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518
Chapter 5
5.38.
IDENTIFY:
r
f
n
μ
=
. Apply
m
=
∑
F
a
!
!
to the tire.
SET UP:
nm
g
=
and
f
ma
=
.
EXECUTE:
22
0
x
vv
a
L
−
=
, where
L
is the distance covered before the wheel&s speed is reduced to half its original
speed and
0
/2
=
.
2
1
00
4
r
3
8
avv
v
g
Lg
Lg
Lg
−
−
==
=
=
.
Low pressure,
18.1 m
L
=
and
2
2
3
(3.50 m/s)
0.0259
8 (18.1 m)(9.80 m/s )
=
.
High pressure,
92.9 m
L
=
and
2
2
3 (3.50 m/s)
0.00505
8 (3.50 m/s)
=
.
EVALUATE:
r
is inversely proportional to the distance
L
, so
r1
2
r2
1
L
L
=
.
5.39.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the box. Use the information about sliding to calculate the mass of the box.
SET UP:
kk
f
n
=
,
rr
f
n
=
and
g
=
.
EXECUTE:
Without the dolly:
g
=
and
k
0
Fn
−
=
(0
x
a
=
since speed is constant).
2
k
160 N
34.74 kg
(0.47) (9.80 m s )
F
m
g
=
With the dolly: the total mass is 34.7 kg
5.3 kg
40.04 kg
+
=
and friction now is rolling friction,
.
f
mg
=
r
Fm
g
m
a
−=
.
2
r
3.82 m s
g
a
m
−
.
EVALUATE:
160 N
fm
g
and
4.36 N
g
, or,
f
f
=
. The rolling friction force is much less
than the kinetic friction force.
5.40.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to the truck. For constant speed,
0
a
=
and
horiz
r
Ff
=
.
SET UP:
r
f
g
μμ
. Let
21
1.42
mm
=
and
r2
r1
0.81
=
.
EXECUTE:
Since the speed is constant and we are neglecting air resistance, we can ignore the 2.4 m/s, and
net
F
in
the horizontal direction must be zero. Therefore
r
r
horiz
200 N
fn
F
=
before the weight and pressure changes
are made. After the changes,
r
horiz
(0.81 ) (1.42 )
,
nF
=
because the speed is still constant and
net
0
F
=
. We can
simply divide the two equations
:
r
horiz
r
(0.81 )(1.42 )
200 N
μ
n
=
and
horiz
(0.81) (1.42) (200 N)
230 N
F
.
EVALUATE:
The increase in weight increases the normal force and hence the friction force, whereas the decrease
in
r
reduces it. The percentage increase in the weight is larger, so the net effect is an increase in the friction force.
5.41.
IDENTIFY:
Apply
m
=
∑
F
a
!
!
to each block. The target variables are the tension
T
in the cord and the
acceleration
a
of the blocks. Then
a
can be used in a constant acceleration equation to find the speed of each block.
The magnitude of the acceleration is the same for both blocks.
SET UP:
The system is sketched in Figure 5.41a.
For each block take a positive
coordinate direction to be the
direction of the block&s acceleration.
Figure 5.41a
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View Full DocumentApplying Newton&s Laws
519
block on the table
: The freebody is sketched in Figure 5.41b.
EXECUTE:
y
y
Fm
a
=
∑
0
A
nmg
−
=
A
nm
g
=
kk
k
A
f
g
μμ
==
Figure 5.41b
x
x
a
=
∑
k
A
Tf m
a
−=
k
AA
Tm
g
m
a
μ
SET UP:
hanging block
: The freebody is sketched in Figure 5.41c.
EXECUTE:
y
y
a
=
∑
B
B
mg T ma
B
B
gm
a
=−
Figure 5.41c
(a)
Use the second equation in the first
k
BB
mg ma
−−
=
k
()
(
)
AB
B
A
mm
am
m
g
+=
−
2
2
k
(
)
(1.30 kg
(0.45)(2.25 kg))(9.80 m/s )
0.7937 m/s
2.25 kg 1.30 kg
BA
g
a
=
++
SET UP:
Now use the constant acceleration equations to find the final speed. Note that the blocks have the same
speeds.
0
0.0300 m,
xx
2
0.7937 m/s ,
x
a
=
0
0,
x
v
=
?
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

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