Download Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25

Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
1
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Central time.
Chapters 3 and 4 problems.
2.44 × 108 m. It takes 26.4 days for the Moon
to complete one revolution about the planet.
Find the mean orbital speed of the Moon.
Correct answer: 672.128 m/s.
Explanation:
Dividing the length C = 2πr of the trajectory
of the Moon by the time
001 (part 1 of 1) 0 points
A ball on the end of a string is whirled around
in a horizontal circle of radius 0.454 m. The
plane of the circle is 1.49 m above the ground.
The string breaks and the ball lands 2.71 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9.8 m/s2 .
Find the centripetal acceleration of the ball
during its circular motion.
Correct answer: 53.1977 m/s2 .
Explanation:
In order to find the centripetal acceleration
of the ball, we need to find the initial velocity
of the ball. Let y be the distance above the
ground. After the string breaks, the ball has
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
T = 26.4 days = 2.28096 × 106 s
1
y = gt2
2
Solving for t,
⇒t=
r
2y
g
Let d be the distance traveled by the ball.
Then
d
d
vx = = q
2y
t
g
Hence, the centripetal acceleration of the ball
during its circular motion is
ac =
vx2
d2 g
=
= 53.1977 m/s2
r
2yr
002 (part 1 of 2) 0 points
The orbit of a Moon about its planet is approximately circular, with a mean radius of
of one revolution (in seconds!), we obtain that
the mean orbital speed of the Moon is
C
2πr
v=
=
T
T
2 π (2.44 × 108 m )
=
2.28096 × 106 s
= 672.128 m/s .
003 (part 2 of 2) 0 points
Find the Moon’s centripetal acceleration.
Correct answer: 0.00185146 m/s2 .
Explanation:
Since the magnitude of the velocity is constant, the tangential acceleration of the Moon
is zero. For the centripetal acceleration we
use the formula
v2
ac =
r
(672.128 m/s )2
=
2.44 × 108 m
= 0.00185146 m/s2 .
004 (part 1 of 3) 0 points
A train slows down at a constant rate as it
rounds a sharp circular horizontal turn. Its
initial speed is not known. It takes 16.4 s to
slow down from 78 km/h to 30 km/h. The
radius of the curve is 180 m.
As the train goes around the turn, what is
the magnitude of the tangential component of
the acceleration?
Correct answer: 0.813008 m/s2 .
Explanation:
Basic concepts
The components of normal and tangential acceleration are given by
∆v
at =
∆t
Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
2
¶¸2
·
µ
v2
1000
ar =
.
(50 km/h)
r
3600
=
Solution: Since the tangential acceleration
180 m
changes the magnitude of the velocity and the
= 1.07167 m/s2 .
normal acceleration changes the direction of
the velocity, with the given initial and final veTherefore, the magnitude of the total accelerlocities vi = 78 km/h and vf = 30 km/h with
ation is
the time interval ∆t = 16.4 s, the tangential
q
acceleration is
a = a2t + a2r
∆v
~at =
q
∆t
=
(0.813008 m/s2 )2 + (1.07167 m/s2 )2
vf − v i
=
= 1.34516 m/s2 .
¸
· ∆t
(78 km/h) − (30 km/h)
=
16.4 s
¶µ
¶
µ
1h
1000 m
007 (part 1 of 2) 5 points
×
1 km
3600 s
Two masses m1 and m2 are connected in the
2
manner
shown.
= −0.813008 m/s
k~at k = 0.813008 m/s2 .
T1
m1
g
a=
005 (part 2 of 3) 0 points
T2
2
As the train goes around the turn, what is
m2
the sign of the tangential component of the
acceleration?
The system is accelerating downward with
g
Take the direction in which the train is
acceleration of magnitude .
2
moving to be the positive direction.
Determine T2 .
¶
µ
1. − correct
1
m2 − m 1 g
1. T2 =
2
2. 0
3
2. T2 = m2 g
2
3. +
4. Not enough information
Explanation:
See previous solution.
006 (part 3 of 3) 0 points
At the moment the train’s speed is 50 km/h,
what is the magnitude of the total acceleration?
Correct answer: 1.34516 m/s2 .
Explanation:
When v = 50 km/h, r = 180 m, the normal
acceleration is
v2
at =
r
3. T2 = m2 g
1
(m2 − m1 ) g
2
¶
µ
1
5. T2 = m2 − m1 g
2
1
6. T2 = m2 g correct
2
4. T2 =
Explanation:
T2
m2
m2 g
a=
g
2
Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
From Newton’s Second Law,
g
m2 g − T 2 = m 2 .
2
So
3
3. The horse can pull the buggy forward only
if the horse weighs more than the buggy.
4. The horse pulls before the buggy has time
to react so they move forward.
1
g
T2 = m 2 g − m 2 = m 2 g .
2
2
008 (part 2 of 2) 5 points
Determine T1 .
Explanation:
According to Newton’s third law, the force
of the buggy on the horse is as strong as the
force of the horse on the buggy.
1
1. T1 = (m1 + m2 ) g correct
2
2. None of these
010 (part 1 of 1) 10 points
The pulleys are frictionless and massless.
The acceleration of gravity is 9.8 m/s2 .
The system is in equilibrium.
5
(m1 + m2 ) g
2
3
4. T1 = (m1 + m2 ) g
2
3. T1 =
5. T1 = (m1 + m2 ) g
6. T1 = 2 (m1 + m2 ) g
Explanation:
T1
m1 + m 2
a=
g
2
7.2 kg
(m1 + m2 ) g
11 kg
Consider m1 and m2 as a whole object with
mass (m1 + m2 ), then
(m1 + m2 ) g − T1 = (m1 + m2 )
So
T1 =
g
.
2
T
37 kg
1
(m1 + m2 ) g .
2
009 (part 1 of 1) 0 points
Consider a buggy being pulled by a horse.
Which is correct?
Find the tension T .
Correct answer: 1483.72 N.
Explanation:
1. The force on the buggy is as strong as the
force on the horse. The horse is joined to the
Earth by flat hoofs, while the buggy is free to
roll on its round wheels. correct
2. The horse pulls forward slightly harder
than the buggy pulls backward on the horse,
so they move forward.
Let :
m1 = 37 kg ,
m2 = 7.2 kg ,
m3 = 11 kg .
and
Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
4
constant speed, the scale reads U . When
the elevator is moving down with the same
constant speed, the scale reads D.
T3
m2
T4
T2
m3
T1
T
m1
Scale
The mass m1 defines the tension T1
T1 = m 1 g .
At the right-hand pulley, T1 acts down on
either side of the pulley and T2 acts up, so
T2 = 2 T 1 = 2 m 1 g .
At the mass m2 , T3 acts up, and m2 g and T2
act down, so
T3 = m 2 g + T 2 = m 2 g + 2 m 1 g .
At the left-hand pulley, T3 acts up on either
side of the pulley and T4 acts down, so
T4 = 2 T 3
= 2 m 2 g + 4 m 1 g 2 T3
= 2 m2 g + 4 m1 g .
At the mass m3 ,
Choose the correct relationship between the
three scale readings.
1. U > W > D
2. U = W = D correct
3. U = D,
U >W
4. U < W < D
5. U = D,
U <W
Explanation:
Consider the free body diagram for each
the case where the elevator is accelerating
down (left) and up (right). The man is represented as a sphere and the scale reading is
represented as T .
T
a
T
a
T4 = T + m 3 g
=⇒ T = T4 − m3 g = (2 m2 + 4 m1 − m3 ) g .
011 (part 1 of 2) 0 points
A man stands on a scale in an elevator. When
the elevator is at rest, the scale reads W.
When the elevator is moving upward with a
mg
mg
Note: The acceleration in all three cases
is zero, so according to Newton’s Second Law
the force measured by the scale is Fscale = W.
Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
012 (part 2 of 2) 0 points
Suppose the elevator now accelerates downward at a constant rate of 0.37 g.
What is the ratio of the new scale reading
to the value W of the scale reading when the
elevator is at rest?
Fscale
= 1.37
W
Fscale
2.
= 2.37
W
Fscale
= 1.0
3.
W
F
4. scale = 0.63 correct
W
F
5. scale = 0.37
W
Fscale
= 1.63
6.
W
Fscale
7.
= 2.0
W
Explanation:
Newtons’ Second Law in this case reads
Explanation:
Given :
where W is the weight of the man. Therefore,
Fscale
= 1 − 0.37 = 0.63 .
W
013 (part 1 of 3) 4 points
A block of mass 1.39231 kg lies on a frictionless table, pulled by another mass 4.42328 kg
under the influence of Earth’s gravity.
The acceleration of gravity is 9.8 m/s2 .
1.39231 kg
µ=0
4.42328 kg
What is the magnitude of the net external
force F acting on the two masses?
Correct answer: 43.3481 N.
m1 = 1.39231 kg ,
m2 = 4.42328 kg ,
µ = 0.
a
1.
W − Fscale = 0.37 W ,
5
T
T
m1
N
and
m1 g
m2
a
m2 g
The net force on the system is simply the
weight of m2 .
Fnet = m2 g
= (4.42328 kg) (9.8 m/s2 )
= 43.3481 N .
014 (part 2 of 3) 3 points
What is the magnitude of the acceleration a
of the two masses?
Correct answer: 7.45378 m/s2 .
Explanation:
From Newton’s second law,
Fnet = m2 g = (m1 + m2 ) a .
Solving for a,
m2
a=
g
m1 + m 2
4.42328 kg
(9.8 m/s2 )
=
1.39231 kg + 4.42328 kg
= 7.45378 m/s2 .
015 (part 3 of 3) 3 points
What is the magnitude of the tension T of the
rope between the two masses?
Correct answer: 10.378 N.
Explanation:
Analyzing the horizontal forces on block
m1 , we have
X
Fx : T = m 1 a
= (1.39231 kg) (7.45378 m/s2 )
= 10.378 N .
Answer, Key – Homework 4 – David McIntyre – 45123 – Mar 25, 2004
016 (part 1 of 3) 4 points
A 4.14 kg block slides down a smooth, frictionless plane having an inclination of 35◦ .
The acceleration of gravity is 9.8 m/s2 .
6
along that axis. So, by Newton’s second law,
X
Fx = m a = m g sin θ
Thus
a = g sin θ
With our particular value of θ,
8
2.9
a = (9.8 m/s2 ) sin 35◦ = 5.62105 m/s2
g
4k
m
4.1
µ
=
0
35◦
017 (part 2 of 3) 3 points
What is the block’s speed when, starting from
rest, it has traveled a distance of 2.98 m along
the incline.
Correct answer: 5.78804 m/s.
Explanation:
Since v0 = 0, a = 5.62105 m/s2 and L =
2.98 m,
Find the acceleration of the block.
Correct answer: 5.62105 m/s2 .
Explanation:
Given :
vf2 = v02 + 2 a (x − x0 )
m = 4.14 kg ,
θ = 35◦ , and
µs = 0 .
= 2 (5.62105 m/s2 ) (2.98 m)
vf = 5.78804 m/s .
Consider the free body diagram for the
block
N
nθ
m
i
gs
N
s
co
µ Nm g
=
θ
W = mg
Basic Concepts:
Fx,net = F cos θ − W|| = 0
W|| = m g sin θ = m a
Motion has a constant acceleration. Recall
the kinematics of motion with constant acceleration.
Solution: Because the block slides down
along the plane of the ramp, it seems logical
to choose the x-axis in this direction. Then
the y-axis must emerge perpendicular to the
ramp, as shown.
Let us now examine the forces in the xdirection. Only the weight has a component
It is very important to note at this point that
neither of these values depended on the mass
of the block. This may seem odd at first, but
recall what Galileo discovered 300 years ago –
objects of differing mass fall at the same rate.
018 (part 3 of 3) 3 points
What is the magnitude of the perpendicular
force that the block exerts on the surface of
the plane at a distance of 2.98 m down the
incline?
Correct answer: 33.2346 N.
Explanation:
By examining the free body diagram again,
we see that the force in the y direction is given
by
X
F = −m g cos θ = FN .
=⇒
| FN | = m g cos θ
= 4.14 kg (9.8 m/s2 ) cos 35◦
= 33.2346 N.