Bell, Evan – Homework 3 – Due: Feb 2 2005, 11:00 pm – Inst: Furlan
1
This
printout
should
have
32
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 2) 10 points
Sue and Jenny kick a soccer ball at exactly
the same time.
Sue’s foot exerts a force of
65
.
9 N to the north.
Jenny’s foot exerts a
force of 98
.
1 N to the east.
a) What is the magnitude of the resultant
force on the ball?
Correct answer: 118
.
18 N.
Explanation:
f
j
F
f
s
θ
The magnitude of the resultant can be cal
culated using the Pythagorean Theorem:
F
=
q
f
2
s
+
f
2
j
=
q
(65
.
9 N)
2
+ (98
.
1 N)
2
= 118
.
18 N
002
(part 2 of 2) 10 points
b) What is the direction of the resultant force
(measured from East)?
Correct answer: 33
.
8918
◦
.
Explanation:
Jenny’s force is directed east, so
f
s
is the
side opposite
θ
and
f
j
is the side adjacent,
and
tan
θ
=
f
s
f
j
θ
= arctan
f
s
f
j
¶
= arctan
65
.
9 N
98
.
1 N
¶
= 33
.
8918
◦
003
(part 1 of 1) 10 points
Beth, a construction worker, attempts to pull
a stake out of the ground by pulling on a rope
that is attached to the stake. The rope makes
an angle of 59
.
9
◦
with the horizontal.
If Beth exerts a force of 137
.
2 N on the
rope, what is the magnitude of the upward
component of the force acting on the stake?
Correct answer: 118
.
699 N.
Explanation:
θ
θ
F
F
f
v
θ
is the angle the rope makes with the
horizontal, so the upward component of the
force is the side opposite
θ
, and
sin
θ
=
f
v
F
f
v
=
F
sin
θ
= (137
.
2 N) sin 59
.
9
◦
= 118
.
699 N
004
(part 1 of 1) 10 points
A force is applied to a 2
.
2 kg mass and pro
duces 2
.
2 m
/
s
2
acceleration.
What acceleration would be produced by
the same force applied to a 7
.
9 kg mass?
Correct answer: 0
.
612658 m
/
s
2
.
Explanation:
The force acting on both masses is the
same, so
F
=
m
1
a
1
=
m
2
a
2
a
2
=
m
1
a
1
m
2
=
(2
.
2 kg) (2
.
2 m
/
s
2
)
7
.
9 kg
= 0
.
612658 m
/
s
2
005
(part 1 of 1) 10 points
A person weighing 0
.
8 kN rides in an elevator
that has a downward acceleration of 1
.
9 m
/
s
2
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
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Bell, Evan – Homework 3 – Due: Feb 2 2005, 11:00 pm – Inst: Furlan
2
What is the magnitude of the force of the
elevator floor on the person?
Correct answer: 0
.
644898 kN.
Explanation:
Basic Concepts:
X
~
F
=
m~a
~
W
=
m~g
Solution:
Since
W
=
m g
,
F
net
=
ma
=
W 
f
f
=
W 
ma
=
W
1

a
g
¶
= (0
.
8 kN)
1

1
.
9 m
/
s
2
9
.
8 m
/
s
2
¶
= 0
.
644898 kN
.
006
(part 1 of 3) 10 points
The following 3 questions refer to a toy car
which is given a quick push so that it rolls up
an inclined ramp. After it is released, it rolls
up, reaches its highest point and rolls back
down again.
Friction is so small it can be
ignored.
Indicate the
net force
acting on the
car for each of the cases described below.
The car is moving up the ramp after it is
released.
1.
Net
constant
force
down
the ramp.
cor
rect
2.
Net
decreasing
force
up
the ramp.
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 Spring '07
 GALLIS,MICHAELRO
 Force, Friction, Work, Correct Answer, kg

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