USING NEWTON’S LAWS
E
XERCISES
Section 5.1 Using Newton’s Second Law
12.
Newton’s second law for this mass says
net
1
2
,
F
F
F
ma
+
=
v
v
v
v
=
where we assume no other significant forces are
acting. Since the acceleration and the first force are given, one can solve for the second,
2
2
1
(1.5 kg)(7.3 m/s
)
F
ma
F
=
v
v
v
=

ˆ
ˆ
ˆ
ˆ
ˆ
(
cos30
sin30
)
(6.8 N)
(2.68
5.48 )N.
i
j
i
i
j
°
° 
+
=
+
This has magnitude
6.10 N and direction 63.9
°
CCW from the
x
axis.
13.
I
NTERPRET
In this problem, two forces are exerted on the object of interest and produce an acceleration. With the
mass of the object and one force given, we are asked to find the other force.
D
EVELOP
Newton’s second law for this mass says
net
1
2
,
F
F
F
ma
=
+
=
r
r
r
r
where we assume no other significant
forces are acting. The second force is given by
2
1
F
ma
F
=

r
r
r
E
VALUATE
Substituting the expressions given in the problem statement, we obtain
2
2
2
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(3.1 kg)[(0.91 m/s
)
(0.27 m/s ) ]
[( 1.2 N)
(2.5 N) ]
(4.02 N)
(1.66 N)
F
ma
F
i
j
i
j
i
j
=

=




=
+
r
r
r
A
SSESS
To show that the answer is correct, let’s add the two forces together. The net force is
net
1
2
ˆ
ˆ
ˆ
ˆ
[( 1.2 N)
(2.5 N) ]
[(4.02 N)
(1.66 N) ]
ˆ
ˆ
ˆ
ˆ
(2.82 N)
(
0.84 N)
(3.1)[(0.91 N)
( 0.27 N) ]
F
F
F
i
j
i
j
i
j
i
j
=
+
=


+
+
=
+ 
=
+ 
r
r
r
From the expression, it is clear that
net
F
r
points in the same direction as
.
a
r
14.
The horizontal forces on the barge are the two tensions and the resistance of the water, as shown in the figure. The
net force is in the
x
direction, so
res
2
cos25
,
x
T
F
ma
° 
=
v
v
since
1
2
.
T
T
T
=
=
(a)
If
res
0,
2(1100 N
)cos25
1.99 kN.
x
a
F
=
=
° =
v
(b)
If
2
2
res
0.16 m/s
,
1.99 kN
(3700 kg)(0.16 m/s
)
1.40 kN.
x
a
F
=
=

=
v
15.
I
NTERPRET
In this problem, we are asked to find the tilt angle of the table such that the acceleration of an object
sliding on the surface of the table is the same as the gravitational acceleration near the surface of the Moon.
D
EVELOP
In Example 5.1 it was shown that the acceleration down an incline is
sin .
a
g
θ
=
This equation is what
we shall use to solve for the tilt angle
.
θ
5.1
5
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5.2
Chapter 5
E
VALUATE
Given that the gravitational acceleration near the surface of the Moon is
2
1.6 m/s
,
g
=
the angle of tilt
should be
2
1
1
2
1.6 m/s
sin
sin
9.40
9.8 m/s
a
g
θ


=
=
=
°
A
SSESS
A tilt angle of
9.40
°
would give an acceleration that is the same as that of the Moon’s surface. On the
other hand, if we want to simulate the motion near Earth’s surface, then
,
a
g
=
and
90 .
θ
=
°
The greater the value
of
a
, the larger the tilt angle.
16.
The acceleration down a frictionless incline is
sin
a
g
θ
=
(see Example 5.1), and the distance traveled down the
incline, starting from rest
0
(
0)
v
=
at the top
0
(
0),
x
=
is
2
1
2
.
x
at
=
Therefore
2
2
/
sin
2(1.3 km)/(9.8 m/s
)sin 24
25.5 s.
t
x g
θ
=
=
° =
17.
I
NTERPRET
In this problem, the physical quantity of interest is the tension force in the cable. To compute the
tension, we analyze the motion of the car.
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 Spring '09
 Prof.Kim
 Physics, Acceleration, Force, Friction, Mass, Sin

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