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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
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* 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 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)
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 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.
#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 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 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
#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 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
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 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
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 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 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:
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
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:
TEOREMA DO TRABALHO E ENERGIA CINÉTICA
∆K = Kf −Ki =W
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 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:
TEOREMA DO TRABALHO E ENERGIA CINÉTICA
Kf = Ki +W.∆K = Kf −Ki =W
segunda-feira, 13 de maio de 2013
segunda-feira, 13 de maio de 2013
PARA CALCULAR O TRABALHO QUE UMA FORÇA REALIZADA SOBRE
UM OBJETO QUANDO ESTE SOFRE UM DESLOCAMENTO, USAMOS
APENAS A COMPONENTE DA FORÇA EM RELAÇÃO AO DESLOCAMENTO
DO OBJETO. A COMPONENTE DA FORÇA PERPENDICULAR AO
DESLOCAMENTO NÃO REALIZA TRABALHO.
segunda-feira, 13 de maio de 2013
UMA FORÇA REALIZA TRABALHO POSITIVO SE POSSUI UMA
COMPONENTE VETORIAL NO MESMO SENTIDO DO DESLOCAMENTO.
PARA CALCULAR O TRABALHO QUE UMA FORÇA REALIZADA SOBRE
UM OBJETO QUANDO ESTE SOFRE UM DESLOCAMENTO, USAMOS
APENAS A COMPONENTE DA FORÇA EM RELAÇÃO AO DESLOCAMENTO
DO OBJETO. A COMPONENTE DA FORÇA PERPENDICULAR AO
DESLOCAMENTO NÃO REALIZA TRABALHO.
segunda-feira, 13 de maio de 2013
UMA FORÇA REALIZA TRABALHO POSITIVO SE POSSUI UMA
COMPONENTE VETORIAL NO MESMO SENTIDO DO DESLOCAMENTO.
UMA FORÇA REALIZA UM TRABALHO NEGATIVO QUANDO POSSUI
UMA COMPONENTE VETORIAL NO SENTIDO OPOSTO.
PARA CALCULAR O TRABALHO QUE UMA FORÇA REALIZADA SOBRE
UM OBJETO QUANDO ESTE SOFRE UM DESLOCAMENTO, USAMOS
APENAS A COMPONENTE DA FORÇA EM RELAÇÃO AO DESLOCAMENTO
DO OBJETO. A COMPONENTE DA FORÇA PERPENDICULAR AO
DESLOCAMENTO NÃO REALIZA TRABALHO.
segunda-feira, 13 de maio de 2013
UMA FORÇA REALIZA TRABALHO POSITIVO SE POSSUI UMA
COMPONENTE VETORIAL NO MESMO SENTIDO DO DESLOCAMENTO.
UMA FORÇA REALIZA UM TRABALHO NEGATIVO QUANDO POSSUI
UMA COMPONENTE VETORIAL NO SENTIDO OPOSTO.
A FORÇA REALIZA UM TRABALHO NULO QUANDO NÃO POSSUI
UMA COMPONENTE E VETORIAL NA DIREÇÃO DO DESLOCAMENTO.
PARA CALCULAR O TRABALHO QUE UMA FORÇA REALIZADA SOBRE
UM OBJETO QUANDO ESTE SOFRE UM DESLOCAMENTO, USAMOS
APENAS A COMPONENTE DA FORÇA EM RELAÇÃO AO DESLOCAMENTO
DO OBJETO. A COMPONENTE DA FORÇA PERPENDICULAR AO
DESLOCAMENTO NÃO REALIZA TRABALHO.
segunda-feira, 13 de maio de 2013
EXEMPLO:
DURANTE UMA TEMPESTADE, UMA CAIXA DESLIZA POR UM PISO
E S C O R R E G A D I O D E E S T A C I O N A M E N T O , S O F R E N D O U M
DESLOCAMENTO AO SER EMPURRADO PELO VENTO COM
UMA FORÇA A SITUAÇÃO E OS EIXOS DE COOR-
DENADAS ESTÃO REPRESENTADOS NA FIGURA ABAIXO.
�d = (−3, 0m)ˆi
�F = (2, 0N )ˆi+ (−6, 0N)jˆ.
(A)QUAL É O TRABALHO REALIZADO PELO VENTO SOBRE O CAIXOTE?
(B)SE O CAIXOTE TEM UMA ENERGIA CINÉTICA DE 10J NO INÍCIO DO
DESLOCAMENTO, QUAL É A SUA ENERGIA AO FINAL DO
DESLOCAMENTO?
7-5 I Work and Kinetic Energy
Figure 7-4a shows two industrial spies sliding_,an initially
stationary 225 kg floor safe a displacement d of mlgni-
tude 8.50 m, straight toward their truck. The push F1 of
spy 001 is I2.0 \ directed at an angle of 30.0" down-
ward from the horizontal; the pull F2 of spy 002 is
10.0 N, directed at 40.0" above the horizontal. The mag-
nitudes and directions of these forces do not change as
the safe moves, and the floor and safe make frictionless
contact.
(u) What is the net work done on the safe by for""t F,
and F, dutitlg the displac ementd?
(1) The net work W'done on the safe by the
two forces is the sum of the works they do individually.
(2) Because we can treat the safe as a particle and the
forces are constant in both magnitude and directior,
we can use either Eq. 7 -7 (W : Fd cos il or Eq. 7 -8
(W : F.i) to calculale those works. Since we know the
magnitudes and directions of the forces, we choose
Eq.7-7.
Calculafions; From Eq. 7 -7 and the free-body diagram
for the safe in Fig. 7 -4bJtr. work done Uy F, is
Wt: Frd cos Qt: Q2.0 N)(8.50 m)(cos 30.0'): 88.33 J,
and the work done by F, is
Wz: Frd cos
Qz: (10.0 NX8.50 m)(cos 40.0')
: 65.11 J.
F2
40.0"
(a) (b)
F$G" ?$ (a) Two spies move a floor safe through a displace-
ment d.(b) A free-body diagram for the safe.
Thus, the net work W'is
W - Wr+ Wz : 88.33 J + 65.11 J: 153.4 J : 1.53 J. (Answer)
During the 8.50 m displacement, therefore, the spies
transfer 153 J of energy to the kinetic energy of the safe.
(b) During the displacement, what is the work W, done
on the safe by the gravitational force F, and *hel is the
work W N done on the safe by the normal force F1,. from
the floor?
Because these forces are constant in both
magnitude and directior, we can find the work they do
with Eq. 7 -7 .
Calcula ns; Thus, with mg as the magnitude of the
gravitational force, we write
Wr: mgd cos 90" - mgd(O) - 0 (Answer)
WN : Frvd cos 90" : FNd(0) : 0. (Answer)
We should have known this result. Because these forces
are perpendicular to the displacement of the safe, they
do zero work on the safe and do not transfer any energy
to or from it.
(r) The safe is initially stationary.What is its speed vyat
the end of the 8.50 m displacement?
and
speed of
changed
the safe changes because its
when energy is transferred
combining Eqs. 7 -I0 and 7 -1.:
w-Kf-Ki -;mv? - +mv?.
rN
-->F,
b
d
The initial speed v; is zero, and
work done is 153 .4 J. Solving for
known data,we find that
we now know that the
v7 and then substituting
: 1,.I7 m/s. (Answer)
Spy 002
30.0"
2(rs3.4 J)
225 kg
During a storffi, a qate of crepe is sliding_,,across a slick,
oily parking lot through a displacement d _ (-3.0 m)i
dinate axes are shown in Fig. 7 -5.
(u) How much work does this force do on the crate
during the displacement? F{ffi. ?-S Force F slows a crate during displa cementi.
segunda-feira, 13 de maio de 2013
TRABALHO REALIZADO PELA FORÇA GRAVITACIONAL
PELO QUE JÁ FOI VISTO ATÉ AQUI, TEMOS:
TRABALHO REALIZADO PELA FORÇA GRAVITACIONAL
Chapter 7 I Kinetic Energy and Work
Because we can treat the crate as a particle
and because the wind force is constant ("steady") in
both magnitude and direction during the displacement,
we can use either Eq. 7 -7 (W : Fd cos il or Eq. 7 -8
(W -_, F'd.) to calculate the work. Since we know F
and d tn unit-vector notation, we choose Eq. 7 -8.
Cafculatiens; We write
w : F-d : lQ.lN)i + (-6.0 N)il .[(-3.0 m)i].
Of the possible unit-vector dot products, only i'i, j 'i,
and k.k are nonzero (see Appendix E). Here we obtain
W : (2.0 NX-3.0 -)i.i + (-6.0 N)(-3.0 -)j'i
Tlhus, the force does a negative 6.0 J of work on the
erate, transferring 6.0 J of energy trom the kinetic
energy of the crate.
(b) If the crate has a kinetic energy of 10 J at the
beginnitrg ofdisplaceme nt i, what is its kinetic energy
at the end of d?
Because the force does negative work on
the crate,it reduces the crate's kinetic energy.
Cafcu$atfon: [Jsing the work-kinetic energy theorem
in the form of Eq. 7 -II, we have
Kf : Ki t W -- 10 J + (-6.0 J) : 4.0 J. (Answer)
Less kinetic energy means that the crate has been slowed.
7* x rk Done by the Gravitational Force
We next examine the work done on an object by the gravitational force acting on
it. Figure 7-6 shows a pafiicle-like tomato of mass m that is thrown upward with
initial speed v0 and thus with initial kinetic energy Ki - tmvfi. As the tomato
rises, it is slowed b11u gravitational force Fr; that is, the tomato's kinetic energy
decreases because F, does work on the tomato as it rises. Because we can tteat
the tomato as a particle, we can use Eq. 7 -7 (W : Fd cos Q) to express the work
done during a diiplacementd.For the force magnitude F,we use mgas the mag-
nitude of 4.Thus, the work W, done by the gravitational force 4 it
W r : mgd cos @ (work done by gravitational force). (7 -I2)
For a rising obj ect,force 4 ir directed opposite the displacement 7,ut indi-
cated in Fig. 7 -6.Thus, 0 - 180' and
Wr: mgd cos 180" - mgd(-1) - -mgd.
The minus sign tells us that during the object's rise, the gravitational force acting
on the obj ect transfers energy in the amount mgd from the kinetic energy of the
object.This is consistent with the slowing of the object as it rises.
After the object has reached its maximum_,height and is falling back down,
the angle ,f between force i and displacement d ,t zero.Thus,
Wr: mgd cos 0" - mgd(+1)
The plus sign tells us that the gravttational force now transfers energy in the amount
mgd to the kinetic energy of the object. This is consistent with the speeding up of the
object as it falls. (Actually, as we shall see in Chapter B, energy transfers associated
with lifting and lowering an object involve the full object-Earth system.)
Now suppose we lift a particle-like object by applying a vertical force F to it.
During the upward displacement, our applied force does positive work Woon the
object while the gravitational force does negative work Ws on it. Our applied
force tends to transfer energy to the object while the gravitational force tends to
transfer energy from it. By Eq.7-I0, the change LK in the kinetic energy of the
object due to these two energy transfers is
(7 -13)
(7 -r4)
F$ffi. 7-6 Because the gravitational
force { acts on it, a partrcle-like
tomato of mass m thrown upward
slows from velocity Tgto velocity 7
during displacement d. A kinetic
energy gauge indicates the resulting
change in the kinetic energy of the
tomato, from K,(: )mvfl) to
Kr G j.*r'). AK:Kf-K-Wo*W* (7-1s)
Chapter 7 I Kinetic Energy and Work
Because we can treat the crate as a particle
and because the wind force is constant ("steady") in
both magnitude and direction during the displacement,
we can use either Eq. 7 -7 (W : Fd cos il or Eq. 7 -8
(W -_, F'd.) to calculate the work. Since we know F
and d tn unit-vector notation, we choose Eq. 7 -8.
Cafculatiens; We write
w : F-d : lQ.lN)i + (-6.0 N)il .[(-3.0 m)i].
Of the possible unit-vector dot products, only i'i, j 'i,
and k.k are nonzero (see Appendix E). Here we obtain
W : (2.0 NX-3.0 -)i.i + (-6.0 N)(-3.0 -)j'i
Tlhus, the force does a negative 6.0 J of work on the
erate, transferring 6.0 J of energy trom the kinetic
energy of the crate.
(b) If the crate has a kinetic energy of 10 J at the
beginnitrg ofdisplaceme nt i, what is its kinetic energy
at the end of d?
Because the force does negative work on
the crate,it reduces the crate's kinetic energy.
Cafcu$atfon: [Jsing the work-kinetic energy theorem
in the form of Eq. 7 -II, we have
Kf : Ki t W -- 10 J + (-6.0 J) : 4.0 J. (Answer)
Less kinetic energy means that the crate has been slowed.
7* x rk Done by the Gravitational Force
We next examine the work done on an object by the gravitational force acting on
it. Figure 7-6 shows a pafiicle-like tomato of mass m that is thrown upward with
initial speed v0 and thus with initial kinetic energy Ki - tmvfi. As the tomato
rises, it is slowed b11u gravitational force Fr; that is, the tomato's kinetic energy
decreases because F, does work on the tomato as it rises. Because we can tteat
the tomato as a particle, we can use Eq. 7 -7 (W : Fd cos Q) to express the work
done during a diiplacementd.For the force magnitude F,we use mgas the mag-
nitude of 4.Thus, the work W, done by the gravitational force 4 it
W r : mgd cos @ (work done by gravitational force). (7 -I2)
For a rising obj ect,force 4 ir directed opposite the displacement 7,ut indi-
cated in Fig. 7 -6.Thus, 0 - 180' and
Wr: mgd cos 180" - mgd(-1) - -mgd.
The minus sign tells us that during the object's rise, the gravitational force acting
on the obj ect transfers energy in the amount mgd from the kinetic energy of the
object.This is consistent with the slowing of the object as it rises.
After the object has reached its maximum_,height and is falling back down,
the angle ,f between force i and displacement d ,t zero.Thus,
Wr: mgd cos 0" - mgd(+1)
The plus
sign tells us that the gravttational force now transfers energy in the amount
mgd to the kinetic energy of the object. This is consistent with the speeding up of the
object as it falls. (Actually, as we shall see in Chapter B, energy transfers associated
with lifting and lowering an object involve the full object-Earth system.)
Now suppose we lift a particle-like object by applying a vertical force F to it.
During the upward displacement, our applied force does positive work Woon the
object while the gravitational force does negative work Ws on it. Our applied
force tends to transfer energy to the object while the gravitational force tends to
transfer energy from it. By Eq.7-I0, the change LK in the kinetic energy of the
object due to these two energy transfers is
(7 -13)
(7 -r4)
F$ffi. 7-6 Because the gravitational
force { acts on it, a partrcle-like
tomato of mass m thrown upward
slows from velocity Tgto velocity 7
during displacement d. A kinetic
energy gauge indicates the resulting
change in the kinetic energy of the
tomato, from K,(: )mvfl) to
Kr G j.*r'). AK:Kf-K-Wo*W* (7-1s)
DE MANEIRA QUE PODEMOS TER DUAS POSSIBILIDADES PARA O
RESULTADO DA EQUAÇÃO ACIMA: POSITIVA, SE AS GRANDEZAS FOREM
PARALELAS OU ANTI-PARALELAS!!!
Wg = mgd
φ = 00 Cosφ = 1
Wg = −mgd
φ = 1800 Cosφ = −1
segunda-feira, 13 de maio de 2013
A FORÇA ELÁSTICA É DADA POR:
X:0
4=0
Chapter 7 I Kinetic Energy and Work
Block
attached
to spnng
0
(")
force acts to restore the relaxed state, it is sometimes said to be a restoring force.)If we compress the spring by pushing the block to the left as in Fig. 7 -l.Ic, the
spring now pushes on the block toward the right.
___>
To a good approximation for many springs, the force F, from a spring is pro-
portional to the displacemen td ofthe free end from its position when tG rpiitrg
is in the relaxed state. The spring force is given by
F, (7 -20)
X
(r)
FBffi. ?-{ ,$ (o) A spring in its relaxed
state. The origin of an x axis has been
placed at the end of the spring that is
attached to atlock .(b) The block is
displaced by d, and the spring is
stretched by a positive amount x.
Note the restoring force { exerted
by the spring. (c) The spring is com-
pressed by a negative amounl x.
Again, note the restoring force.
which is known as Hooke's law after Robert Hooke, an English scientist of the
late 1600s. The minus sign in Eq. 7 -20 indicates that the direction of the spring
force is always opposite the direction of the displacement of the spring's free end.
The constant k is called the spring constant (or force constant) and is a measure
of the stiffness of the spring. The larger k is, the stiffer the spring; that is, the larger
k is, the stronger the spring's pull or push for a given displacement. The SI unit for
k is the newton per meter.
In Fig. 7 -II an x axis has been placed parallel to the length of the spring, with
the origin (* - 0) at the position of the free end when the spring is in its relaxed
state. For this common arrangement, we can write Eq. 7 -20 as
F* : - kx (Hooke's law), (7 -2r)
where we have changed the subscript. If. x is positive (the spring is stretched
toward the right on the x axis), then F" is negative (it is a pull toward the left). If
x is negative (the spring is compressed toward the left), then F" is positive (it is a
push toward the right). Note that a spring force is a variable forcebecause it is a
function of x,the position of the free end. Thus F* canbe symbolized as F(*). Also
note that Hooke's law is a linear relationship betwe en F* and x.
To find the work done by the spring force as the block in Fig. 7 -IIa moves, let us
make two simplifying assumptions about the spring. (1) It is massless; that is, its
mass is negligible relative to the block's mass. (2) It is an ideal spring; that is, it
obeys Hooke's law exactly. Let us also assume that the contact between the block
and the floor is frictionless and that the block is particle-like.
We give the block a rightward jerk to get it moving and then leave it alone.
As the block moves rightward, the spring force F* does work on the block,
decreasing the kinetic energy and slowing the block. However, we cannol find
this work by using Eq. 7 -7 (W : Fd cos ,f) because that equation assumes a con-
stant force. The spring force is a variable force.
To flnd the work done by the spring, we use calculus. Let the block's initial
position be xi and its later position xy.Then divide the distance between those two
positions into many segments, each of tiny length Lx. Label these segments, start-
ing from xi,zs segments 1,2,and so on.As the block moves through a segment,
the spring force hardly varies because the segment is so short that x hardly varies.
Thus, we can approximate the force magnitude as being constant within the seg-
ment. Label these magnitudes as 4r in segment I, F*zin segment 2, and so on.
With the force now constant in each segment, we can find the work done
within each segment by using Eq.7-7. Here 0 - 180", and so cos O - 4.Then
the work done is - Fs Lx in segment L, -F*2 Lx in segment 2,and so on.The net
work W, done by the spring, from x;to xJ, is the sum of all these works:
(7 -22)
zero,Eq. 7 -22 becomeswhere 7 labels the segments. In the lim s to
w-
posrtrve
negatle
negatrve
posrtrve
(7 -23)
TRABALHO REALIZADO PELA FORÇA VARIÁVEL
ESTA É CONHECIDA COMO A LEI DE HOOKE
�Fm = k�d
A CONSTANTE ELÁSTICA K DA MOLA E SUA
UNIDADE!! [K] = 1N/m
PARA ESTA CONFIGURAÇÃO, PODEMOS ESCREVER:
Fx = −Kx
VAMOS CALCULAR O TRABALHO REALIZADO
POR UMA FORÇA DESSA NATUREZA.
segunda-feira, 13 de maio de 2013
DA FIGURA ABAIXO, TEMOS S ECT I O N 7. 4 • Work Done by a Varying Force 189
Example 7.4 Calculating Total Work Done from a Graph
A force acting on a particle varies with x, as shown in
Figure 7.8. Calculate the work done by the force as the parti-
cle moves from x ! 0 to x ! 6.0 m.
Solution The work done by the force is equal to the area
under the curve from xA ! 0 to xC ! 6.0 m. This area is
equal to the area of the rectangular section from ! to "
plus the area of the triangular section from " to #. The
area of the rectangle is (5.0 N)(4.0 m) ! 20 J, and the area
of the triangle is . Therefore, the to-
tal work done by the force on the particle is 25 J.
1
2(5.0 N)(2.0 m) ! 5.0 J
Example 7.5 Work Done by the Sun on a Probe
Graphical Solution The negative sign in the equation for
the force indicates that the probe is attracted to the Sun.
Because the probe is moving away from the Sun, we expect
to obtain a negative value for the work done on it. A spread-
sheet or other numerical means can be used to generate a
graph like that in Figure 7.9b. Each small square of the grid
corresponds to an area (0.05 N)(0.1 " 1011 m) ! 5 " 108 J.
The work done is equal to the shaded area in Figure 7.9b. Be-
cause there are approximately 60 squares shaded, the total
The interplanetary probe shown in Figure 7.9a is attracted
to the Sun by a force given by
in SI units, where x is the Sun-probe separation distance.
Graphically and analytically determine how much work is
done by the Sun on the probe as the probe–Sun separation
changes from 1.5 " 1011 m to 2.3 " 1011 m.
F ! #
1.3 " 1022
x 2
1 2 3 4 5 6
x(m)0
5
Fx(N)
#
! "
Figure 7.8 (Example 7.4) The force acting on a particle is con-
stant for the first 4.0 m of motion and then decreases linearly
with x from xB ! 4.0 m to xC ! 6.0 m. The net work done by
this force is the area under the curve.
If the size of the displacements is allowed to approach zero, the number of terms in
the sum increases without limit but the value of the sum approaches a definite value
equal to the area bounded by the Fx curve and the x axis:
Therefore, we can express the work done by Fx as the particle moves from xi to xf as
(7.7)
This equation reduces to Equation 7.1 when the component Fx ! F cos $ is constant.
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 just the work done by the net force. If we express the
net force in the x direction as %Fx, then the total work, or net work, done as the particle
moves from xi to xf is
(7.8)
If the system cannot be modeled as a particle (for example, if the system consists of
multiple particles that can move with respect to each other), we cannot use Equation
7.8. This is because different forces on the system may move through different dis-
placements. In this case, we must evaluate the work done by each force separately and
then add the works algebraically.
!W ! Wnet ! "xf
xi
#! Fx$
dx
W ! "xf
xi
Fx dx
lim
&x:0!
xf
xi
Fx &x ! "xf
xi
Fxdx
(a)
Fx
Area = ∆A = Fx ∆x
Fx
xxfxi
∆x
(b)
Fx
xxfxi
Work
Figure 7.7 (a) The work done by
the force component Fx for the
small displacement &x is Fx &x,
which equals the area of the shaded
rectangle. The total work done for
the displacement from xi to xf is ap-
proximately equal to the sum of the
areas of all the rectangles. (b) The
work done by the component Fx of
the varying force as the particle
moves from xi to xf is exactly equal to
the area under this curve.
Wi ≈ Fx∆x
W ≈
xf�
xi
Wi = Fx1∆x+ Fx2∆x+ ...+ Fxn∆x
TRABALHO PARA CADA INTERVALO : ∆x
PARA O INTERVALO DE ATÉ , TEMOS: xi xj
SE O NÚMERO DE DE RETÂNGULOS SE TORNAR INFINITAMENTE
GRANDE,TEMOS
TRABALHO DE UMA FORÇA
VARIÁVEL
S ECT I O N 7. 4 • Work Done by a Varying Force 189
Example 7.4 Calculating Total Work Done from a Graph
A force acting on a particle varies with x, as shown in
Figure 7.8. Calculate the work done by the force as the parti-
cle moves from x ! 0 to x ! 6.0 m.
Solution The work done by the force is equal to the area
under the curve from xA ! 0 to xC ! 6.0 m. This area is
equal to the area of the rectangular section from ! to "
plus the area of the triangular section from " to #. The
area of the rectangle is (5.0 N)(4.0 m) ! 20 J, and the area
of the triangle is . Therefore, the to-
tal work done by the force on the particle is 25 J.
1
2(5.0 N)(2.0 m) ! 5.0 J
Example 7.5 Work Done by the Sun on a Probe
Graphical Solution The negative sign in the equation for
the force indicates that the probe is attracted to the Sun.
Because the probe is moving away from the Sun, we expect
to obtain a negative value for the work done on it. A spread-
sheet or other numerical means can be used to generate a
graph like that in Figure 7.9b. Each small square of the grid
corresponds to an area (0.05 N)(0.1 " 1011 m) ! 5 " 108 J.
The work done is equal to the shaded area in Figure 7.9b. Be-
cause there are approximately 60 squares shaded, the total
The interplanetary probe shown in Figure 7.9a is attracted
to the Sun by a force given by
in SI units, where x is the Sun-probe separation distance.
Graphically and analytically determine how much work is
done by the Sun on the probe as the probe–Sun separation
changes from 1.5 " 1011 m to 2.3 " 1011 m.
F ! #
1.3 " 1022
x 2
1 2 3 4 5 6
x(m)0
5
Fx(N)
#
! "
Figure 7.8 (Example 7.4) The force acting on a particle is con-
stant for the first 4.0 m of motion and then decreases linearly
with x from xB ! 4.0 m to xC ! 6.0 m. The net work done by
this force is the area under the curve.
If the size of the displacements is allowed to approach zero, the number of terms in
the sum increases without limit but the value of the sum approaches a definite value
equal to the area bounded by the Fx curve and the x axis:
Therefore, we can express the work done by Fx as the particle moves from xi to xf as
(7.7)
This equation reduces to Equation 7.1 when the component Fx ! F cos $ is constant.
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 just the work done by the net force. If we express the
net force in the x direction as %Fx, then the total work, or net work, done as the particle
moves from xi to xf is
(7.8)
If the system cannot be modeled as a particle (for example, if the system consists of
multiple particles that can move with respect to each other), we cannot use Equation
7.8. This is because different forces on the system may move through different dis-
placements. In this case, we must evaluate the work done by each force separately and
then add the works algebraically.
!W ! Wnet ! "xf
xi
#! Fx$
dx
W ! "xf
xi
Fx dx
lim
&x:0!
xf
xi
Fx &x ! "xf
xi
Fxdx
(a)
Fx
Area = ∆A = Fx ∆x
Fx
xxfxi
∆x
(b)
Fx
xxfxi
Work
Figure 7.7 (a) The work done by
the force component Fx for the
small displacement &x is Fx &x,
which equals the area of the shaded
rectangle. The total work done for
the displacement from xi to xf is ap-
proximately equal to the sum of the
areas of all the rectangles. (b) The
work done by the component Fx of
the varying force as the particle
moves from xi to xf is exactly equal to
the area under this curve.
W =
W ≈
xf�
xi
Fx∆x
segunda-feira, 13 de maio de 2013
PARA O CASO DA FORÇA VARIÁVEL SER A FORÇA ELÁSTICA
TEMOS:
Wm = (−1
2
K)[x2]
xf
xi = (−12K)(x
2
f − x2i )
POR FIM,
POTÊNCIA
Wm =
� xf
xi
Fxdx =
� xf
xi
−Kxdx = −K
� xf
xi
xdx
POTÊNCIA MÉDIA POTÊNCIA INSTANTÂNEA
P =
dW
dt
P =
dW
dt
= Fvcosφ P =
dW
dt
= �F · �v
Pmed =
W
∆t
Ghapten 7 | Kinetic Energy and Work
and
steam engines could do work. In the British system, the unit of power is the foot-
pound per second. Often the horsepower is used. These are related by
1 watt - 1 W : 1 J/s : 0.738 ft' lb/s
t horsepower - I hp - 550 ft'lb/s :746W.
Inspection of Eq. 7 -42 shows that work can be expressed as power multiplied
by time, as in the common unit kilowatt-hour. Thus,
1 kilowatt-hour : 1 kW.h - (10' W)(3600 s)
- 3.60 x 106 J - 3.60 MJ.
Perhaps because they appear on our utility bills, the watt and the kilowatt-hour
have become identifled as electrical units. They can be used equally well as units
for other examples of power and energy. Thus, if you pick up a book from the
floor and put it on a tabletop, you are free to report the work that you have done
as, say, 4 x 10-6 kW'h (or more conveniently as 4 mW'h).
We can also express the rate at which a force does work on a particle (ot
particle-like object) itr terms of that force and the particle's velocity. For a par-
ticle that is moving along a straight line (ruy, an x axis) and is acted on by a
constant force F directed at some angle Q to that line, Eq. 7 -43 becomes
(7 -44)
(7 -4s)
(7 -46)
ffififfi" 3-1S The power due to the
truck's applied force on the trailing
load is the rate at which that force
does work on the load. (REGLAIN
F R E D E RI C/ G amma- P res s e, Inc. )
Fcos Qdx
dt
P-Fvcosor (7 -47)
Reorg anrzing the right side of Eq. 7 -47 as the dot produ ct F. 7, we may also write
the equation as
P_F.i (instantaneous power). (7 -48)
For example, the truck in Fig. 7 -L5 exerts a force F onthe trailing load, which
has velocity i at some instant. The instantaneous power due to F is the rate at
which F do"s work on the load at thatinstant and is given by Eqs. 7 -47 and 7 -48.
Saying that this power is "the power of the truck" is often acceptable, but keep in
mind what is meant: Power is the rate at which the applie d force does work.
on the block from the cord positive, negative, or zero?
Figure 7 -16 shows constant forc"t Fr and F 2 acting on a
box as the box slides rightward across a frictionless
floor. Force F, is horizontal, with magnitude 2.0 N;
force F, tt angled upward by 60' to the floor and has
magnitude 4.0 N. The speed v of the box at a certain
instant is 3.0 m/s. What is the power due to each force
acting on the box at that instant, and what is the net
power? Is the net power changing at that instant?
We want an instantaneous power, not an
average power over a time period. Also, we know the
box's velocity (rather than the work done on it).
Ff;ffi" 7-1S Two forces F1 and F2act on a box that slides right-
ward across a frictionless floor. The velocity of the box is n.
Calculatfon: We use
F r,at angle Q,
Pr -- F! cos ,f1
Eq. 7 -47 for each force. For force
to velocity 7, we have
_ (2.0 N)(3.0 m/s) cos 1.80"
(Answer)
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