position 𝑥𝑖 = 0, what is its final position?
Now you are given an initial position of 0 m and asked to solve its final position.
Using the same parameters as (a), you’re going to use a formula that describes the
jet’s position in terms of speed and time.
1
𝑥𝑓 = 𝑥𝑖 + (𝑣𝑖 + 𝑣𝑓)(𝑡)
2
1
𝑚
𝑚
𝑥𝑓 = 0 𝑚 + (63 + 0 ) (2.0) = 𝟔𝟑 𝒎
2
𝑠
𝑠
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Example 2
A car traveling at a constant speed of 45.0 m/s passes a trooper on a motorcycle hidden
behind a billboard. One second after the speeding car passes the billboard, the trooper sets
out from the billboard to catch the car, accelerating at a constant rate of 3.00 m/s 2. How
long does it take the trooper to overtake the car?
A pictorial representation is shown in Fig. 6 to better understand the problem,
Fig. 6
First, examine the given.
Car:
𝑚
𝑣 = 45.0
𝑣𝑖
𝑥𝑖 = =
𝑡
𝑠
45.0
𝑚
𝑠
1.00 𝑠
Trooper:
𝑥𝑖 = 0 𝑚
𝑚
constant acceleration, 𝑎 = 3.00
𝑠2
= 45.0 𝑚
constant speed, 𝑣 = 𝑣 , 𝑎 = 0
𝑖
𝑓
𝑚
𝑠2
If you notice trooper’s initial position is set as the reference point and since the car
started ahead of 1 s, the initial position for the car is 45.0 𝑚. You are then asked to
solve the final time at which the trooper overtakes the car. That is the time when
𝑥𝑓(𝑐𝑎𝑟) = 𝑥𝑓(𝑡𝑟𝑜𝑜𝑝𝑒𝑟). With all the given values, identify the suited formula to use.
𝑥𝑓(𝑐𝑎𝑟) = 𝑥𝑓(𝑡𝑟𝑜𝑜𝑝𝑒𝑟)
1
𝑥𝑖 + (𝑣𝑖 + 𝑣𝑓)(𝑡) = 𝑥𝑖 +
2
38
1
(𝑣𝑖 )(𝑡) + 𝑎𝑡2
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𝑚
)(𝑡)2
45.0 𝑚 + 1 (45.0 𝑚 + 45.0 𝑚) (𝑡) = 0 𝑚 + (0 𝑚 ( ) 1
)1𝑠
+ (3.00 2
2
𝑠
𝑠
𝑠
2
𝑠
𝑚
𝑚 2
45.0 𝑚 + (45.0 ) 𝑡 = (1.50 ) 𝑡
𝑠
𝑠2
𝑚 2
𝑚
(1.50 ) 𝑡 − (45.0 ) 𝑡 − 45.0𝑚 = 0
𝑠2
𝑠
Solving the quadratic equation would give you,
𝒕 = 𝟑𝟎. 𝟗 𝒔 and 𝑡 = −0.96 𝑠 (𝑎𝑏𝑠𝑢𝑟𝑑)
Example 3
A typical jetliner lands at a speed of 1.60 x102 mi/h and decelerates at the rate of 10.0
𝑚𝑖/ℎ𝑟
𝑠
.
If the plane travels at a constant speed of 1.60 x102 mi/hr for 1.00 s after landing before
applying the brakes, what is the total displacement of the aircraft between touchdown on
the runway and coming to rest?
Notice the inconsistencies of the units, thus the need for unit conversion.
𝑣𝑖 = 1.60𝑥102
𝑎 = −10.0
𝑚𝑖 1609 𝑚
1 ℎ𝑟
𝑚
(
)(
) = 71.5
ℎ𝑟 1 𝑚𝑖
3600 𝑠
𝑠
𝑚𝑖⁄
ℎ𝑟 (1609 𝑚) ( 1 ℎ𝑟 ) = −4.47 𝑚
𝑠
1 𝑚𝑖
3600 𝑠
𝑠2
A picture representation is shown in Fig. 7, to better understand the problem. You
are asked to solve for the final displacement of the jet as it comes to rest. Meaning,
𝑚
𝑥𝑓 at 𝑣𝑓 = 0 .
𝑠
`
Fig. 7
During coating, the jet is at constant velocity thus acceleration is zero for 1 s. At
braking distance, the jet decelerates to stop thus a negative acceleration and zero
final velocity. You have to realize that the final position at coasting is the initial
position at braking.
𝑥𝑓(𝑐𝑜𝑎𝑠𝑡𝑖𝑛𝑔) = 𝑥𝑖(𝑏𝑟𝑎𝑘𝑖𝑛𝑔)
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𝑚
1 𝑚
𝑥𝑖(𝑏𝑟𝑎𝑘𝑖𝑛𝑔) = 𝑥𝑓(𝑐𝑜𝑎𝑠𝑡𝑖𝑛𝑔) = 0 𝑚 + (71.5 ) (1.00 𝑠) + (0 2) (1.00 𝑠) = 𝟕𝟏. 𝟓 𝒎
𝑠
2 𝑠
Use the obtained value to solve the final position as it brakes to stop.
𝑣2 = 𝑣2 + 2𝑎(𝑥 − 𝑥 )
𝑓
(0
𝑥
𝑓
𝑖
𝑚
𝑚2
𝑚2
) = (71.5 ) + 2(−4.47 ) (𝑥𝑓 − 71.5 𝑚)
𝑠
𝑠
𝑠2
Solving for the final position we get, 𝒙𝒇 = 𝟔𝟒𝟑 𝒎
Free Falling Objects
You might already hear about an experiment of simultaneously dropping a coin and a
feather from the same height. Neglecting the effects of air resistance, both will hit the floor
at the same time. In the idealized case, in which air resistance is absent, such motion is
referred to as free-fall motion. This behavior of free-falling objects was introduced by
Galileo Galilei. A freely falling object is any object moving freely under the influence of
gravity alone, regardless of its initial motion.
Free-falling objects experience acceleration due to gravity. Earth’s gravity denoted as 𝑔 is
approximately 9.80 m/s2 and which will be used throughout this module. Freely falling
objects moving vertically is equivalent to the motion of a particle under constant
acceleration in one dimension. Therefore, the formulas introduced before for particle under
constant acceleration can also be used in free-fall analysis. The only difference is that, since
free-falling objects fall under constant acceleration due to gravity, we then use 𝑎 = 𝑔 =
−9.80 𝑚/𝑠2. The negative sign indicates the direction of fall which is downwards. (Also,
we’re going to use variable 𝑦 for the position to indicate vertical distance.)
Example 1
A ball is thrown from the top of a building with an initial velocity of 20.0 m/s straight
upward, at an initial height of 50.0 m above the ground. The ball just misses the edge of the
roof on its way down, as shown in Fig. 8 and determine the following:
a. the time needed for the ball to reach its maximum height.
b. the maximum height.
c. the time needed for the ball to return to the height from which it was thrown and the
velocity of the ball at that instant,
d. the time needed for the ball to reach the ground
e. the velocity and position of the ball at t = 5.00 s.
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a. At maximum height, the velocity is always zero.
Examine between points A and B,
𝑣𝐵 = 𝑣𝐴 + 𝑔 𝑡
𝑚
0
𝑚
𝑚
= 20.0 + (−9.80 2)(𝑡)
𝑠
𝑠
𝑠
𝒕 = 𝟐. 𝟎𝟒 𝒔
b. Using the time obtained in (a),
1 2
𝑦𝐵 = 𝑦𝐴 + 𝑣𝐴𝑡 + 𝑔𝑡
2
𝑚
𝑦𝐵 = 0 𝑚 + (20.0 )(2.04 𝑠)
1𝑠
𝑚
+ (−9.80 )(2.04 𝑠)2
2
𝑠
𝒚𝑩 = 𝟐𝟎. 𝟒 𝒎
c. Examine between points A to C,
1
𝑦𝐶 = 𝑦𝐴 + 𝑣𝐴𝑡 + 𝑔𝑡
2
𝒕 = 𝟒. 𝟎𝟖 𝒔
𝑚
2
𝑣𝐶 = 𝑣𝐴 + 𝑔 𝑡 𝑚
𝒗𝒄 = −𝟏𝟗. 𝟗𝟖𝑚⁄𝑠
d. Examine between points A to E,
1
d. 𝑦𝐸 = 𝑦𝐴 + 𝑣𝐴 2 𝑔𝑡2
𝒕 = 𝟓. 𝟖𝟑 𝒔
e. at t = 5.00 s
𝑚
𝑣 = 𝑣𝐴 + 𝑔𝑡 = 20.0
𝑠
+ (−9.80
𝑚
𝒎
(5.00𝑠)
)
=
−𝟐𝟗.
𝟎
𝑠2
𝒔
𝑥 = 𝑥𝐴 + 𝑣𝐴 𝑡 + 1 𝑔𝑡2 = 0 𝑚 + (20.0 𝑚 ) (5.00 𝑠) + 1 (−9.80 𝑚 ) (5.00 𝑠)2 = −𝟐𝟐. 𝟓 𝒎
2
𝑠
2
𝑠2
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Example 2
A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2. It
runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum
height before falling back to Earth.
a. Find the rocket’s velocity and position at the end of 4.00 s.
The rocket starts with an acceleration of +29.4 m/s2 then the fuel runs out at t=4.00s.
𝑚
𝑚
𝒎
𝑣𝑓 = 𝑣𝑖 + 𝑔𝑡 = 0 + (29.4 2) (4.00𝑠) = 𝟏𝟏𝟖
𝑠
𝑠
𝒔
1 2
𝑚
1
𝑚
𝑦 = 𝑦 + 𝑣 𝑡 + 𝑔𝑡 = 0𝑚 + (0 ) (4.00 𝑠) + (29.4 ) (4.00𝑠)2 = 𝟐𝟑𝟓 𝒎
𝑖
𝑓
𝑖
2
𝑠
2
𝑠2
b. Find the maximum height the rocket reaches.
At maximum height 𝑣𝑓 = 0.
𝑣2 = 𝑣2 + 2𝑔(𝑦𝑓 − 𝑦𝑖)
𝑓
1
𝑚 2
𝑚
0 = (118 ) + 2(−9.80 )(𝑦𝑓 − 235 𝑚)
𝑠
𝑠2
𝒚𝒇 = 𝟗𝟒𝟓 𝒎
c. Find the velocity the instant before the rocket crashes on the ground.
𝑣2 = 𝑣2 + 2𝑔(𝑦𝑓 − 𝑦𝑖)
𝑓
1
𝑣2 = 0 𝑚 + 2(−9.80
𝑓
𝑚
)(0 − 945 𝑚)
𝑠2
𝒎
𝒗𝒇 = ±𝟏𝟑𝟔
𝒔
Taking the negative sign to indicate the direction of going down.
Motion in Two-Dimensions
This section includes the study of particles moving in both the x- and y- direction
simultaneously under constant acceleration. Generally, motion in two dimensions can be
modeled as two independent motions in each of the two perpendicular directions
associated with the x and y axes. That is, any influence in the y-direction does not affect the
motion in the x-direction and vice versa. The component vectors learned from ULOa is now
applied.
The position vector for a particle moving in the xy plane can be written as
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𝑟
= 𝑥𝒊 + 𝑦𝒋
If the position vector is known, the velocity of the particle can be written as
𝑣
𝑑𝑟 𝑑𝑥
𝑑𝑦
=
=
𝒊 + 𝒋 = 𝒗𝒙𝒊 + 𝒗𝒚𝒋
𝑑𝑡
𝑑𝑡
𝑑𝑡
Example 1
A particle moves in the XY plane, starting from the origin at t = 0s with an initial velocity
having an x component of 20 m/s and a y component of –15 m/s. The particle experiences
an acceleration in the x-direction, given by 𝑎𝑥 = 4.0 𝑚/𝑠2.
a. Determine the total velocity vector at any time.
𝑚
𝑚
0 𝑠2
𝑚
𝑚
To begin with, let us identify the given. 𝑣𝑥𝑖 = 20 𝑠 , 𝑣 𝑦𝑖 = −15 𝑠 , 𝑎 𝑥 = 4.0 𝑠2, 𝑎𝑦=
And substitute the given to the vector components,
𝑚
𝑚
𝑚
𝑚
+ 4.0 2 𝑡) 𝒊 + (−15 + 0 2 ) 𝒋
𝑠
𝑠
𝑠
𝑠
𝒗𝒇 = [(𝟐𝟎 + 𝟒. 𝟎𝒕)𝒊 + (−𝟏𝟓)𝒋]
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 = (𝑣𝑥𝑖 + 𝑎𝑥𝑖 𝑡)𝒊 + (𝑣𝑦𝑖 + 𝑎𝑦𝑖 𝑡)𝒋 = (20
b. Calculate the velocity and speed of the particle at t = 5.0 s and the angle the velocity vector
makes with the x-axis.
Substitute the given time to the velocity vector obtained in (a),
𝒎
𝑣𝑓 = [(20 + 4.0𝑡)𝒊 + (−15)𝒋] = [(20 + (4.0)(5.0))𝒊 + (−15)𝒋] = [𝟒𝟎𝒊 − 𝟏𝟓 𝒋]
Remember that the speed is the magnitude of the velocity,
𝒎
𝑠𝑝𝑒𝑒𝑑 𝑣𝑓 = | 𝑣 𝑓 | = √402 + (−15)2 = 𝟒𝟑
𝒔
𝑚
𝑣𝑓𝑦 −15
𝑠
=
tan 𝜃 =
𝑚
𝒗𝒇𝒙
40 𝑠
𝜽 = 𝒕𝒂𝒏−𝟏 (−
43
𝟏𝟓
𝟒𝟎
) = −𝟐𝟏°
𝒔
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Projectile Motion
Analysis in projectile motion requires the following assumptions:
1. the free-fall acceleration is constant over the range of motion and is directed downward
2. the effect of air resistance is negligible
Fig. 9 shows the trajectory and the component vectors of projectile motion. The curved path
of the motion is called the trajectory. The most important experimental fact about projectile
motion in two dimensions is that the horizontal and vertical motions are completely
independent of each other. In general, the equations of constant acceleration developed
earlier follow separately for both the x-direction and the y-direction. Also, the y-component
of velocity is zero at the highest point of the trajectory and notice how the x-component of
velocity remains constant. (In this context, 𝑣0 is the same as 𝑣𝑖.)
Fig. 9
If the velocity vector makes an angle 𝜃 with the horizontal, where 𝜃 is called the projection
angle, then from the definitions of the cosine and sine functions and Fig. 9, we have,
𝒗𝒙𝒊 = 𝒗𝒊 𝐜𝐨𝐬 𝜽
and
𝒗𝒚𝒊 = 𝒗𝒊 𝐬𝐢𝐧 𝜽
The same is true for the equations we had earlier about particle under constant velocity and
acceleration, that we obtain the following for a two-dimensional case:
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In the y-direction:
𝒗𝒚 = 𝒗𝒚𝒊 + 𝒈𝒕
𝟏
𝒚𝒇 = 𝒚𝒊 + 𝒗𝒚𝒊𝒕 + 𝒈𝒕𝟐
𝟐
𝟐 = 𝒗𝟐 + 𝟐𝒈(𝒚 − 𝒚 )
𝒗
𝒚
𝒇
𝒊
𝒚𝒊
In the x-direction:
𝒗𝒙 = 𝒗𝒙𝒊 + 𝒂𝒙𝒕
𝟏
𝒙𝒇 = 𝒙𝒊 + 𝒗𝒙𝒊𝒕 + 𝒂𝒙𝒕𝟐
𝟐
𝒗𝟐 = 𝒗𝟐 + 𝟐𝒂 (𝒙 − 𝒙 )
𝒙
𝒙
𝒙𝒊
𝒇
𝒊
The object’s velocity vector is written as 𝒗 = 𝒗𝒙𝒊 + 𝒗𝒚𝒋. The speed is the magnitude of the
vector and is solved using the Pythagorean theorem:
|𝒗| = √𝒗𝟐 + 𝒗𝟐
𝒙
𝒚
And the projection angle is the included angle of the component vectors:
𝒗𝒚
𝜽 = 𝒕𝒂𝒏−𝟏 ( )
𝒗𝒙
An important fact in projectile motion is that the acceleration of the x-component of the
motion is always zero (𝑎𝑥 = 0 𝑚/𝑠2) to which if substituted to the above formulas for the
x-direction, would give you 𝒗𝒙 = 𝒗𝒙𝒊. That is why the x-component velocity remains constant.
Example 1
An Alaskan rescue plane drops a package of emergency rations to stranded hikers. The
plane is traveling horizontally at 40.0 m/s at a height of 100. m above the ground.
a. Where does the package strike the ground relative to the point at which it was released?
𝑚
First, identify that we have 𝑣 = 40.0 and 𝑦 = 100. 𝑚
𝑥
𝑠
𝑖
1
2
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑡 + 𝑎𝑥𝑡
2 𝑡, we solve time 𝑡 from:
Since 𝑎𝑥 = 0 𝑚 always and we don’t have time
𝑠2
1 2
𝑦𝑓 = 𝑦𝑖 + 𝑣𝑦𝑖𝑡 + 𝑔𝑡
2
𝑡 = 4.52 𝑠
Substituting the time to solve 𝑥𝑓 we have,
1
2
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑡 + 𝑎𝑥𝑡
2
𝒙𝒇 = 𝟏𝟖𝟏 𝒎
b. What are the horizontal and vertical components of the velocity of the package just before
it hits the ground?
𝒎
𝑣𝑥 = 𝑣𝑥𝑖 = 𝟒𝟎. 𝟎
𝒔
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𝑚
𝒎
𝑣𝑦 = 𝑣𝑦𝑖 + 𝑔𝑡 = 𝑣𝑖 sin 𝜃 + 𝑔𝑡 = 0 + (−9.80 ) (4.52 𝑠) = −𝟒𝟒. 𝟑
𝑠
𝒔
c. Find the angle of the impact
𝑚
−44.3
𝑣𝑦
𝑠
−1
𝜃 = 𝑡𝑎𝑛−1 ( ) = tan (
𝑚 ) = −𝟒𝟖. 𝟎°
𝑣𝑥
40.0
𝑠
Horizontal Range and Maximum Height
Two points along the trajectory are of sometimes special interest, those are the horizontal
range 𝑅 and the maximum height ℎ. By doing appropriate derivations from the rectilinear
formulas discussed, we obtain the following equations for range and maximum height
respectively:
𝑅=
𝑉𝑖 2 𝑠𝑖𝑛2𝜃
𝑔
ℎ=
𝑉𝑖 2 𝑠𝑖𝑛2 𝜃
2𝑔
where 𝑔 = +9.80 𝑚/𝑠2. Remember that these formulas are only applicable to projectile
motions of which the initial and final position has the same horizontal level. Fig. 10
illustrates various trajectories for a projectile having a given initial speed but launched at
different angles. As you can see, the range is a maximum at 𝜃 = 45°.
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Fig. 10
Example 1
A long jumper leaves the ground at an angle of 20.0° above the horizontal and at a speed
of 11.0 m/s.
a. How far does he jump in the horizontal direction?
𝑅=
𝑚 2
2𝜃 (11.0 𝑠 ) sin(2)(20 °)
= 𝟕. 𝟗𝟒 𝒎
=
𝑚
𝑔
2(9.80 𝑠2)
𝑣𝑖2 sin
b. What is the maximum height reached?
ℎ=
𝑣𝑖2 sin2
2𝑔
𝜃
=
(11.0
𝑚 2
2
(sin 20.0°)
)
𝑠
= 𝟎. 𝟕𝟐𝟐 𝒎
𝑚
2 (9.80 𝑠 )
It is important to make sure that the particle returns to the same horizontal level to
use the formulas.
Self-Help: You can also refer to the sources below to help you further understand the
lesson:
*Serway, R. (2014). Physics for Scientist and Engineers with Modern Physics (9th ed)
Australia: Cengage Learning..
*Katz, D. (2017). Physics for Scientist and Engineers: Foundations and Connections.
Australia: Cengage Learning.
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Let’s Check
Activity 1. Practice Problems
Problem 1. Speed and Velocity
The speed of a nerve impulse in the human body is about 100 m/s. If you accidentally stub
your toe in the dark, estimate the time it takes the nerve impulse to travel to your brain.
Problem 2: Speed and Velocity
A motorist drives north for 35.0 minutes at 85.0 km/h and then stops for 15.0 minutes. He
then continues north, traveling 130 km in 2.00 h.
a. What is his total displacement?
b. What is his average velocity?
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Problem 3: Acceleration
A certain car is capable of accelerating at a rate of 0.60 m/s2. How long does it take for this
car to go from a speed of 55 mi/h to a speed of 60 mi/h?
Problem 4: Particle Under Constant Acceleration
A space capsule was fired from a 220-m-long cannon with final speed of 10.97 km/s. What
would have been the unrealistically large acceleration experienced by the space travelers
during their launch?
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Problem 5: Free-falling Objects
A ball is thrown vertically upward with a speed of 25.0 m/s.
a. How high does it rise?
b. How long does it take to reach its highest point?
c. How long does the ball take to hit the ground after it reaches its highest point?
d. What is its velocity when it returns to the level from which it started?
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Problem 6: Components of a Vector
A person walks 25.0° north of east for 3.10 km. How far due north and how far due east
would she have to walk to arrive at the same location?
Problem 7: Projectile Motion
A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and
with an initial speed of 15 m/s. If the brick is in flight for 3.0 s, how tall is the building?
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Let’s Analyze
Problem 1. Speed and Velocity
To qualify for the finals in a racing event, a race car must achieve an average speed of 250
km/h on a track with a total length of 1 600 m. If a particular car covers the first half of the
track at an average speed of 230 km/h, what minimum average speed must it have in the
second half of the event to qualify
Problem 2: Particle Under Constant Acceleration
An object moves with a constant acceleration of 4.00 m/s2 and over a time interval reaches
a final velocity of 12.0 m/s.
a. If its original velocity is 6.00 m/s, what is its displacement during the time interval?
b. What is the distance it travels during this interval?
c. If its original velocity is 26.00 m/s, what is its displacement during this interval?
d. What is the total distance it travels during the interval in part (c)?
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Problem 3: Free-falling Objects
A baseball is hit so that it travels straight upward after being struck by the bat. A fan
observes that it takes 3.00 s for the ball to reach its maximum height. Find the ball’s initial
velocity and the height it reaches.
Problem 4: Projectile Motion
A fireman d = 50.0 m away from a burning building directs a stream of water from a groundlevel fire hose at an angle of 𝜃 = 30.0° above the horizontal. If the speed of the stream as it
leaves the hose is 𝑣 𝑖 = 40.0 m/s, at what height will the stream of water strike the building?
Problem 5: Projectile Motion
A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the
horizontal. The projectile lands on a hillside 4.00 s later. Neglect air friction.
a. What is the projectile’s velocity at the highest point of its trajectory?
b. What is the straight-line distance from where the projectile was launched to where it hits
its target?
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In a Nutshell
Activity 1.
Under what circumstances would a vector have components that are equal in magnitude?
Activity 2.
Can the instantaneous velocity of an object at an instant of time ever be greater in
magnitude than the average velocity over a time interval containing that instant? Can it ever
be less?
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Big Picture in Focus: ULO-1c. Recall and apply the concept of force and systems of
forces
Metalanguage
In the ULOb we discussed principles of rectilinear motion while ignoring the
interactions affecting the motion. In this section, the motion of an object is described
considering the influences and causes of its movement. You will be able to answer the
following questions:
1. Why does the motion of an object change?
2. What might cause one object to remain at rest and another object to accelerate?
3. Why is it generally easier to move a small object than a large object?
For you to demonstrate ULOc, you will need to have an operational understanding
of the main factors to be considered in motion – that is, the forces acting on an object and
the mass of the object.
Thefollowing Newton’s Lawof Forcesarethekeyprincipleintheanalysisofsystemsofforces
and the achievement of this ULO-1c.
1. First law states that an object moves at constant velocity unless acted on by a force.
2. Second law states that the acceleration of an object is directly proportional to the net force
acting on it and inversely proportional to its mass. The net force acting on an object equals the
product of its mass and acceleration.
3. Third law states that in every applied force there is a always a reaction opposite to the
exerted force.
Please note that you will also be required to refer to the previous principles found in
ULO-1b.
Essential Knowledge
This section discusses on Newton’s three laws of motion and his law of gravity, the
concept of force on a more fundamental level. Dynamics is the branch of classical
mechanics concerned with the study of forces and their effects on motion. Isaac Newton
defined the fundamental physical laws which govern dynamics in physics. As long as the
system under study doesn’t involve objects comparable in size to an atom or traveling close
to the speed of light, classical mechanics provides an excellent description of nature.
Keywords
Force
inertia
equilibrium
Action force
Reaction force
Frictional Force
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Force
The basic understanding of force refers to an interaction with an object that causes it to
move. A physical contact between two objects referred to as contact forces that causes the
object’s velocity to change. However, forces do not always cause motion. For example,
when you are sitting, a gravitational force acts on your body and yet you remain stationary.
You can push on a large boulder and not be able to move it. Another applied force called
field forces doesn’t involve any direct physical contact such as gravitational force between
to masses, electric force between two charges, and magnetic force. These forces act
through empty space. Fig. 12 shows examples of forces applied to various objects.
Fig. 12
Newton’s First Law
Imagine a heavy slab lying on the floor, if you do nothing the slab remains in its position
unmoved. But if you try to push it, you may move the slab because of the applied force.
Now consider, a smooth and waxed floor, moving the slab would be easier and requires
less force. Newton’s Law describes these phenomena into three Laws of Motion.
Newton’s first law of motion sometimes called the law of inertia states that an object moves
with a velocity that is constant in magnitude and direction unless a non-zero net force acts
on it. This law explains what happens to an object that has no net force acting on it. The net
force on an object is defined as the vector sum of all external forces exerted on the object.
In the absence of external forces, an object at rest remains at rest and an object in motion
continues in motion with a constant velocity. In other words, when no force acts on an
object, the acceleration of the object is zero. The tendency of an object to resist any attempt
to change its velocity is called inertia.
Mass
Inertia is the tendency of an object to continue its motion in the absence of a force. On the
other hand, mass is a measure of the object’s resistance to changes in its motion due to a
force. The greater the mass of a body, the less it accelerates under the action of a given
applied force. From that definition, we can say that the acceleration is inversely
proportional to the mass under a given force,
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𝒎=
𝑭
𝒂
Newton’s Second Law
The second law answers the question of what happens to an object that does have a net
force acting on it. Simply stated, Newton’s Second Law states that the acceleration of an
object is directly proportional to the net force acting on it and inversely proportional to its
mass. Mathematically, we can describe the second law as
∑ 𝑭
𝒂=
𝒎
where 𝒂 is the acceleration of the object, 𝑚 is its mass, and ∑ 𝑭is the vector sum of all
forces acting on it. Force is a vector quantity, in terms of its components we write as
∑ 𝑭𝒙 = 𝒎𝒂𝒙 and ∑ 𝑭𝒚 = 𝒎𝒂𝒚
When there is no net force on an object, its acceleration is zero, which means the velocity
is constant (Newton’s First Law).
By dimension analysis, the SI standard unit of force is
∑𝐹
𝑎
=
𝑚
𝒎
∑ 𝐹 = 𝑚𝑎 = 𝒌𝒈 ∙ = 𝟏 𝑵𝒆𝒘𝒕𝒐𝒏
𝒔𝟐
In the U.S. customary system, the unit of force is the pound (lb) which is 𝑠𝑙𝑢𝑔 ∙ 𝑓𝑡/𝑠2 The
conversion from newtons to pounds is given by 1 N = 0.225 lb.
Example 1
An airboat with mass 3.50 x102 kg, including the passenger, has an engine that produces a
net horizontal force of 7.70 x102 N, after accounting for forces of resistance.
a. Find the acceleration of the airboat.
Apply Newton’s second law and solve for the acceleration,
𝐹𝑛𝑒𝑡 7.70 𝑥 102 𝑁
=
𝐹𝑛𝑒𝑡 = 𝑚𝑎 → 𝑎 =
= 𝟐. 𝟐𝟎 𝒎/𝒔𝟐
3.50 𝑥102 𝑁
𝑚
b. Starting from rest, how long does it take the airboat to reach a speed of 12.0 m/s?
Apply kinematics velocity equation from ULOb,
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡
𝑚
𝑚
𝑚
12.0 = 0 + (2.20 2 ) 𝑡
𝑠
𝑠
𝑠
𝒕 = 𝟓. 𝟒𝟓 𝒔
c. After reaching that speed, the pilot turns off the engine and drifts to a stop over a distance
of 50.0 m. Find the resistance force, assuming it’s constant.
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Let 12.0 m/s of (b) be the initial velocity. Also, know that the 50.0 m is the
displacement ∆𝑥,
𝑣2 = 𝑣2 + 2𝑎(𝑥𝑓 − 𝑥𝑖)
𝑓
𝑖
𝑚
𝑚 2
0 = (12.0 ) + 2𝑎(50.0𝑚)
𝑠
𝑠
𝑎 = −1.44 𝑚/𝑠2
Substitute the acceleration into Newton’s second law, finding the resistance force:
𝐹𝑟𝑒𝑠𝑖𝑠𝑡 = 𝑚𝑎 = (3.50𝑥102𝑘𝑔)(−1.44𝑚/𝑠2) = −𝟓𝟎𝟒 𝑵
Example 2
Two horses are pulling a barge with mass 2.00 x103 kg along a canal, as shown in Fig. 13.
The cable connected to the first horse makes an angle of 𝜃 = 30.0° with respect to the
direction of the canal, while the cable connected to the second horse makes an angle of 𝜃 =
− 45.0°. Find the initial acceleration of the barge, starting at rest, if each horse exerts a
force of magnitude 6.00x102 N on the barge. Ignore forces of resistance on the barge.
Fig. 13
Compute the total x- and y- component forces exerted by the horses,
𝐹1𝑥 = 𝐹1𝑐𝑜𝑠𝜃 = (6.00𝑥102𝑁)(cos 30.0°) = 5.20𝑥102𝑁
𝐹2𝑥 = 𝐹2𝑐𝑜𝑠𝜃 = (6.00𝑥102𝑁)(cos −45.0°) = 4.24𝑥102𝑁
∑ 𝐹 = 𝐹1𝑥 + 𝐹2𝑥 = 9.44 𝑥102𝑁
𝑥
𝐹1𝑦 = 𝐹1𝑠𝑖𝑛𝜃 = (6.00𝑥102𝑁)(sin 30.0°) = 3.00𝑥102𝑁
𝐹2𝑦 = 𝐹2𝑠𝑖𝑛𝜃 = (6.00𝑥102𝑁)(sin −45.0°) = −4.24𝑥102𝑁
∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = −1.24 𝑥102𝑁
Obtain the components of the acceleration,
∑ 𝐹𝑥
9.44𝑥102𝑁
𝑚
=
=
0.472
𝑎𝑥 =
𝑚 2.00 𝑥103𝑘𝑔
𝑠2
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∑ 𝐹𝑦 −1.24 𝑥102𝑁
𝑚
=
=
−0.0620
𝑚 2.00 𝑥103𝑘𝑔
𝑠2
Calculate the magnitude of the acceleration,
𝑎𝑦 =
𝑎 = √𝑎2 + 𝑎2 = √(0.472 𝑚 )2 + (−0.0620 𝑚 )2 = 𝟎. 𝟒𝟕𝟔 𝒎/𝒔𝟐
𝑥
𝑦
𝑠2
𝑠2
Calculate the direction,
𝜃 = tan−1
𝑎𝑦
𝑎𝑥
=
−0.0620𝑚/𝑠2
= −𝟕. 𝟒𝟔 °
0.472 𝑚/𝑠2
Example 3
A hockey puck having a mass of 0.30 kg slides on the frictionless, horizontal surface of an
ice rink. Two hockey sticks strike the puck simultaneously, exerting the forces on the puck
shown in Fig. 14. The force 𝐹1 has a magnitude of 5.0 N, and is directed at 𝜃 = 20° below
the x-axis. The force 𝐹2 has a magnitude of 8.0 N and its direction is 𝜙 = 60° above the xaxis. Determine both the magnitude and the
direction of the puck’s acceleration.
Compute the total x- and y- component forces
exerted by the sticks,
𝐹1𝑥 = 𝐹1𝑐𝑜𝑠𝜃 = (5.0𝑁)(cos −20°) = 4.7𝑁
𝐹2𝑥 = 𝐹2𝑐𝑜𝑠𝜃 = (8.0𝑁)(cos 60°) = 4.0𝑁
∑ 𝐹𝑥 = 𝐹1𝑥 + 𝐹2𝑥 = 8.7𝑁
𝐹1𝑦 = 𝐹1𝑠𝑖𝑛𝜃 = (5.0𝑁)(sin −20°) = −1.7𝑁
𝐹2𝑦 = 𝐹2𝑠𝑖𝑛𝜃 = (8.0𝑁)(sin 60°) = 6.9𝑁
∑ 𝐹𝑦 = 𝐹1𝑦 + 𝐹2𝑦 = 5.2𝑁
Obtain the components of the acceleration,
∑ 𝐹𝑥
8.7𝑁
𝑚
=
= 29 2
𝑎𝑥 =
𝑚 0.30𝑘𝑔
𝑠
∑ 𝐹𝑦
6.9𝑁
𝑚
=
= 23 2
𝑎𝑦 =
𝑚 0.30𝑘𝑔
𝑠
Calculate the magnitude of the acceleration,
𝑎 = √𝑎2 + 𝑎2 = √(29 𝑚 )2 + (23 𝑚 )2 = 𝟑𝟒𝒎/𝒔𝟐
𝑥
𝑦
𝑠2
𝑠2
Calculate the direction,
𝑎𝑦 23𝑚/𝑠2
−1
𝜃 = tan
=
= 𝟑𝟏 °
𝑎𝑥 29𝑚/𝑠2
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Gravitational Force
The gravitational force is the mutual force of attraction between any two objects in the
Universe, as shown in Fig. 15. Newton’s law of universal gravitation states that every
particle in the Universe attracts every other particle with a force that is directly
proportional to the product of the masses of the particles and inversely proportional to
the square of the distance between them. That is,
𝑚1𝑚2
𝐹𝑔 = 𝐺 2
𝑟
Where G = 6.67x10-11 N m2/kg2 is the universal gravitation constant.
Fig. 15
The magnitude of the gravitational force acting on an object of mass 𝑚 is called the weight
𝑤 of the object, given by
𝑊 = 𝑚𝑔
Where 𝑔 is the acceleration due to gravity. The SI standard unit for weight is Newton 𝑁.
Newton’s Third Law
Newton recognized, however, that a single isolated force couldn’t exist. Instead, forces in
nature always exist in pairs. He stated on his third law that, if two objects interact, the force
𝐹12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the
force 𝐹21 exerted by object 2 on object 1:
𝐹12 = −𝐹21
The force that object 1 exerts on object 2 is popularly called the action force, and the force
of object 2 on object 1 is called the reaction force. The negative sign indicates that the
reaction force is acted in the opposite direction of equal magnitude.
There are many kinds of reaction forces such as friction force and normal force. Frictional
Force refers to the force generated by two surfaces that contacts and slide against each
other. Example of which is a slab on a rough surface, a book sliding on a smooth surface,
and such. The normal force is the support force exerted upon an object that is in contact
with another stable object or surface, it is acted perpendicular to the action force. It is the
upward force that opposes the weight of an object.
Example 1
A man of mass 75.0 kg and woman of mass 55.0 kg stand facing each other on an ice rink,
both wearing ice skates. The woman pushes the man with a horizontal force of 85.0 N in
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the positive x-direction. Assume the ice is frictionless.
a. What is man’s acceleration?
𝐹 = 𝑚𝑎 → 𝑎 =
𝐹
=
85.0𝑁
= 𝟏. 𝟏𝟑 𝒎/𝒔𝟐
𝑚 75.0 𝑘𝑔
b. What is the reaction force acting on the woman?
Apply Newton’s third law of motion, finding that the reaction force R acting on the
woman:
𝑅 = −𝐹 = −𝟖𝟓. 𝟎𝑵
c. Calculate the woman’s acceleration.
𝐹 −85.0𝑁
𝐹 = 𝑚𝑎 → 𝑎 =
=
= −1.55 𝒎/𝒔𝟐
𝑚 55.0 𝑘𝑔
Example 2
A traffic light weighing 1.00 x102 N hangs from a vertical cable tied to two other cables that
are fastened to a support, as in Fig. 16a. The upper cables make angles of 37.0° and 53.0°
with the horizontal. Find the tension in each of the three cables.
Tension is described as the pulling force transmitted through a rope, string or wire.
Because we are interested only at the body and the acting forces, a force diagram
called free-body diagram (FBD) would be helpful in the analysis. Identify all the
action and reaction forces, in this case, we have the weight (Fg) of the traffic light
and the tension forces of the strings. The construction of a correct free-body
diagram is an essential step in applying Newton’s laws. An incorrect diagram will
most likely lead to incorrect answers! Fig. 16 is an example of an FBD which shows
the external forces acting on the body necessary for the analysis. Fig. 16b shows the
forces acting on the traffic light and Fig.16c shows the forces acting on the cable
knot.
Fig. 16
Objects that are either at rest or moving with constant velocity are said to be in
equilibrium. Because of 𝑎 = 0 𝑚/𝑠2, Newton’s second law applied to an object in
equilibrium gives
∑𝐹 = 0
Find T3 from Fig. 16b, using the condition of equilibrium:
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∑ 𝐹 = 𝑇3 − 𝐹𝑔 = 0 → 𝑇3 = 𝐹𝑔
𝑻𝟑 = 𝟏. 𝟎𝟎𝒙𝟏𝟎𝟐𝑵
Note the direction of the force, positive are those applied upward and rightwards
while negative are those downward and leftwards. 𝑇1 and 𝑇2 are two-dimensional
forces which means you need to solve for the x- and y- components.
For the summation of forces along the x-direction, there are two component forces
acted by 𝑇1 and 𝑇2:
𝑇1𝑥 = 𝑇1𝑐𝑜𝑠𝜃 = −𝑇1(cos 37.0°)
𝑇2𝑥 = 𝑇2𝑐𝑜𝑠𝜃 = 𝑇2(cos 53.0°)
∑ 𝑇𝑥 = −𝑇1(cos 37.0°) + 𝑇2(cos 53.0°) = 0
→ 𝑒𝑞. 1
While in the y-direction, there are three component forces acted by 𝑇1, 𝑇2 and
weight 𝐹𝑔:
𝑇1𝑦 = 𝑇1𝑠𝑖𝑛𝜃 = 𝑇1(sin 37.0°)
𝑇2𝑦 = 𝑇2𝑠𝑖𝑛𝜃 = 𝑇2(sin 53.0°)
𝑇3 = −1.00𝑥102𝑁
∑ 𝑇𝑦 = 𝑇1(sin 37.0°) + 𝑇2(sin 53.0°) − 1.00𝑥102𝑁 = 0
→ 𝑒𝑞. 2
Equating eq. 1 and 2 to solve 𝑇1 and 𝑇2
−𝑇1(cos 37.0°) + 𝑇2(cos 53.0°) = 0
𝑇1(sin 37.0°) + 𝑇2(sin 53.0°) − 1.00𝑥102𝑁 = 0
𝑻𝟏 = 𝟔𝟎. 𝟏𝑵 and 𝑻𝟐 = 𝟕𝟗. 𝟗 𝑵
Example 3
A sled is tied to a tree on a frictionless, snow-covered hill shown in Fig. 17. If the sled
weighs 77.0 N, find the magnitude of the tension force exerted by the rope on the sled
and that of the normal force 𝑛 exerted by the hill on the sled.
Fig. 17
Fig. 17b is the FBD for this force analysis. The object is at equilibrium, therefore,
apply Newton’s second law.
For the summation of forces along the x-axis,
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∑ 𝐹𝑥 = +𝑇 − 𝐹𝑔𝑠𝑖𝑛30 = 0
𝑇 − 77.0𝑁 𝑠𝑖𝑛30° = 0
𝑻 = 𝟑𝟖. 𝟓 𝑵
Notice that sine function is used to solve for the x-component, that is because of the
angle 30° is opposite to the x-component of the weight force.
For the summation of forces along the y-axis,
∑ 𝐹𝑦 = +𝑛 − 𝐹𝑔𝑐𝑜𝑠30 ° = 0
𝑛 − 77.0𝑁 𝑐𝑜𝑠30 ° = 0
𝒏 = 𝟔𝟔. 𝟕 𝑵
Example 4
A car of mass m is on an icy driveway inclined at an angle 𝜃 = 20.0°, as in Fig. 18a.
Fig. 18
a. Determine the acceleration of the car, assuming the incline is frictionless.
Since the force causes the car to move, the summation of forces is equal to 𝑚𝑎. The
acceleration happens only along x-direction, hence
∑ 𝐹𝑥= 𝐹𝑔sin 20.0 ° = 𝑚𝑎𝑥
(𝑚𝑔) sin 20.0 ° = 𝑚𝑎𝑥
𝑚
(9.80 ) sin 20.0 ° = 𝑎𝑥
𝑠2
𝒂𝒙 = 𝟑. 𝟑𝟓 𝒎/𝒔𝟐
b. If the length of the driveway is 25.0 m and the car starts from rest at the top, how long
does it take to travel to the bottom?
1 2
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑖𝑡 + 𝑎𝑡
1 2 𝑚
25.0 𝑚 = 0 + 0 + (3.35 ) (𝑡2)
2
𝑠2
𝒕 = 𝟑. 𝟖𝟔 𝒔
c. What is the car’s speed at the bottom?
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𝑚
𝑚
𝒎
𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡 = 0 + (3.35 2) (3.86 𝑠) = 𝟏𝟐. 𝟗
𝑠
𝑠
𝒔
Forces of Friction
A moving object on a surface or through a viscous medium experiences an opposite
resistance force called friction. There are two types of friction: static 𝑓𝑠 and kinetic 𝑓𝑘. A
frictional force of an object exerted by its surface is proportional to its normal force 𝑛.
𝑓 ≤𝜇𝑛
Where 𝜇 is the proportionality constant called the coefficient of static friction 𝜇𝑠 or the
coefficient of kinetic friction 𝜇𝑘, depending on the frictional force. 𝜇𝑘 is generally less than
𝜇𝑠 and these values are constants depending on the surface or medium. Table 1.6 lists the
most common mediums and their coefficient of friction. These values are approximated,
however, normally the coefficients to be used in the problem are given.
Table 1.6 Coefficients of Friction
Steel on steel
Aluminum on steel
Rubber on concrete
Wood on wood
Glass on glass
𝜇𝑠
0.74
0.61
1.0
0.25-0.5
0.94
𝜇𝑘
0.57
0.47
0.8
0.2
0.4
Example 1
Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static
friction between the block and ramp is 0.350, what maximum angle can the ramp make
with the horizontal before the block starts to slip down?
Identifying the forces we only have the weight. Next, identify whether the object is
at equilibrium, constant velocity, or with acceleration. In this case, the block is at
equilibrium before it starts to slip down, therefore, use Newton’s second law for the
x- and y- components. Draw an FBD as shown,
Now take note of the presence of the frictional force opposite to the applied force
which is the x-component of the weight. We know that 𝜇𝑠 = 𝑓𝑠/𝑛, meaning, we
have to solve for the normal force first along the y-axis.
∑ 𝐹𝑦 = 𝐹𝑔𝑐𝑜𝑠𝜃 − 𝑛 = (𝑚𝑔)(cos 𝜃) − 𝑛 = 0
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𝑚
equation for the normal force → (2.50 𝑘𝑔) (9.80 2) cos 𝜃 = 𝑛
𝑠
∑ 𝐹𝑥 = 𝐹𝑔𝑠𝑖𝑛𝜃 − 𝑓𝑠 = (𝑚𝑔)(sin 𝜃) − 𝜇𝑠𝑛 = 0
𝑚
𝑚
(2.50 𝑘𝑔) (9.80 2) sin 𝜃 − (0.350) [(2.50𝑘𝑔) (9.80 2) cos 𝜃] = 0
𝑠
𝑠
Solving for the angle we get,
𝜃 = tan−1 0.350
𝜽 = 𝟏𝟗. 𝟑 °
Example 2
The hockey puck struck by a hockey stick is given an initial speed of 20.0 m/s on a frozen
pond. The puck remains on the ice and slides 1.20 x102 m, slowing down steadily until it
comes to rest. Determine the coefficient of kinetic friction between the puck and the ice.
Identify the acting forces, in this case, we have the weight, normal, 𝑓𝑘, and the
applied force that caused the motion. The object is at motion at the x-axis and
equilibrium at y-axis, therefore, use Newton’s third and second law respectively.
Draw the FBD,
We know that 𝜇𝑘 = 𝑓𝑘/𝑛, meaning, we have to solve for the normal and frictional
force first. Identify the given, we have the puck’s initial and final speed and distance.
We cannot go directly with the summation of forces because the given values are
not sufficient. Utilizing the given to solve the unknown, we have,
𝑣2 = 𝑣2 + 2𝑎(𝑥𝑓 − 𝑥𝑖)
𝑓
𝑖
𝑚
𝑚 2
0 = (20.0 ) + 2𝑎(1.20 𝑥102 − 0 𝑚)
𝑠
𝑠
𝑎 = −1.67 𝑚/𝑠2
Now we can proceed to Newton’s second law for the y-axis,
∑ 𝐹𝑦 = 𝑛 − 𝐹𝑔 = 0
𝑛 = 𝑚𝑔
Then Newton’s third law for the x-axis,
∑ 𝐹𝑥 = − 𝑓𝑘 = 𝑚𝑎
−𝜇𝑘𝑛 = 𝑚𝑎
𝑚
𝑚
−𝜇𝑘(𝑚) (9.80 2) = 𝑚(−1.67 2)
𝑠
𝑠
𝝁𝒌 = 𝟎. 𝟏𝟕𝟎
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Self-Help: You can also refer to the sources below to help you further understand the
lesson:
*Serway, R. (2014). Physics for Scientist and Engineers with Modern Physics (9th ed)
Australia: Cengage Learning..
*Katz, D. (2017). Physics for Scientist and Engineers: Foundations and Connections.
Australia: Cengage Learning.
*Young, H. D. (2016). Sears and Zemanky's University Physics with Modern Physics (14th
ed.). Harlow, England: Pearson
Let’s Check
Activity 1. Practice Problems
Problem 1. Laws of Motion
The heaviest invertebrate is the giant squid, which is estimated to have a weight of about 2
tons spread out over its length of 70 feet. What is its weight in newtons?
Problem 2: Laws of Motion
A 6.0-kg object undergoes an acceleration of 2.0 m/s2.
a. What is the magnitude of the resultant force acting on it?
b. If this same force is applied to a 4.0-kg object, what acceleration is produced?
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Problem 3: Laws of Motion
A 75-kg man standing on a scale in an elevator notes that as the elevator rises, the scale
reads 825 N. What is the acceleration of the elevator?
Problem 4: Laws of Motion
A dockworker loading crates on a ship finds that a 20-kg crate, initially at rest on a
horizontal surface, requires a 75-N horizontal force to set it in motion. However, after the
crate is in motion, a horizontal force of 60 N is required to keep it moving with a constant
speed. Find the coefficients of static and kinetic friction between crate and floor.
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Let’s Analyze
Problem 1: Laws of Motion
A 5.0-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. What force (assumed
constant) is exerted on the bullet while it is traveling down the 0.82-m-long barrel of the
rifle?
Problem 2: Laws of Motion
Two forces are applied to a car to move it, as shown below.
a. What is the resultant of these two forces?
b. If the car has a mass of 3 000 kg, what acceleration does it have? Ignore friction.
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Problem 3: Laws of Motion
A 150-N bird feeder is supported by three cables as shown. Find the tension in
each cable.
Problem 4: Laws of Motion
A 1 000-N crate is being pushed across a level floor at a constant speed by a force of 300 N
at an angle of 20.0° below the horizontal, as shown in Figure a.
a. What is the coefficient of kinetic friction between the crate and the floor?
b. If the 300-N force is instead pulling the block at an angle of 20.0° above the horizontal,
as shown in Figure b, what will be the acceleration of the crate? Assume that the coefficient
of friction is the same as that found in part (a).
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In a Nutshell
Activity 1
A space explorer is moving through space far from any planet or star. He notices a large
rock, taken as a specimen from an alien planet, floating around the cabin of the ship. Should
he push it gently, or should he kick it toward the storage compartment? Explain.
Activity 2:
If
only
one
force
acts
on
an
object,
70
can
it
be
in
equilibrium?
Explain.
71