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hinojosa (jlh3938) – homework 13 – Turner – (58185)
1
This printout should have 11 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
There is Friction between the block and the
table.
The suspended 2 kg mass on the leFt is
moving up, the 4 kg mass slides to the right
on the table, and the suspended mass 4 kg on
the right is moving down.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
2 kg
4 kg
4 kg
μ
= 0
.
11
What is the magnitude oF the acceleration
oF the system?
Correct answer: 1
.
5288 m
/
s
2
.
Explanation:
m
1
m
2
m
3
μ
a
Let :
m
1
= 2 kg
,
m
2
= 4 kg
,
m
3
= 4 kg
,
and
μ
= 0
.
11
.
Basic Concepts:
The acceleration
a
oF
each mass is the same, but the tensions in the
two strings will be di±erent.
F
net
=
ma
n
= 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
²or 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)
²or 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
−
μm
2
g .
(2)
²or 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
−
μm
2
g
−
m
1
g
a
=
m
3
−
μm
2
−
m
1
m
1
+
m
2
+
m
3
g
=
4 kg
−
(0
.
11) (4 kg)
−
2 kg
2 kg + 4 kg + 4 kg
×
(9
.
8 m
/
s
2
)
=
1
.
5288 m
/
s
2
.
002
10.0 points
The suspended 2
.
1 kg mass on the right is
moving up, the 1
.
5 kg mass slides down the
ramp, and the suspended 8
.
5 kg mass on the
leFt is moving down. There is Friction between
the block and the ramp.
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View Full Documenthinojosa (jlh3938) – homework 13 – Turner – (58185)
2
The acceleration of gravity is 9
.
8 m
/
s
2
.
The
pulleys are massless and frictionless.
1
.
5 kg
μ
= 0
2
25
◦
8
.
5 kg
2
.
1 kg
What is the tension in the cord connected
to the 8
.
5 kg block?
Correct answer: 36
.
7481 N.
Explanation:
Let :
m
1
= 2
.
1 kg
,
m
2
= 1
.
5 kg
,
m
3
= 8
.
5 kg
,
and
θ
= 25
◦
.
Basic Concept:
F
net
=
ma
n
= 0
Solution:
The acceleration
a
of each mass is
the same, but the tensions in the two strings
will be diFerent.
Let
T
1
be the tension in
the right string and
T
3
the tension in the left
string.
Consider the free body diagrams for each
mass
T
3
m
3
g
a
T
1
m
1
g
a
T
3
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
F
net
1
=
m
1
a
=
T
1
−
m
1
g
(1)
±or the mass on the table, the parallel compo
nent of its weight is
mg
sin
θ
and the perpen
dicular component of its weight is
mg
cos
θ
.
(
N
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 Spring '08
 Turner
 Physics, Friction, Work

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