194
CHAPTE R 8 POTE NTIAL E N E RGY AN D CONSE RVATION OF ENERGY
Checkpoint 5
In three trials, a block is pushed
by a horizontal applied force
across a floor that is not frictionless, as in Fig. 8-13a.The magnitudes F of the applied force and
the results of the pushing on the
block’s speed are given in the
table. In all three trials, the block is pushed through the same distance d. Rank the
three trials according to the change in the thermal energy of the block and floor that
occurs in that distance d, greatest first.
Trial
F
Result on Block’s Speed
a
5 . 0N
d e c r e a s e s
b
7 . 0N
r e m a i n s c o n s t a n t
c
8 . 0N
i n c r e a s e s
be friction and a change ⌬E th in thermal energy of the crate
and the floor. Therefore, the system on which the work is
done is the crate – floor system, because both energy
changes occur in that system.
(b) What is the increase ⌬E th in the thermal energy of the
crate and floor?
KEY IDEA
We can relate ⌬E th to the work W done by with the energy
statement of Eq. 8-33 for a system that involves friction:
W ϭ ⌬E mec ϩ ⌬E th .
( 8 - 3 4 )
Calculations: We know the value of W from (a). The
change ⌬E mec in the crate’s mechanical energy is just the
change in its kinetic energy because no potential energy
changes occur, so we have
Substituting this into Eq. 8-34 and solving for ⌬E th ,we find
(Answer)
Without further experiments, we cannot say how much of
this thermal energy ends up in the crate and how much in
the floor. We simply know the total amount.
ϭ 22.2 J Ϸ 22 J.
ϭ 20 J Ϫ
1
2 (14 kg)[(0.20 m/s)
2 Ϫ (0.60 m/s)
2
]
⌬E th ϭ W Ϫ (
1
2 mv
2 Ϫ
1
2 mv 0
2
) ϭ W Ϫ
1
2 m(v
2 Ϫ v 0
2
)
⌬E mec ϭ ⌬K ϭ
1
2 mv
2 Ϫ
1
2 mv 0
2
.
F
:
Sample Problem 8.05 Work, friction, change in thermal energy, cabbage heads
A food shipper pushes a wood crate of cabbage heads (total
mass m 14 kg) across a concrete floor with a constant
horizontal force of magnitude 40 N. In a straight-line displacement of magnitude d 0.50 m, the speed of the crate
decreases from v 0 0.60 m/s to v 0.20 m/s.
(a) How much work is done by force , and on what system
does it do the work?
KEY IDEA
Because the applied force
is constant, we can calculate
the work it does by using Eq. 7-7 (
).
Calculation: Substituting given data, including the fact that
force and displacement are in the same direction, we
find
W ϭ Fd cos f ϭ (40 N)(0.50 m) cos 0Њ
ϭ 20 J.
(Answer)
Reasoning: To determine the system on which the work is
done, let’s check which energies change. Because the crate’s
speed changes, there is certainly a change ⌬K in the crate’s
kinetic energy. Is there friction between the floor and the
crate, and thus a change in thermal energy? Note that and
the crate’s velocity have the same direction. Thus, if there is
no friction, then
should be accelerating the crate to a
greater speed. However, the crate is slowing, so there must
F
:
F
:
d
:
F
:
W ϭ Fd cos
F
:
F
:
ϭ
ϭ
ϭ
F
:
ϭ
Additional examples, video, and practice available at WileyPLUS
changes, and the thermal energies of the block and floor also change. Therefore, the
work done by force is done on the block – floor system.That work is
W ϭ ⌬E mec ϩ ⌬E th (work done on system, friction involved).
(8-33)
This equation, which is represented in Fig. 8-13b, is the energy statement for the
work done on a system by an external force when friction is involved.
F
:
CHAPTE R 8 POTE NTIAL E N E RGY AN D CONSE RVATION OF ENERGY
Checkpoint 5
In three trials, a block is pushed
by a horizontal applied force
across a floor that is not frictionless, as in Fig. 8-13a.The magnitudes F of the applied force and
the results of the pushing on the
block’s speed are given in the
table. In all three trials, the block is pushed through the same distance d. Rank the
three trials according to the change in the thermal energy of the block and floor that
occurs in that distance d, greatest first.
Trial
F
Result on Block’s Speed
a
5 . 0N
d e c r e a s e s
b
7 . 0N
r e m a i n s c o n s t a n t
c
8 . 0N
i n c r e a s e s
be friction and a change ⌬E th in thermal energy of the crate
and the floor. Therefore, the system on which the work is
done is the crate – floor system, because both energy
changes occur in that system.
(b) What is the increase ⌬E th in the thermal energy of the
crate and floor?
KEY IDEA
We can relate ⌬E th to the work W done by with the energy
statement of Eq. 8-33 for a system that involves friction:
W ϭ ⌬E mec ϩ ⌬E th .
( 8 - 3 4 )
Calculations: We know the value of W from (a). The
change ⌬E mec in the crate’s mechanical energy is just the
change in its kinetic energy because no potential energy
changes occur, so we have
Substituting this into Eq. 8-34 and solving for ⌬E th ,we find
(Answer)
Without further experiments, we cannot say how much of
this thermal energy ends up in the crate and how much in
the floor. We simply know the total amount.
ϭ 22.2 J Ϸ 22 J.
ϭ 20 J Ϫ
1
2 (14 kg)[(0.20 m/s)
2 Ϫ (0.60 m/s)
2
]
⌬E th ϭ W Ϫ (
1
2 mv
2 Ϫ
1
2 mv 0
2
) ϭ W Ϫ
1
2 m(v
2 Ϫ v 0
2
)
⌬E mec ϭ ⌬K ϭ
1
2 mv
2 Ϫ
1
2 mv 0
2
.
F
:
Sample Problem 8.05 Work, friction, change in thermal energy, cabbage heads
A food shipper pushes a wood crate of cabbage heads (total
mass m 14 kg) across a concrete floor with a constant
horizontal force of magnitude 40 N. In a straight-line displacement of magnitude d 0.50 m, the speed of the crate
decreases from v 0 0.60 m/s to v 0.20 m/s.
(a) How much work is done by force , and on what system
does it do the work?
KEY IDEA
Because the applied force
is constant, we can calculate
the work it does by using Eq. 7-7 (
).
Calculation: Substituting given data, including the fact that
force and displacement are in the same direction, we
find
W ϭ Fd cos f ϭ (40 N)(0.50 m) cos 0Њ
ϭ 20 J.
(Answer)
Reasoning: To determine the system on which the work is
done, let’s check which energies change. Because the crate’s
speed changes, there is certainly a change ⌬K in the crate’s
kinetic energy. Is there friction between the floor and the
crate, and thus a change in thermal energy? Note that and
the crate’s velocity have the same direction. Thus, if there is
no friction, then
should be accelerating the crate to a
greater speed. However, the crate is slowing, so there must
F
:
F
:
d
:
F
:
W ϭ Fd cos
F
:
F
:
ϭ
ϭ
ϭ
F
:
ϭ
Additional examples, video, and practice available at WileyPLUS
changes, and the thermal energies of the block and floor also change. Therefore, the
work done by force is done on the block – floor system.That work is
W ϭ ⌬E mec ϩ ⌬E th (work done on system, friction involved).
(8-33)
This equation, which is represented in Fig. 8-13b, is the energy statement for the
work done on a system by an external force when friction is involved.
F
: