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2.
REASONING AND SOLUTION When the birdfeeder is hanging freely and no one is
pulling on the dangling (lower) cord, there is a tension in the cord between the birdfeeder
and the tree limb (the upper cord), because the upper cord supports the weight of the
birdfeeder. When the lower cord is pulled down with a slow continuous pull, the tension in
both cords increases slowly. Since the upper cord has a larger tension to begin with, it
always has the greater tension as the lower cord is pulled. Thus, the upper cord snaps first.
On the other hand, when the child gives the lower cord a sudden, downward pull, the
tension in the lower cord increases suddenly. However, the tension in the upper cord does
not increase as suddenly. The reason is that the birdfeeder has a large mass, so it accelerates
very slowly. Thus, the upper cord is stretched slowly and, consequently, the tension in the
upper cord rises slowly. Since the tension rises much faster in the lower cord, it is the first
to snap.
3.
REASONING AND SOLUTION If the net external force acting on an object is zero, it is
possible for the object to be traveling with a nonzero velocity. According to Newton’s
second law, ΣF = ma, if the net external force ΣF is zero, the acceleration a is also zero. If
the acceleration is zero, the velocity must be constant, both in magnitude and in direction.
Thus, an object can move with a constant nonzero velocity when the net external force is
zero.
4.
REASONING AND SOLUTION
According to Newton's second law, a net force is
required to give an object a non-zero acceleration.
a. If an object is moving with a constant acceleration of 9.80 m/s2, we can conclude that
there is a net force on the object.
b. If an object moves with a constant velocity of 9.80 m/s, its acceleration is zero; therefore,
we can conclude that the net force acting on the object is zero.
1.
REASONING AND SOLUTION According to Newton’s second law, the acceleration is
a = ΣF/m. Since the pilot and the plane have the same acceleration, we can write
§ ΣF ·
§ ΣF ·
=¨
¨
¸
¸
© m ¹ PILOT © m ¹ PLANE
or
( ΣF )PILOT
§ ΣF ·
= mPILOT ¨
¸
© m ¹PLANE
Therefore, we find
§ 3.7 × 104 N ·
¸ = 93 N
4
© 3.1 × 10 kg ¹
( ΣF )PILOT = ( 78 kg ) ¨
2.
REASONING Newton’s second law of motion gives the relationship between the net force
ȈF and the acceleration a that it causes for an object of mass m. The net force is the vector
sum of all the external forces that act on the object. Here the external forces are the drive
force, the force due to the wind, and the resistive force of the water.
SOLUTION We choose the direction of the drive force (due west) as the positive direction.
Solving Newton’s second law ( ΣF = ma ) for the acceleration gives
a=
ΣF +4100 N − 800 N − 1200 N
=
= +0.31 m/s 2
m
6800 kg
The positive sign for the acceleration indicates that its direction is due west .
3.
REASONING According to Newton’s second law, Equation 4.1, the average net force ΣF
is equal to the product of the object’s mass m and the average acceleration a . The average
acceleration is equal to the change in velocity divided by the elapsed time (Equation 2.4),
where the change in velocity is the final velocity v minus the initial velocity v0.
SOLUTION The average net force exerted on the car and riders is
¦ F = ma = m
(
)
v − v0
45 m/s − 0 m/s
= 5.5 × 103 kg
= 3.5 × 104 N
t − t0
7.0 s
4.
REASONING AND SOLUTION Using Equation 2.4 and assuming that t0 = 0 s, we have
for the required time that
v –v0
t=
a
Since ΣF = ma, it follows that
t=
v – v0
ΣF / m
=
m (v – v 0 )
ΣF
=
(5.0 kg )[(4.0 × 10 3
)
]=
m/s – (0 m/s )
4.9 × 10 N
5
4.1 × 10 –2 s
5.
SSM REASONING The net force acting on the ball can be calculated using Newton's
second law. Before we can use Newton's second law, however, we must use Equation 2.9
from the equations of kinematics to determine the acceleration of the ball.
SOLUTION According to Equation 2.9, the acceleration of the ball is given by
a=
v 2 − v20
2x
Thus, the magnitude of the net force on the ball is given by
§ v 2 − v02 ·
ª (45 m/s)2 – (0 m/s) 2 º
¦ F = ma = m ¨
¸ = (0.058 kg) «
» = 130 N
2(0.44 m)
¬
¼
© 2x ¹
6.
REASONING AND SOLUTION The acceleration required is
v2 - v2
a=
0
2x
2
- (15.0 m/s)
=
= - 2.25 m/s2
2(50.0 m)
Newton's second law then gives the magnitude of the net force as
F = ma = (1580 kg)(2.25 m/s2) = 3560 N
7.
SSM REASONING According to Newton's second law of motion, the net force applied
to the fist is equal to the mass of the fist multiplied by its acceleration. The data in the
problem gives the final velocity of the fist and the time it takes to acquire that velocity. The
average acceleration can be obtained directly from these data using the definition of average
acceleration given in Equation 2.4.
SOLUTION The magnitude of the average net force applied to the fist is, therefore,
§ ∆v ·
§ 8.0 m/s – 0 m/s ·
¦ F = ma = m ¨ ¸ = ( 0.70 kg ) ¨
¸ = 37 N
0.15 s
© ∆t ¹
©
¹
8.
REASONING AND SOLUTION From Equation 2.9,
v 2 = v 20 + 2ax
Since the arrow starts from rest, v0 = 0 m/s. In both cases, x is the same so
v12 2a1 x a1
=
=
v 22 2a 2 x a 2
or
v1
=
v2
a1
a2
Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and
v1
F
= 1
v2
F2
or
v 2 = v1
F2
F1
= v1
2F1
F1
= ( 25.0 m/s ) 2 = 35.4 m/s
9.
SSM WWW
REASONING Let due east be chosen as the positive direction. Then,
when both forces point due east, Newton's second law gives
FA + FB = ma1
(1)
ΣF
where a1 = 0.50 m/s . When FA points due east and FB points due west, Newton's second
law gives
2
FA – FB = ma2
(2)
ΣF
2
where a2 = 0.40 m/s . These two equations can be used to find the magnitude of each force.
SOLUTION
a. Adding Equations 1 and 2 gives
FA =
m ( a1 + a2 )
2
=
( 8.0 kg ) ( 0.50 m / s 2 + 0.40 m / s 2 )
2
= 3.6 N
b. Subtracting Equation 2 from Equation 1 gives
FB =
m ( a1 − a2 )
2
=
( 8.0 kg ) ( 0.50 m / s 2 − 0.40 m / s 2 )
2
= 0.40 N