Physics I course – Winter
2021
Dr. Yehia Eissa
Lecturer of Physics, Faculty of Engineering
German International University (GIU)
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Learning outcomes
Objective:
The major objectives of this course
are for students to learn the
fundamental principles of classical
mechanics, to develop solid and
systematic problem solving skills, and
to lay the foundations for further
studies in physics, physical sciences,
and engineering.
Topics:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Physics and measurement
Motion in one dimension
Vectors
Motion in two dimensions
The laws of motion
Energy and energy transfer
Potential energy
Linear momentum and collisions
Rotation of a rigid object about a
fixed axis
Angular momentum
Static equilibrium and elasticity
Fluid mechanics
Introduction to oscillatory motion
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References
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Chapters 7 & 8: Energy
Dec. 18, 2021
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Doing work, transferring energy
• The force exerted by the weightlifter transfers energy from her
to the weights. We know that the
weights have gained energy
because, when the athlete
releases them, they come
crashing down to the ground.
• ‘Doing work’ is a way of
transferring energy from one
object to another.
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Work
The work, W, done on a system by an agent exerting a constant force on the
system is the product of the magnitude F of the force, the magnitude Dr of the
displacement of the point of application of the force, and cos q, where q is the
angle between the force and the displacement vectors.
 The meaning of the term work is distinctly different in physics than in
everyday meaning.
 Work is done by some part of the environment that is interacting directly with
the system.
 Work is done on the system.
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Work, cont.
W = F Dr cos q
 The displacement is that of the
point of application of the force.
 A force does no work on the object
if the force does not move through
a displacement.
 The work done by a force on a
moving object is zero when the
force applied is perpendicular to
the displacement of its point of
application.
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Work Example
The normal force and the gravitational
force do no work on the object.
 cos q = cos 90° = 0
The force F is the only force that does
work on the object.
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More About Work
The sign of the work depends on the direction of the force relative to the
displacement.
 Work is positive when projection of
as the displacement.
F onto Dr
is in the same direction
 Work is negative when the projection is in the opposite direction.
The work done by a force can be calculated, but that force is not necessarily
the cause of the displacement.
Work is a scalar quantity.
The unit of work is a joule (J)
 1 joule = 1 newton . 1 meter = kg ∙ m² / s²
 J=N·m
 The joule is defined as the amount of work done when a force of 1
newton moves a distance of 1 meter in the direction of the force.
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Work Is An Energy Transfer
This is important for a system approach to solving a problem.
If the work is done on a system and it is positive, energy is transferred to the
system.
If the work done on the system is negative, energy is transferred from the
system.
If a system interacts with its environment, this interaction can be described as a
transfer of energy across the system boundary.
 This will result in a change in the amount of energy stored in the system.
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Example 7.1
• A man cleaning a floor pulls a
vacuum cleaner with a force of
magnitude F = 50.0 N at an
angle of 30.0° with the
horizontal.
• Calculate the work done by the
force on the vacuum cleaner as
the vacuum cleaner is displaced
3.00 m to the right.
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Example 7.1 – Solution
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Scalar Product of Two Vectors
The scalar product of two vectors is
written as A  B .
 It is also called the dot product.
A  B  A B cos q
 q is the angle between A and B
Applied to work, this means
W  F Dr cosq  F  Dr
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Scalar Product, cont
The scalar product is commutative.
 A B  B  A
The scalar product obeys the distributive law of multiplication.


 A  B  C  A B  A  C
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Dot Products of Unit Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  1
ˆi  ˆj  ˆi  kˆ  ˆj  kˆ  0
Using component form with vectors:
A  Ax ˆi  Ay ˆj  Azkˆ
B  Bx ˆi  By ˆj  Bzkˆ
A B  Ax Bx  Ay By  Az Bz
In the special case where
A  B;
A  A  Ax2  Ay2  Az2  A2
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Example 7.2
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Example 7.2 – Solution
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Example 7.3
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Example 7.3 – Solution
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Work Done by a Varying Force
To use W = F Δr cos θ, the force must
be constant, so the equation cannot be
used to calculate the work done by a
varying force.
Assume that during a very small
displacement, Dx, F is constant.
For that displacement, W ~ F Dx
For all of the intervals,
xf
W   Fx Dx
xi
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Work Done by a Varying Force, cont.
Let the size of the small displacements
approach zero .
Since
lim
Dx 0
xf
 F Dx  
x
xi
xf
xi
Fx dx
Therefore,
xf
W   Fx dx
xi
The work done is equal to the area
under the curve between xi and xf.
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Work Done By Multiple Forces
If more than one force acts on a system and the system can be modeled as a
particle, the total work done on the system is the work done by the net force.
W  W
ext

xf
xi
  F dx
x
In the general case of a net force whose magnitude and direction may vary.
W  W
ext

xf
xi
 Fdr
The subscript “ext” indicates the work is done by an external agent on the
system.
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Example 7.4
• A force acting on a particle
varies with x, as shown in the
figure.
• Calculate the work done by the
force as the particle moves from
x = 0 to x = 6.0 m.
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Example 7.4 – Solution
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Example 7.5
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Example 7.5 – Solution
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Work Done By A Spring
A model of a common physical system
for which the force varies with position.
The block is on a horizontal, frictionless
surface.
Observe the motion of the block with
various values of the spring constant.
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Spring Force (Hooke’s Law)
The force exerted by the spring is
Fs = - kx
 x is the position of the block with respect to the equilibrium position (x = 0).
 k is called the spring constant or force constant and measures the stiffness of the spring.
 k measures the stiffness of the spring.
This is called Hooke’s Law.
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Hooke’s Law, final
The force exerted by the spring is always directed opposite to the displacement
from equilibrium.
The spring force is sometimes called the restoring force.
If the block is released it will oscillate back and forth between –x and x.
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Work Done by a Spring
Identify the block as the system.
Calculate the work as the block moves
from xi = - xmax to xf = 0.
0
𝑊𝑠 = න 𝐹𝑆 𝑑𝑥 = න
−𝑥𝑚𝑎𝑥
1 2
−𝑘𝑥 𝑑𝑥 = 𝑘𝑥𝑚𝑎𝑥
2
The net work done as the block moves
from -xmax to xmax is zero
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Work Done by a Spring, cont.
Assume the block undergoes an arbitrary displacement from x = xi to x = xf.
The work done by the spring on the block is
Ws  
xf
xi
 kx  dx 
1 2 1 2
kxi  kxf
2
2
 If the motion ends where it begins, W = 0
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Example 7.6
• A common technique used to measure
the force constant of a spring is
demonstrated by the setup in the
figure.
• The spring is hung vertically, and an
object of mass m is attached to its
lower end. Under the action of the
“load” mg, the spring stretches a
distance d from its equilibrium
position.
• If a spring is stretched 2.0 cm by a
suspended object having a mass of
0.55 kg, what is the force constant of
the spring?
• How much work is done by the spring
as it stretches through this distance?
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Example 7.6 – Solution
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Kinetic Energy
One possible result of work acting as an influence on a system is that the system
changes its speed.
The system could possess kinetic energy.
Kinetic Energy is the energy of a particle due to its motion.
 K = ½ mv2
 K is the kinetic energy
 m is the mass of the particle
 v is the speed of the particle
A change in kinetic energy is one possible result of doing work to transfer energy
into a system.
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Kinetic Energy, cont
Calculating the work:
Wext  
xf
xi
 F dx  
xf
xi
vf
Wext   mv dv
vi
Wext
Wext
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1
1
2
 mv f  mv i2
2
2
 K f  K i  DK
ma dx
Work-Kinetic Energy Theorem
The Work-Kinetic Energy Theorem states Wext = Kf – Ki = ΔK
When work is done on a system and the only change in the system is in its speed, the net
work done on the system equals the change in kinetic energy of the system.
 The speed of the system increases if the work done on it is positive.
 The speed of the system decreases if the net work is negative.
The work-kinetic energy theorem is not valid if other changes (besides its speed) occur in
the system or if there are other interactions with the environment besides work.
The work-kinetic energy theorem applies to the speed of the system, not its velocity.
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Work-Kinetic Energy Theorem – Example
The block is the system and three
external forces act on it.
The normal and gravitational forces do
no work since they are perpendicular to
the direction of the displacement.
Wext = DK = ½ mvf2 – 0
The answer could be checked by
modeling the block as a particle and
using the kinematic equations.
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Work-Kinetic Energy Theorem – Example
• A 6.0-kg block initially at rest is
pulled to the right along a
horizontal, frictionless surface
by a constant horizontal force of
12 N.
• Find the speed of the block after
it has moved 3.0 m.
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Work-Kinetic Energy Theorem – Example
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Any questions?
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