Energia Cinética e Trabalho

Prévia do material em texto

ENERGIA CINÉTICA E 
TRABALHO
Chapter 7
Energy and Energy Transfer
! On a wind farm, the moving air does work on the blades of the windmills, causing the
blades and the rotor of an electrical generator to rotate. Energy is transferred out of the sys-
tem of the windmill by means of electricity. (Billy Hustace/Getty Images)
CHAPTE R OUTL I N E
7.1 Systems and Environments
7.2 Work Done by a Constant
Force
7.3 The Scalar Product of Two
Vectors
7.4 Work Done by a Varying
Force
7.5 Kinetic Energy and the 
Work–Kinetic Energy
Theorem
7.6 The Nonisolated System—
Conservation of Energy
7.7 Situations Involving Kinetic
Friction
7.8 Power
7.9 Energy and the Automobile
181
segunda-feira, 13 de maio de 2013
segunda-feira, 13 de maio de 2013
* CONTEÚDO DA AULA
segunda-feira, 13 de maio de 2013
ENERGIA CINÉTICA
TRABALHO
TRABALHO E ENERGIA CINÉTICA
TRABALHO DE UMA FORÇA CONSTANTE
TRABALHO DE UMA FORÇA VARIÁVEL
POTÊNCIA
* CONTEÚDO DA AULA
segunda-feira, 13 de maio de 2013
segunda-feira, 13 de maio de 2013
* ENERGIA CINÉTICA
segunda-feira, 13 de maio de 2013
* ENERGIA CINÉTICA
- A ENERGIA CINÉTICA K É A ENERGIA ASSOCIADA AO ESTADO DE 
MOVIMENTO DE UM OBJETO. 
segunda-feira, 13 de maio de 2013
* ENERGIA CINÉTICA
- A ENERGIA CINÉTICA K É A ENERGIA ASSOCIADA AO ESTADO DE 
MOVIMENTO DE UM OBJETO. 
PARA UM OBJETO DE MASSA CUJA VELOCIDADE V É MUITO MENOR 
QUE A VELOCIDADE DA LUZ, A ENERGIA CINÉTICA É DADO POR:
m
k =
1
2
mv2.
segunda-feira, 13 de maio de 2013
* ENERGIA CINÉTICA
- A ENERGIA CINÉTICA K É A ENERGIA ASSOCIADA AO ESTADO DE 
MOVIMENTO DE UM OBJETO. 
PARA UM OBJETO DE MASSA CUJA VELOCIDADE V É MUITO MENOR 
QUE A VELOCIDADE DA LUZ, A ENERGIA CINÉTICA É DADO POR:
m
k =
1
2
mv2.
A UNIDADE DE ENERGIA CINÉTICA NO SI É O JOULE( J ). 
7-3 | Kinetic Energy
One of the fundamental goals of physics is to investi gate something that every-
one talks about: energy. The topic is obviously important. Indeed, our civrhzation
is based on acquiritrg and effectively using energy.
For example, everyone knows that any type of motion requires energy:
Flying across the Pacific Ocean requires it. Lifting material to the top floor of an
office building or to an orbiting space station requires it. Throwing a fastball
requires it. We spend a tremendous amount of money to acquire and use energy.
Wars have been started because of energy resources. Wars have been ended
because of a sudden, overpowering use of energy by one side. Everyone knows
many examples of energy and its use, but what does the term energy really mean?
7-Z $ hat Is Energy?
The term energy is so broad that a clear deflnition is difflcult to write. Technically,
energy is a scalar quantity associated with the state (or condition) of one or more
objects. However, this deflnition is too vague to be of help to us now.
A looser definition might at least get us started. Energy is a number that we
associate with a system of one or more objects. If a force changes one of the
objects by, say, making it move, then the energy number changes.After countless
experiments, scientists and engineers reahzed that if the scheme by which we
assign energy numbers is planned carefully, the numbers can be used to predict
the outcomes of experiments and, even more important, to build machines, such
as flying machines. This success is based on a wonderful property of our universe:
Energy can be transformed from one type to another and transferred from one
object to another, but the total amount is always the same (energy is conserved).
No exception to this principle of energy conservation has ever been found.
Think of the many types of energy as being numbers representing money in
many types of bank accounts. Rules have been made about what such money
numbers mean and how they can be changed. You can transfer money numbers
from one account to another or from one system to another, perhaps electroni-
cally with nothing material actually movirg.However, the total amount (the total
of all the money numbers) can always be accounted for: It is always conserved.
In this chapter we focus on only one type of energy (kinetic energy) and on
only one way in which energy can be transferred (work). In the next chapter we
examine a few other types of energy and how the principle of energy conserva-
tion can be written as equations to be solved.
7-3&Klnetic Energy
Kinetic energy K is energy associated with the state of motion of an object. The
faster the object moves, the greater is its kinetic energy. When the object is
station zty,its kinetic energy is zero.
For an object of mass m whose speed v is well below the speed of light,
K - **r' (kinetic energy). (7 -r)
For example, a 3.0 kg duck flying past us at 2.0 m/s has a kinetic energy of
6.0 kg 'm2lsz;that is, we associate that number with the duck's motion.
The SI unit of kinetic energy (and every other type of energy) is the joule (J),
named for James Prescott Joule, an English scientist of the 1800s. It is defined
directly from Eq. 7 -I tn terms of the units for mass and velocity:
l joule - L J: 1kg-m2ls2
Thus, the flying duck has a kinetic energy of 6.0 J.
(7 -2)
segunda-feira, 13 de maio de 2013
* ENERGIA CINÉTICA
- A ENERGIA CINÉTICA K É A ENERGIA ASSOCIADA AO ESTADO DE 
MOVIMENTO DE UM OBJETO. 
PARA UM OBJETO DE MASSA CUJA VELOCIDADE V É MUITO MENOR 
QUE A VELOCIDADE DA LUZ, A ENERGIA CINÉTICA É DADO POR:
m
k =
1
2
mv2.
A UNIDADE DE ENERGIA CINÉTICA NO SI É O JOULE( J ). 
7-3 | Kinetic Energy
One of the fundamental goals of physics is to investi gate something that every-
one talks about: energy. The topic is obviously important. Indeed, our civrhzation
is based on acquiritrg and effectively using energy.
For example, everyone knows that any type of motion requires energy:
Flying across the Pacific Ocean requires it. Lifting material to the top floor of an
office building or to an orbiting space station requires it. Throwing a fastball
requires it. We spend a tremendous amount of money to acquire and use energy.
Wars have been started because of energy resources. Wars have been ended
because of a sudden, overpowering use of energy by one side. Everyone knows
many examples of energy and its use, but what does the term energy really mean?
7-Z $ hat Is Energy?
The term energy is so broad that a clear deflnition is difflcult to write. Technically,
energy is a scalar quantity associated with the state (or condition) of one or more
objects. However, this deflnition is too vague to be of help to us now.
A looser definition might at least get us started. Energy is a number that we
associate with a system of one or more objects. If a force changes one of the
objects by, say, making it move, then the energy number changes.After countless
experiments, scientists and engineers reahzed that if the scheme by which we
assign energy numbers is planned carefully, the numbers can be used to predict
the outcomes of experiments and, even more important, to build machines, such
as flying machines. This success is based on a wonderful property of our universe:
Energy can be transformed from one type to another and transferred from one
object to another, but the total amount is always the same (energy is conserved).
No exception to this principle of energy conservation has ever been found.
Think of the many types of energy as being numbers representing money in
many types of bank accounts. Rules have been made about what such money
numbers mean and how they can be changed. You can transfer money numbers
from one account to another or from one system to another, perhaps electroni-
cally with nothing material actually movirg.However, the total amount (the total
of all the money numbers) can always be accounted for: It is alwaysconserved.
In this chapter we focus on only one type of energy (kinetic energy) and on
only one way in which energy can be transferred (work). In the next chapter we
examine a few other types of energy and how the principle of energy conserva-
tion can be written as equations to be solved.
7-3&Klnetic Energy
Kinetic energy K is energy associated with the state of motion of an object. The
faster the object moves, the greater is its kinetic energy. When the object is
station zty,its kinetic energy is zero.
For an object of mass m whose speed v is well below the speed of light,
K - **r' (kinetic energy). (7 -r)
For example, a 3.0 kg duck flying past us at 2.0 m/s has a kinetic energy of
6.0 kg 'm2lsz;that is, we associate that number with the duck's motion.
The SI unit of kinetic energy (and every other type of energy) is the joule (J),
named for James Prescott Joule, an English scientist of the 1800s. It is defined
directly from Eq. 7 -I tn terms of the units for mass and velocity:
l joule - L J: 1kg-m2ls2
Thus, the flying duck has a kinetic energy of 6.0 J.
(7 -2)
EXEMPLIFICAR !
segunda-feira, 13 de maio de 2013
segunda-feira, 13 de maio de 2013
* TRABALHO
segunda-feira, 13 de maio de 2013
* TRABALHO
- TRABALHO(W) É A ENERGIA TRANSFERIDA PARA UM OBJETO OU 
DE UM OBJETO ATRAVÉS DE UMA FORÇA QUE AGE SOBRE O OBJETO. 
QUANDO A ENERGIA É TRANSFERIDA PARA O OBJETO, O TRABALHO 
É POSITIVO; QUANDO A ENERGIA ASSOCIADA É TRANSFERIDA DO 
OBJETO, O TRABALHO É NEGATIVO.
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
segunda-feira, 13 de maio de 2013
* TRABALHO
- TRABALHO(W) É A ENERGIA TRANSFERIDA PARA UM OBJETO OU 
DE UM OBJETO ATRAVÉS DE UMA FORÇA QUE AGE SOBRE O OBJETO. 
QUANDO A ENERGIA É TRANSFERIDA PARA O OBJETO, O TRABALHO 
É POSITIVO; QUANDO A ENERGIA ASSOCIADA É TRANSFERIDA DO 
OBJETO, O TRABALHO É NEGATIVO.
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
(PARA UMA FORÇA 
CONSTANTE). 
segunda-feira, 13 de maio de 2013
A UNIDADE DE TRABALHO NO SI É O JOULE. 
* TRABALHO
- TRABALHO(W) É A ENERGIA TRANSFERIDA PARA UM OBJETO OU 
DE UM OBJETO ATRAVÉS DE UMA FORÇA QUE AGE SOBRE O OBJETO. 
QUANDO A ENERGIA É TRANSFERIDA PARA O OBJETO, O TRABALHO 
É POSITIVO; QUANDO A ENERGIA ASSOCIADA É TRANSFERIDA DO 
OBJETO, O TRABALHO É NEGATIVO.
#hmpten 7 | Kinetic Energy and Work
which says that
We can also
of it must move together, in the same direction. In this chapter we consider onlyparticle-like objects, such as the bed and its occupant being pushed in Fig. 7 -3.
Signs fo, work. The work done on an object by a force-can be either positive
work or negative work. For example, if the angle Q inEq. 7 -7 rsless than 90", then cos
d is positive and thus so is the work. It Ois greater than 90" (.rp to 180"), then cos d isnegative and thus so is the work. (Can you see that the work is zero when O -- 90.?)These results lead to a simple rule. To flnd the sign of the work done by a force, con-
sider the force vector component that is parallel to the displacement:
opposite direction. It does zero work when it has no such vector component.
Units fo, work. Work has the SI unit of the joule, the same as kinetic energy.
However, from Eqs. 7 -6 and 7 -7 we can see that an equivalent unit is the newton-
meter (N'm). The corresponding unit in the British system is the foot-pound
(ft . lb). Extending Eq. i -Z,we have
1J : 1kg.m2ls2 - 1N.rrr - 0.738 ft.lb.
Equation 7 -5 relates the change in kinetic energy of the bead (from an initial[(' - ]*r'o to a later Kf : ]*r') to the work W7-- F"d) done on the bead. For
such particle-like objects, we can generahze that equation. Let AK be the change
in the kinetic energy of the object, and IetW be the net work done on it.Then
LK:Kf-K-W,
(7 -e)
(7-10)
(7 -11)
(change in the kinetic\
\ energy of a parti cle )
write
Kf : Kir W,
- ( kinetic energy \ r ( rhe net \\before the net work/ ' \work done/'
( kinetic energy after \
\ttre net work is done/
(net work done on\
\ the particle )
Ffiffi" P-# A contestant in a bed race.
We can approximate the bed and its
occupant as being a particle for the
purpose of calculating the work done
on them by the force applied by the
student.
which says that
These statements are known traditionally as the work - kinetic energy theorernfor particles.Th"y hold for both positive and negative work: If the net work done
on a particle is positive, then the particle's kinetic energy increases by the amount
of the work. If the net work done is negative, then the particle's kinetic energy
decreases by the amount of the work.
For example, if the kinetic energy of a particle is initially 5 J and there is a
net transfer of 2 J to the particle (positive net work), the final kinetic energy is
7 J-If-,instead, there is a net transfer of 2J from the particle (negative net work),
the flnal kinetic energy is 3 J.
(a) from -3 m/s to - 2 mls and (b) from - 2 mls to 2 m/s? (c) In each situation, is the
work done on the particle positive, negative, or zero?
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energygauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
(PARA UMA FORÇA 
CONSTANTE). 
segunda-feira, 13 de maio de 2013
A UNIDADE DE TRABALHO NO SI É O JOULE. 
* TRABALHO
- TRABALHO(W) É A ENERGIA TRANSFERIDA PARA UM OBJETO OU 
DE UM OBJETO ATRAVÉS DE UMA FORÇA QUE AGE SOBRE O OBJETO. 
QUANDO A ENERGIA É TRANSFERIDA PARA O OBJETO, O TRABALHO 
É POSITIVO; QUANDO A ENERGIA ASSOCIADA É TRANSFERIDA DO 
OBJETO, O TRABALHO É NEGATIVO.
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and theforce F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
#hmpten 7 | Kinetic Energy and Work
which says that
We can also
of it must move together, in the same direction. In this chapter we consider onlyparticle-like objects, such as the bed and its occupant being pushed in Fig. 7 -3.
Signs fo, work. The work done on an object by a force-can be either positive
work or negative work. For example, if the angle Q inEq. 7 -7 rsless than 90", then cos
d is positive and thus so is the work. It Ois greater than 90" (.rp to 180"), then cos d isnegative and thus so is the work. (Can you see that the work is zero when O -- 90.?)These results lead to a simple rule. To flnd the sign of the work done by a force, con-
sider the force vector component that is parallel to the displacement:
opposite direction. It does zero work when it has no such vector component.
Units fo, work. Work has the SI unit of the joule, the same as kinetic energy.
However, from Eqs. 7 -6 and 7 -7 we can see that an equivalent unit is the newton-
meter (N'm). The corresponding unit in the British system is the foot-pound
(ft . lb). Extending Eq. i -Z,we have
1J : 1kg.m2ls2 - 1N.rrr - 0.738 ft.lb.
Equation 7 -5 relates the change in kinetic energy of the bead (from an initial[(' - ]*r'o to a later Kf : ]*r') to the work W7-- F"d) done on the bead. For
such particle-like objects, we can generahze that equation. Let AK be the change
in the kinetic energy of the object, and IetW be the net work done on it.Then
LK:Kf-K-W,
(7 -e)
(7-10)
(7 -11)
(change in the kinetic\
\ energy of a parti cle )
write
Kf : Kir W,
- ( kinetic energy \ r ( rhe net \\before the net work/ ' \work done/'
( kinetic energy after \
\ttre net work is done/
(net work done on\
\ the particle )
Ffiffi" P-# A contestant in a bed race.
We can approximate the bed and its
occupant as being a particle for the
purpose of calculating the work done
on them by the force applied by the
student.
which says that
These statements are known traditionally as the work - kinetic energy theorernfor particles.Th"y hold for both positive and negative work: If the net work done
on a particle is positive, then the particle's kinetic energy increases by the amount
of the work. If the net work done is negative, then the particle's kinetic energy
decreases by the amount of the work.
For example, if the kinetic energy of a particle is initially 5 J and there is a
net transfer of 2 J to the particle (positive net work), the final kinetic energy is
7 J-If-,instead, there is a net transfer of 2J from the particle (negative net work),
the flnal kinetic energy is 3 J.
(a) from -3 m/s to - 2 mls and (b) from - 2 mls to 2 m/s? (c) In each situation, is the
work done on the particle positive, negative, or zero?
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angleQ to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
(PARA UMA FORÇA 
CONSTANTE). 
segunda-feira, 13 de maio de 2013
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
DA SEGUNDA LEI DE NEWTON, TEMOS:
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as Fcos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
DA SEGUNDA LEI DE NEWTON, TEMOS:
POR OUTRO LADO, DA EQUAÇÃO DE TORRICELLI
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
africtionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
DA SEGUNDA LEI DE NEWTON, TEMOS:
POR OUTRO LADO, DA EQUAÇÃO DE TORRICELLI
UTILIZANDO A 2ª LEI DE NEWTON NA EQUAÇÃO DE TORRICELLI, 
PODEMOS CHEGAR A:
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement.Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the objectmust be particle-like.This means that the object must be rigid; allparts
DA SEGUNDA LEI DE NEWTON, TEMOS:
POR OUTRO LADO, DA EQUAÇÃO DE TORRICELLI
UTILIZANDO A 2ª LEI DE NEWTON NA EQUAÇÃO DE TORRICELLI, 
PODEMOS CHEGAR A:
TEOREMA DO TRABALHO E ENERGIA CINÉTICA
segunda-feira, 13 de maio de 2013
* TRABALHO E ENERGIA CINÉTICA
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along the wire, changing the
velocity of the bead from i,toi.
A "kinetic energy gauge" indicates
the resulting change in the kinetic
energy of the bead, from the value K,
to the value K1.
Solving this equation for A", substituting into Eq. 7 -3, and reaffanging then give us
(7 -s)
The first term on the left side of the equation is the kinetic energy Ky of the bead
at the end of the displacement d,and the second term is the kinetic energy Ktof
the bead at the start of the displacement. Thus, the left side of Eq. 7 -5 tells us
the kinetic energy has been changed by the force, and the right side tells us the
change is equal to F*d.Therefore, the work W' done on the bead by the force
(the energy transfer due to the force) is
w - F*d.
If we know values for F* and d,we can use this equation to calculate the work W
done on the bead by the force.
To calculate the work a force he object moves through some
placernento we use only the fo the object's displacement. The
force component perpendicular to the displacement does zero work.
From Fig. 7-2, we see that we can write F, as F cos @,rvhere d is the angle
between the directions of the displacement d and the force F.Thus,
W - Fd cos $ (work done by a constant force).
(7 -3)
(7 -4)
(7 -6)
(7 -7)
(7-8)
Because the right side of this equation is equivalent to the scalar (dot) product
F .A,we can also write
W : F' d (work donebyaconstantforce),
where F is the magnitude of F. (You may wish to review the discussion of scalar
products in Section 3-8.) Equation 7-B is especially useful for calculating the work
when F and d are given in unit-vector notation.
Cautions; There are two restrictions to using Eqs. 7-6 through 7-B to calculate
work done on an object by a force. First, the force must be a constant force; that
is, it must not change in magnitude or direction as the object moves. (Later, we
shall discuss what to do with a variable force that changes in magnitude.) Secotrd,
the object must be particle-like.This means that the object must be rigid; allparts
I{owever, such effort does not cause an energy transfer to or from the wall and
thus is not work done on the wall as defined here.
To avoid confusion in this chapter, we shall use the symbol W only for work
and shall represent a weight with its equivalent mg.
# '* rE< and Klnetic Energy
Let us find an expression for work by considering a bead that can slide along
a frictionless wire that is stretched along a horrzontal x axis (Fig. I -2). A constant
force F, dit"cted at an angle Q to the wire, accelerates the bead along the wire.
We can relate the force and the acceleration with Newton's second laW written
for components along the x axis:
F*: max)
where m is the bead's mass. As the bead moves through a displacement d, the
force changes the bead's velocity from an initial value 7s to some other value 7.
Because the force is constant, we know that the acceleration is also constant.
Thus, we can use Eq. 2-16 to write, for components along the x axis,
v2: vfi + 2a*d.
7-5 I Work and Kinetic Energy
x
F$ffi" ?-ff A constant force F Oi-
rected at angle Q to the displacement
d of a bead on a wire accelerates the
bead along