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Chapter 4
1
Page 1
CHAPTER 4  Dynamics: Newton’s Laws of Motion
1.
We convert the units:
#(
lb = (0.25 lb)(4.45 N/lb)
≈
1 N.
2.
If we select the bike and rider as the object, we apply Newton’s second law to find the mass:
∑
F
=
ma
;
255 N =
m
(2.20 m/s
2
), which gives
m
= 116 kg
.
3.
We apply Newton’s second law to the object:
∑
F
=
ma
;
F
= (7.0
×
10
–3
kg)(10,000)(9.80 m/s
2
) =
6.9
×
10
2
N
.
4.
Without friction, the only horizontal force is the tension.
We apply Newton’s second law to the car:
∑
F
=
ma
;
F
T
= (1250 kg)(1.30 m/s
2
) =
1.63
×
10
3
N
.
5.
We find the weight from the value of
g
.
(
a
) Earth:
F
G
=
mg
= (58 kg)(9.80 m/s
2
) =
5.7
×
10
2
N
.
(
b
) Moon:
F
G
=
mg
= (58 kg)(1.7 m/s
2
) =
99 N
.
(
c
)
Mars:
F
G
=
mg
= (58 kg)(3.7 m/s
2
) =
2.1
×
10
2
N
.
(
d
) Space:
F
G
=
mg
= (58 kg)(0 m/s
2
) =
0
.
6.
The acceleration can be found from the car’s onedimensional motion:
v
=
v
0
+
at
;
0 = [(90 km/h)/(3.6 ks/h)] +
a
(7.0 s), which gives
a
= – 3.57 m/s
2
.
We apply Newton’s second law to find the required average force
∑
F
=
ma
;
F
= (1050 kg)(– 3.57 m/s
2
) =
– 3.8
×
10
3
N
.
The negative sign indicates that the force is opposite to the velocity.
7.
The required average acceleration can be found from the onedimensional motion:
v
2
=
v
0
2
+ 2
a
(
x
–
x
0
);
(155 m/s)
2
= 0 + 2
a
(0.700 m – 0), which gives
a
= 1.72
×
10
4
m/s
2
.
We apply Newton’s second law to find the required average force
∑
F
=
ma
;
F
= (6.25
×
10
–3
kg)(1.72
×
10
4
m/s
2
) =
107 N
.
8.
(
a
) The weight of the box depends on the value of
g
:
F
G
=
m
1
g
= (30.0 kg)(9.80 m/s
2
) =
294 N
.
We find the normal force from
∑
F
y
=
ma
y
;
F
N
–
m
1
g
= 0, which gives
F
N
=
m
1
g
=
294 N
.
(
b
) We select both blocks as the object and apply Newton’s
second law:
∑
F
y
=
ma
y
;
F
N1
–
m
1
g
–
m
2
g
= 0, which gives
F
N1
= (
m
1
+
m
2
)
g
= (30.0 kg + 20.0 kg)(9.80 m/s
2
) =
490 N
.
If we select the top block as the object, we have
∑
F
y
=
ma
y
;
F
N2
–
m
2
g
= 0, which gives
F
N2
=
m
2
g
= (20.0 kg)(9.80 m/s
2
) =
196 N
.
9.
The required average acceleration can be found from the onedimensional motion:
v
2
=
v
0
2
+ 2
a
(
x
–
x
0
);
F
N1
m
1
g
x
y
m
2
g
m
1
g
F
N2
F
N
m
1
g
(
a
)
(
b
)
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0 = [(100 km/h)/(3.6 ks/h)]
2
+ 2
a
(150 m – 0), which gives
a
= – 2.57 m/s
2
.
We apply Newton’s second law to find the required force
∑
F
=
ma
;
F
= (3.6
×
10
5
kg)(– 2.57 m/s
2
) =
– 9.3
×
10
5
N
.
The weight of the train is
mg
= (3.6
×
10
5
kg)(9.80 m/s
2
) = 3.5
×
10
6
N,
so Superman must apply a force that is
25%
of the weight of the train.
10. The acceleration of the first box is
a
1
=
F
/
m
1
, so the speed after a time
t
is
v
1
=
v
0
+
a
1
t
=
a
1
t
.
For the second box we have
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This note was uploaded on 04/08/2008 for the course PHY 53 taught by Professor Mueller during the Fall '08 term at Duke.
 Fall '08
 Mueller
 Mass

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