Newton°s Laws of Motion
413
E
VALUATE
:
We can also consider the two crates and the rope connecting them as a single object of mass
The freebody diagram is sketched in Figure 4.43c.
1
2
10.0 kg.
m
m
m
=
+
=
x
x
F
ma
=
∑
2
(10.0 kg)(2.50 m/s
)
25.0 N
F
ma
=
=
=
This agrees with our answer in part (d).
Figure 4.43c
4.44.
I
DENTIFY
:
Apply Newton’s second and third laws.
S
ET
U
P
:
Actionreaction forces act between a pair of objects. In the second law all the forces act on the same
object.
E
XECUTE
:
(a)
The force the astronaut exerts on the cable and the force that the cable exerts on the astronaut are
an actionreaction pair, so the cable exerts a force of 80.0 N on the astronaut.
(b)
The cable is under tension.
(c)
2
80.0 N
0.762 m/s
105.0 kg
F
a
m
=
=
=
.
(d)
There is no net force on the massless cable, so the force that the shuttle exerts on the cable must be 80.0 N (this
is
not
an actionreaction pair). Thus, the force that the cable exerts on the shuttle must be 80.0 N.
(e)
4
2
4
80.0 N
8.84
10
m/s
9.05
10
kg
F
a
m
−
=
=
=
×
×
.
E
VALUATE
:
Since the cable is massless the net force on it is zero and the tension is the same at each end.
4.45.
I
DENTIFY
and
S
ET
U
P
:
Take derivatives of
( )
x t
to find
x
v
and
.
x
a
Use Newton°s second law to relate the
acceleration to the net force on the object.
E
XECUTE
:
(a)
3
2
2
4
3
(9.0
10
m/s )
(8.0
10 m/s )
x
t
t
3
=
×
−
×
0
x
=
at
0
t
=
When
0.025 s,
t
=
3
2
2
4
3
3
(9.0
10
m/s
)(0.025 s)
(8.0
10
m/s
)(0.025 s)
4.4 m.
x
=
×
−
×
=
The length of the barrel must be 4.4 m.
(b)
3
2
4
3
(18.0
10
m/s
)
(24.0
10
m/s )
x
dx
v
t
dt
=
=
×
−
×
2
t
At
0
(object starts from rest).
0,
t
=
x
v
=
At
when the object reaches the end of the barrel,
0.025 s,
t
=
2
4
3
2
(18.0
10
m/s
)(0.025 s)
(24.0
10
m/s
)(0.025 s)
300 m/s
x
v
3
=
×
−
×
=
(c)
,
x
x
F
ma
=
∑
so must find
.
x
a
3
2
3
18.0
10
m/s
(48.0
10
m/s )
x
x
dv
a
t
dt
4
=
=
×
−
×
(i) At
and
0,
t
=
3
18.0
10
m/s
x
a
=
×
2
3
2
4
(1.50 kg)(18.0
10
m/s
)
2.7
10
N.
x
F
=
×
=
×
∑
(ii) At
and
0.025 s,
t
=
3
2
4
3
18
10
m/s
(48.0
10
m/s
)(0.025 s)
6.0
10
m/s
x
a
3
=
×
−
×
=
×
2
2
3
(1.50 kg)(6.0
10
m/s
)
9.0
10
N.
x
F
3
=
×
=
×
∑
E
VALUATE
:
The acceleration and net force decrease as the object moves along the barrel.
4.46.
I
DENTIFY
:
Apply
m
=
∑
F
a
!
!
and solve for the mass
m
of the spacecraft.
S
ET
U
P
:
. Let
be upward.
w
mg
=
y
+
E
XECUTE
:
(a)
The velocity of the spacecraft is downward. When it is slowing down, the acceleration is upward.
When it is speeding up, the acceleration is downward.
(b)
In each case the net force is in the direction of the acceleration. Speeding up:
and the net force is
downward. Slowing down:
and the net force is upward.
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 Spring '06
 Buchler
 Physics, Force, Mass, m/s, net force

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