164
CHAPTE R 7 KI N ETIC E N E RGY AN D WOR K
Sample Problem 7.07 Work calculated by graphical integration
In Fig. 7-13b, an 8.0 kg block slides along a frictionless floor
as a force acts on it, starting at x 1 ϭ 0 and ending at x 3 ϭ 6.5 m.
As the block moves, the magnitude and direction of the
force varies according to the graph shown in Fig. 7-13a. For
The work W done by while the particle moves from an initial position r i having
coordinates (x i , y i , z i ) to a final position r f having coordinates (x f , y f , z f ) is then
(7-36)
If has only an x component, then the y and z terms in Eq. 7-36 are zero and the
equation reduces to Eq. 7-32.
Work–Kinetic Energy Theorem with a Variable Force
Equation 7-32 gives the work done by a variable force on a particle in a onedimensional situation. Let us now make certain that the work is equal to the
change in kinetic energy, as the work – kinetic energy theorem states.
Consider a particle of mass m, moving along an x axis and acted on by a
net force F(x) that is directed along that axis. The work done on the particle
by this force as the particle moves from position x i to position x f is given by
Eq. 7-32 as
(7-37)
in which we use Newton’s second law to replace F(x) with ma. We can write the
quantity ma dx in Eq. 7-37 as
(7-38)
From the chain rule of calculus, we have
(7-39)
and Eq. 7-38 becomes
(7-40)
Substituting Eq. 7-40 into Eq. 7-37 yields
(7-41)
Note that when we change the variable from x to v we are required to express the
limits on the integral in terms of the new variable. Note also that because the
mass m is a constant, we are able to move it outside the integral.
Recognizing the terms on the right side of Eq. 7-41 as kinetic energies allows
us to write this equation as
W ϭ K f Ϫ K i ϭ ⌬K,
which is the work – kinetic energy theorem.
ϭ
1
2 mv f
2 Ϫ
1
2 mv i
2
.
W ϭ ͵
v f
v i
mv dv ϭ m ͵
v f
v i
v dv
ma dx ϭ m
dv
dx
v dx ϭ mv dv.
dv
dt
ϭ
dv
dx
dx
dt
ϭ
dv
dx
v,
ma dx ϭ m
dv
dt
dx.
W ϭ ͵
x f
x i
F(x) dx ϭ ͵
x f
x i
ma dx,
F
:
W ϭ ͵
r f
r i
dW ϭ ͵
x f
x i
F x dx ϩ ͵
y f
y i
F y dy ϩ ͵
z f
z i
F z dz.
F
:
example, from x ϭ 0 to x ϭ 1 m, the force is positive (in
the positive direction of the x axis) and increases in magnitude from 0 to 40 N. And from x ϭ 4 m to x ϭ 5 m, the
force is negative and increases in magnitude from 0 to 20 N.
CHAPTE R 7 KI N ETIC E N E RGY AN D WOR K
Sample Problem 7.07 Work calculated by graphical integration
In Fig. 7-13b, an 8.0 kg block slides along a frictionless floor
as a force acts on it, starting at x 1 ϭ 0 and ending at x 3 ϭ 6.5 m.
As the block moves, the magnitude and direction of the
force varies according to the graph shown in Fig. 7-13a. For
The work W done by while the particle moves from an initial position r i having
coordinates (x i , y i , z i ) to a final position r f having coordinates (x f , y f , z f ) is then
(7-36)
If has only an x component, then the y and z terms in Eq. 7-36 are zero and the
equation reduces to Eq. 7-32.
Work–Kinetic Energy Theorem with a Variable Force
Equation 7-32 gives the work done by a variable force on a particle in a onedimensional situation. Let us now make certain that the work is equal to the
change in kinetic energy, as the work – kinetic energy theorem states.
Consider a particle of mass m, moving along an x axis and acted on by a
net force F(x) that is directed along that axis. The work done on the particle
by this force as the particle moves from position x i to position x f is given by
Eq. 7-32 as
(7-37)
in which we use Newton’s second law to replace F(x) with ma. We can write the
quantity ma dx in Eq. 7-37 as
(7-38)
From the chain rule of calculus, we have
(7-39)
and Eq. 7-38 becomes
(7-40)
Substituting Eq. 7-40 into Eq. 7-37 yields
(7-41)
Note that when we change the variable from x to v we are required to express the
limits on the integral in terms of the new variable. Note also that because the
mass m is a constant, we are able to move it outside the integral.
Recognizing the terms on the right side of Eq. 7-41 as kinetic energies allows
us to write this equation as
W ϭ K f Ϫ K i ϭ ⌬K,
which is the work – kinetic energy theorem.
ϭ
1
2 mv f
2 Ϫ
1
2 mv i
2
.
W ϭ ͵
v f
v i
mv dv ϭ m ͵
v f
v i
v dv
ma dx ϭ m
dv
dx
v dx ϭ mv dv.
dv
dt
ϭ
dv
dx
dx
dt
ϭ
dv
dx
v,
ma dx ϭ m
dv
dt
dx.
W ϭ ͵
x f
x i
F(x) dx ϭ ͵
x f
x i
ma dx,
F
:
W ϭ ͵
r f
r i
dW ϭ ͵
x f
x i
F x dx ϩ ͵
y f
y i
F y dy ϩ ͵
z f
z i
F z dz.
F
:
example, from x ϭ 0 to x ϭ 1 m, the force is positive (in
the positive direction of the x axis) and increases in magnitude from 0 to 40 N. And from x ϭ 4 m to x ϭ 5 m, the
force is negative and increases in magnitude from 0 to 20 N.