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Chapter 7
Work and Kinetic Energy
Introduction
The concept of energy is one of the most important topics in science and
engineering.
In everyday life, we think of energy in terms of fuel for transportation and
heating, electricity for lights and appliances, and foods for consumption.
However, these ideas do not really define energy. They merely tell us that fuels
are needed to do a job and that those fuels provide us with something we call
energy.
2
Introduction
In this chapter,
•
we first introduce the concept of work.
•
Work is done by a force acting on an object when the point of application of
that force moves through some distance and the force has a component
along the line of motion.
•
Next, we define kinetic energy, which is energy an object possesses because
of its motion.
•
We shall see that the concepts of work and kinetic energy can be applied to
the dynamics of a mechanical system without resorting to Newton’s laws.
3
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 m e a n i n g o f t h e t e r m w o r k 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.
4
Section 7.2
Work, cont.
W = F Dr cos q
or
W = F d 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.
5
Section 7.2
Displacement in the Work Equation
The displacement is that of the point of application of the force.
If the force is applied to a rigid object that can be modeled as a particle, the
displacement is the same as that of the particle.
For a deformable system, the displacement of the object generally is not the
same as the displacement associated with the forces applied.
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Section 7.2
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.
If an applied force F acts
along the direction of the
displacement, then q =0
and cos 0=1. In this case
W = F d
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Section 7.2
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
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Section 7.2
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.
9
Section 7.2
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 (Fig.) 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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Solutions 7.1
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Scalar Product of Two Vectors
Because of the way the force and displacement vectors are combined in
Equation of work, it is helpful to use a convenient mathematical tool called the
scalar product.
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
or
12
Section 7.3
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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Section 7.3
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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Section 7.3
Example 7.3
A particle moving in the x-y plane undergoes a displacement d = ( 2 . 0 i 3 . 0 j ) m as a constant force F = ( 5 . 0 i - 2 . 0 j ) N acts on the particle. (a)
Calculate the magnitude of the displacement and that of the force. (b)
Calculate the work done by F.
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Solutions
a)
b)
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Work Done by a V a r y i n g F o r c e
To use W = F d 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 or d, F is constant.
For that displacement, W ~ F Dx
For all of the intervals,
xf
W   Fx Dx
xi
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Section 7.4
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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Section 7.4
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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Section 7.4
Work Done by Multiple Forces, cont.
If the system cannot be modeled as a particle, then the total work is equal to the
algebraic sum of the work done by the individual forces.
W  W
ext

 F  d r 
forces
 Remember work is a scalar, so this is the algebraic sum.
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Section 7.4
Example 7.4
A force acting on a particle varies with x, as shown in Figure. Calculate
the work done by the force as the particle moves from x=0 to x=6.0 m.
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Solution
The work done by the force is equal to the area under the curve from x A = 0
t o x C = 6 . 0 m . This area is equal to the area of the rectangular section from
A to B plus the area of the triangular section from B to C.
The area of the rectangle is ( 4 . 0 ) ( 5 . 0 ) N m = 2 0 J , and
the area of the triangle is ½ ( 2 . 0 ) ( 5 . 0 ) N m = 5 0 J .
Therefore, the total work done is 25 J.
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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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Section 7.4
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 s p r i n g c o n s t a n t 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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Section 7.4
Hooke’s Law, cont.
The vector form of Hooke’s Law is
Fs  Fx ˆi  kx ˆi
When x is positive (spring is stretched), F is negative
When x is 0 (at the equilibrium position), F is 0
When x is negative (spring is compressed), F is positive
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Section 7.4
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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Section 7.4
Work Done by a Spring
Identify the block as the system.
Calculate the work as the block
moves from xi = - xmax to xf = 0.
Ws   Fs  d r  
xi
 kx ˆi   dx ˆi 
1 2

kx  dx  kxmax

 xmax
2
0
The net work done as the block
moves from -xmax to xmax is zero
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xf
Section 7.4
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
1 2 1 2
 kx  dx  kxi  kxf
2
2
 If the motion ends where it begins, W = 0
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Section 7.4
Spring with an Applied Force
Suppose an external agent, Fapp, stretches the
spring.
The applied force is equal and opposite to the
spring force.


Fapp  Fapp ˆi  Fs   kx ˆi  kx ˆi
Work done by Fapp as the block moves from –
xmax to x = 0 is equal to - ½ k x 2 m a x
For any displacement, the work done by the
applied force is
Wapp  
xf
xi
1 2 1 2
 kx  dx  kxf  kxi
2
2
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Section 7.4
Example 7.6
A common technique used to measure the force
constant of a spring is described in Figure. The spring is
hung vertically, and an object of mass mis attached to its
lower end. Under the action of the “load” mg, the spring
stretches a distance d from its equilibrium position. What is
the force constant ?
30
Solution
Because the spring force is upward (opposite the displacement), it must balance
the downward force of gravity m g when the system is at rest.
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Energy Review
Kinetic Energy
 Associated with movement of members of a system
Potential Energy
 Determined by the configuration of the system
 Gravitational and Elastic Potential Energies have been studied
Internal Energy
 Related to the temperature of the system
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Introduction
Types of Systems
Non-isolated systems
 Energy can cross the system boundary in a variety of ways.
 Total energy of the system changes
Isolated systems
 Energy does not cross the boundary of the system
 Total energy of the system is constant
Conservation of energy
 Can be used if no non-conservative forces act within the isolated system
 Applies to biological organisms, technological systems, engineering
situations, etc
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Introduction
Ways to Transfer Energy Into or Out of A System
In non-isolated systems, energy crosses the boundary of the system during some
time interval due to an interaction with the environment.
Work – transfers energy by applying a force and causing a displacement of the
point of application of the force.
Mechanical Wave – transfers energy by allowing a disturbance to propagate
through a medium.
Heat – the mechanism of energy transfer that is driven by a temperature
difference between two regions in space.
Matter Transfer – matter physically crosses the boundary of the system, carrying
energy with it.
Electrical Transmission – energy transfer into or out of a system by electric
current.
Electromagnetic Radiation – energy is transferred by electromagnetic waves.
34
Section 8.1
Examples of Ways to Transfer Energy
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Section 8.1
Conservation of Energy
Energy is conserved
 This means that energy cannot be created nor destroyed.
 If the total amount of energy in a system changes, it can only be due to the
fact that energy has crossed the boundary of the system by some method of
energy transfer.
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Section 8.1
Conservation of Energy, cont.
Mathematically, DEsystem = ST
 Esystem is the total energy of the system
 T is the energy transferred across the system boundary by some mechanism
 Established symbols: Twork = W and Theat = Q
 Others just use subscripts
The primarily mathematical representation of the energy version of the analysis model
of the non-isolated system is given by the full expansion of the above equation.
 D K + D U + DEint = W + Q + TMW + TMT + TET + TER
 TMW – transfer by mechanical waves
 TMT – by matter transfer
 TET – by electrical transmission
 TER – by electromagnetic transmission
37
Section 8.1
Isolated System
For an isolated system, DEmech = 0
 Remember Emech = K + U
 This is conservation of energy for an isolated system with no nonconservative forces acting.
If non-conservative forces are acting, some energy is transformed into internal
energy.
Conservation of Energy becomes DEsystem = 0
 Esystem is all kinetic, potential, and internal energies
 This is the most general statement of the isolated system model.
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Section 8.2
Isolated System, cont.
The changes in energy can be written out and rearranged.
Kf + Uf = Ki + Ui
 Remember, this applies only to a system in which conservative forces act.
39
Section 8.2
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 mv
2
 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.
40
Section 7.5
Kinetic Energy, cont
A particle undergoing a displacement d
and a change in velocity under the action of a constant net force SF.
A particle of mass m moving to the right under the action of a constant
net force SF.
Because the force is constant, we know from Newton’s second law that
the particle moves with a constant acceleration a.
If the particle is displaced a distance d, the n e t w o r k done by the
total force SF is
41
Section 7.5
Kinetic Energy, cont
We found that the following relationships are valid when a
particle undergoes constant acceleration:
where vi is the speed at t =0 and vf is the speed at time t. Substituting these
expressions into Equation
42
Kinetic Energy, cont
The quantity ½ m v2 represents the energy associated with the motion
of the particle. This quantity is so important that it has been given a special name
—kinetic energy. The net work done on a particle by a constant net force SF
acting on it equals the change in kinetic energy of the particle.
In general, the kinetic energy K of a particle of mass m moving with a speed v is
defined as
43
Work-Kinetic Energy Theorem
The Work-Kinetic Energy Theorem states W e x t = K f – K i = Δ K
When work is done on a system and the only change in the system is in its speed, the
n e t w o r k done on the system e q u a l s t h e c h a n g e i n k i n e t i c
e n e r g y 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.
 Also valid for changes in rotational speed
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.
44
Section 7.5
Work-Kinetic Energy Theorem, cont
We can apply Newton’s second law, SFx=max
and the net work done as
45
Section 7.5
Situations Involving Kinetic Friction
One way to include frictional forces in analyzing the motion of an object sliding
on a horizontal surface is to describe the kinetic energy lost because of friction.
Suppose a book moving on a horizontal surface is given an initial horizontal velocity
vi and slides a distance d before reaching a final velocity vf as shown in Figure.
46
Situations Involving Kinetic Friction, cont.,
The external force that causes the book to undergo an acceleration in the
negative x direction is the force of kinetic friction fk acting to the left, opposite the
motion. initial kinetic energy of the book is ½ m v i 2 and its final kinetic energy is ½
m v f 2 . N e w t o n ’ s s e c o n d l a w g i v e s - fk =m a x .
When friction — as well as other forces — acts on an object, the work – kinetic
energy theorem reads
47
Example 7.7
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.
48
Solution
Because there is no friction, the net external force acting on the block is the
12-N force. The work done by this force is;
Using the work – kinetic energy theorem and noting that the initial kinetic energy
is zero, we obtain;
49
Example 7.8
A 6.0-kg block initially at rest is pulled to the right along a horizontal, not
frictionless surface by a constant horizontal force of 12 N and has a coefficient of
kinetic friction of 0.15. Find the speed of the block after it has moved 3.0 m.
50
Solution
The applied force does work;
The final speed of the block is;
51
Power
Power is the time rate of energy transfer.
If an external force is applied to an object (which we assume acts as a particle),
and if the work done by this force in the time interval Dt is W, then the average
power expended during this interval is defined as;
Pavg
W

Dt
The instantaneous power is defined as; The work done on the object
contributes to the increase in the energy of the object. Therefore, a more general
definition of power is the time rate of energy transfer.
dE
P
dt
52
Section 8.5
Instantaneous Power and Average Power
The instantaneous power is the limiting value of the average power as Dt
approaches zero.
P
lim
Dt 0
W dW
dr

 F
 F v
Dt
dt
dt
This expression for power is valid for any means of energy transfer.
53
Section 8.5
Units of Power
The SI unit of power is called the watt.
 1 watt = 1 joule / second = 1 kg . m2 / s3
A unit of power in the US Customary system is horsepower.
 1 hp = 746 W
Units of power can also be used to express units of work or energy.
 1 kWh = (1000 W)(3600 s) = 3.6 x106 J
54
Section 8.5
Example 7.12
An elevator car has a mass of 1000 kg and is carrying
passengers having a combined mass of 800 kg. A constant
frictional force of 4000 N retards its motion upward, as shown
in Figure.
(a) What must be the minimum power delivered by the motor
to lift the elevator car at a constant speed of 3.00 m/s?
(b) What power must the motor deliver at the instant its speed
is v if it is designed to provide an upward acceleration of 1.00
m/s2?
55
a)
a=0, and therefore we know from Newton’s second law that S Fy =0.
T is in the same direction as v, we find that;
56
b)
The only change in the setup of the problem is that now a>0. Applying Newton’s
second law to the car gives;
57