seunsom (avs294) – HW5 – bohm – (59970)
1
This
printout
should
have
14
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
5.0 points
Two 15 N forces and a 30 N force act on a
hanging box as shown.
15 N
15 N
30 N
Will the box experience acceleration?
1.
No; it is balanced.
2.
Yes; upward.
3.
Yes; downward.
correct
4.
Unable to determine without the angle.
Explanation:
The horizontal components of the two 15 N
forces cancel, leaving an upward force that is
less than 30 N. Thus, the net force on the box
is down, causing it to accelerate downward.
002
10.0 points
Two forces are the only forces acting on a
6
.
1 kg object which moves with an accelera
tion of 3
.
2 m
/
s
2
in the positive
y
direction.
One of the forces acts in the positive
x
direc
tion and has a magnitude of 13 N.
What is the magnitude of the other force
f
2
?
Correct answer: 23
.
4527 N.
Explanation:
Basic Concepts:
summationdisplay
F
=
m
a
Solution:
f
1
f
2
f
vector
f
2
is the hypotenuse of a right triangle, so
vector
f
2
=
radicalBig
f
2
+
f
1
2
003
(part 1 of 2) 10.0 points
A 32600 kg sailboat experiences an eastward
force 32600 N due to the tide pushing its hull
while the wind pushes the sails with a force of
57900 N directed toward the northwest (45
◦
westward of North or 45
◦
northward of West).
What is the magnitude of the resultant ac
celeration of the sailboat?
Correct answer: 1
.
28167 m
/
s
2
.
Explanation:
According to Newton’s Second Law,
mvectora
=
vector
F
net
=
vector
F
wind
+
vector
F
tide
.
To find the magnitude of the net force, we
draw the parallelogram for vector addition
vector
F
wind
vector
F
tide
vector
F
net
θ
α
NW
W
E
Note that 135
◦
is the angle between the tide
and the wind forces. Use the Law of Cosines:
F
2
net
=
F
2
wind
+
F
2
tide

2
F
wind
F
tide
cos
θ
= (57900 N)
2
+ (32600 N)
2

2 (57900 N)(32600 N) cos 135
◦
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 Spring '08
 Kaplunovsky
 Force, Correct Answer, kg

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