Quach, Phuc – Homework 6 – Due: Sep 29 2006, 5:00 pm – Inst: Mark Rupright
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The due time is Central
time.
001
(part 1 of 1) 10 points
Vector
~
A
has components
A
x
=

5
,
A
y
= 3
.
9
,
A
z
= 2
.
1
,
while vector
~
B
has components
B
x
= 6
.
2
,
B
y
=

5
.
3
,
B
z
= 4
.
9
.
What is the angle
θ
AB
between these vec
tors? (Answer between 0
◦
and 180
◦
.)
Correct answer: 130
.
62
◦
.
Explanation:
Note:
The magnitude of vector
~
X
is
k
X
k
.
Consider two formulæ for the scalar prod
uct
~
A
·
~
B
of two vectors:
~
A
·
~
B
=
A
x
B
x
+
A
y
B
y
+
A
z
B
z
(1)
in terms of the two vectors’ components, and
also
~
A
·
~
B
=
k
~
A
k k
~
B
k
cos
θ
AB
(2)
in term of their magnitudes and the angle be
tween them. Given the data, we immediately
calculate
k
~
A
k
=
q
A
2
x
+
A
2
y
+
A
2
z
= 6
.
67982
,
(3)
k
~
B
k
=
q
B
2
x
+
B
2
y
+
B
2
z
= 9
.
51525
,
(4)
and using eq. (1),
~
A
·
~
B
=

41
.
38
.
(5)
Hence, according to eq. (2),
cos
θ
AB
=
~
A
·
~
B
k
~
A
k k
~
B
k
=

0
.
651037
(6)
and therefore
θ
AB
= arccos(

0
.
651037)
=
130
.
62
◦
.
(7)
Two vectors always lie in a plane.
When
these two vectors are plotted in this plane, we
have
A
B
130
.
62
◦
keywords:
002
(part 1 of 1) 10 points
You drag a(n) 16
.
9 kg steamer trunk over a
rough surface by a constant force of 79
.
2 N
acting at an angle of 32
.
1
◦
above the horizon
tal.
You move the trunk over a distance of
13
.
7 m in a straight line, and the coefficient of
kinetic friction is 0
.
122.
The acceleration of gravity is 9
.
8 m
/
s
2
.
16
.
9 kg
μ
= 0
.
122
79
.
2 N
32
.
1
◦
How much is the work done on the block by
the net force?
Correct answer: 712
.
688 J.
Explanation:
F
θ
m g
n
f
k
Work is
W
=
~
F
·
~s
, where
~s
is the distance
traveled. In this problem
~s
= 5ˆ
x
is only in the
ˆ
x
direction.
W
F
=
~
F
x
·
~s
x
=
F s
x
cos
θ
= (79
.
2 N) (13
.
7 m) cos 32
.
1
◦
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Quach, Phuc – Homework 6 – Due: Sep 29 2006, 5:00 pm – Inst: Mark Rupright
2
= 919
.
161 J
.
To find the frictional force,
F
friction
=
μ
N
,
we need to find
N
from vertical force balance.
Note:
N
is in the same direction as the
y
component of
F
and opposite the force of
gravity. Thus
F
sin
θ
+
N
=
m g ,
so that
N
=
m g

F
sin
θ .
Thus the friction force is
~
F
friction
=

μ
N
ˆ
x
=

μ
(
m g

F
sin
θ
) ˆ
x ,
and the work done by friction is
W
μ
=
~
F
friction
·
~s
=

F
f
 
s

=

μ
(
m g

F
sin
θ
)
s
x
=

0
.
122
h
(16
.
9 kg) (9
.
8 m
/
s
2
)

(79
.
2 N) sin 32
.
1
◦
i
(13
.
7 m)
=

206
.
473 J
.
Since the normal force points in the ˆ
y
direc
tion and
~s
is in the ˆ
x
direction
W
N
=
~
N ·
~s
=
N
ˆ
y
·
ˆ
x
= 0
.
The net work done on the body is the alge
braic sum of the work done by the external
force
F
and the work done by the frictional
force
W
net
=
W
F
+
W
μ
= (919
.
161 J) + (

206
.
473 J)
=
712
.
688 J
.
keywords:
003
(part 1 of 1) 10 points
A force
~
F
=
F
x
ˆ
ı
+
F
y
ˆ
acts on an object with
F
x
= 4 N and
F
y
= 2 N. The angle between
~
F
and the displacement vector
~s
is 33
◦
, and
68
.
3 J of work is done by
~
F
.
Find the magnitude of
~s
.
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