Thevalingam, Donald – Homework 10 – Due: Dec 5 2006, midnight – Inst: Eslami
1
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Multiplechoice questions may continue on
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beFore answering.
The due time is Central
time.
001
(part 1 oF 2) 10 points
A block is released From rest on an inclined
plane and moves 2
.
1 m during the next 4
.
9 s.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
11
kg
μ
k
27
◦
What is the magnitude oF the acceleration
oF the block?
Correct answer: 0
.
174927 m
/
s
2
.
Explanation:
Given :
m
= 11 kg
,
‘
= 2
.
1 m
,
θ
= 27
◦
,
and
t
= 4
.
9 s
.
Consider the Free body diagram For the
block
m
g
sin
θ
N
=
cos
a
mg
The acceleration can be obtained through
kinematics. Since
v
0
= 0,
‘
=
v
0
t
+
1
2
at
2
=
1
2
2
a
=
2
‘
t
2
(1)
=
2(2
.
1 m)
(4
.
9 s)
2
=
0
.
174927 m
/
s
2
.
002
(part 2 oF 2) 10 points
What is the coe±cient oF kinetic Friction
μ
k
For the incline?
Correct answer: 0
.
489493 .
Explanation:
Applying Newton’s Second Law oF Motion
X
F
i
=
ma
and Eq. 1, the sine component
oF the weight acts down the plane and Friction
acts up the plane. The block slides down the
plane, so
=
sin
θ

μ
k
cos
θ
2
‘
t
2
=
g
‡
sin
θ

μ
k
cos
θ
·
2
‘
=
g t
2
‡
sin
θ

μ
k
cos
θ
·
μ
k
=
g t
2
sin
θ

2
‘
g t
2
cos
θ
(2)
= tan
θ

2
‘
g t
2
cos
θ
= tan27
◦

.
1 m)
(9
.
8 m
/
s
2
)(4
.
9 s)
2
cos27
◦
=
0
.
489493
.
keywords:
003
(part 1 oF 1) 10 points
There is Friction between the block and the
table.
The suspended 2 kg mass on the leFt is
moving up, the 3 kg mass slides to the right
on the table, and the suspended mass 5 kg on
the right is moving down.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
2 kg
3 kg
5 kg
μ
= 0
.
15
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View Full DocumentThevalingam, Donald – Homework 10 – Due: Dec 5 2006, midnight – Inst: Eslami
2
What is the magnitude of the acceleration
of the system?
Correct answer: 2
.
499 m
/
s
2
.
Explanation:
m
1
m
2
m
3
μ
a
Let :
m
1
= 2 kg
,
m
2
= 3 kg
,
m
3
= 5 kg
,
and
μ
= 0
.
15
.
Basic Concepts:
The acceleration
a
of
each mass is the same, but the tensions in the
two strings will be diFerent.
F
net
=
ma
6
= 0
Solution:
Let
T
1
be the tension in the left
string and
T
2
be the tension in the right string.
Consider the free body diagrams for each
mass
T
1
m
1
g
a
T
2
m
3
g
a
T
1
T
2
N
μ
N
a
m
2
g
±or the mass
m
1
,
T
1
acts up and the weight
m
1
g
acts down, with the acceleration
a
di
rected upward, so
F
net
1
=
m
1
a
=
T
1

m
1
g .
(1)
±or the mass on the table,
a
is directed to
the right,
T
2
acts to the right,
T
1
acts to the
left, and the motion is to the right so that the
frictional force
μm
2
g
acts to the left and
F
net
2
=
m
2
a
=
T
2

T
1

2
g .
(2)
±or the mass
m
3
,
T
2
acts up and the weight
m
3
g
acts down, with the acceleration
a
di
rected downward, so
F
net
3
=
m
3
a
=
m
3
g

T
2
.
(3)
Adding these equations yields
(
m
1
+
m
2
+
m
3
)
a
=
m
3
g

2
g

m
1
g
a
=
m
3

2

m
1
m
1
+
m
2
+
m
3
g
=
5 kg

(0
.
15)(3 kg)

2 kg
2 kg + 3 kg + 5 kg
×
(9
.
8 m
/
s
2
)
=
2
.
499 m
/
s
2
.
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 Spring '08
 Turner
 Physics, Force, Friction, Work, Correct Answer, kg, Thevalingam

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