hernandez (ejh742) – homework 10 – Turner – (58120)
1
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12
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001
10.0 points
In a certain system of units, 35
.
8 thumbs = 1
meter, 1 pause = 31
.
4 s, and 1 glob = 21
.
7 kg.
The unit of force in this system is a perk (p).
How many kiloperks (kp) are in 64 N?
Correct answer: 104
.
103 kp.
Explanation:
The unit of measure for force in the new
system is 1 perk = 1
glob
·
thumb
pause
2
.
f
=
parenleftbigg
64
kg
·
m
s
2
parenrightbigg
·
1 glob
21
.
7 kg
·
35
.
8 thumbs
1 m
·
parenleftbigg
31
.
4 s
1 pause
parenrightbigg
2
·
1 kp
1000 p
=
104
.
103 kp
.
002
10.0 points
A person weighing 0
.
9 kN rides in an elevator
that has a downward acceleration of 3 m
/
s
2
.
The acceleration of gravity is 9
.
8 m
/
s
2
.
What is the magnitude of the force of the
elevator floor on the person?
Correct answer: 0
.
62449 kN.
Explanation:
Basic Concepts:
summationdisplay
vector
F
=
mvectora
vector
W
=
mvectorg
Solution:
Since
W
=
m g
,
F
net
=
ma
=
W −
f
f
=
W −
ma
=
W
parenleftbigg
1
−
a
g
parenrightbigg
= (0
.
9 kN)
parenleftbigg
1
−
3 m
/
s
2
9
.
8 m
/
s
2
parenrightbigg
= 0
.
62449 kN
.
003
(part 1 of 2) 10.0 points
A pulley is massless and frictionless.
1 kg,
1 kg, and 8 kg masses are suspended as in the
figure.
3
.
2 m
21
.
9 cm
ω
1 kg
1 kg
8 kg
T
2
T
1
T
3
What is the tension
T
1
in the string be
tween the two blocks on the lefthand side
of the pulley?
The acceleration of gravity is
9
.
8 m
/
s
2
.
Correct answer: 15
.
68 N.
Explanation:
Let :
R
= 21
.
9 cm
,
m
1
= 1 kg
,
m
2
= 1 kg
,
m
3
= 8 kg
,
and
h
= 3
.
2 m
.
Consider the free body diagrams
1 kg
1 kg
8 kg
T
1
T
2
T
3
m
1
g
T
1
m
2
g
m
3
g
a
a
For each mass in the system
vector
F
net
=
mvectora .
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hernandez (ejh742) – homework 10 – Turner – (58120)
2
Since the string changes direction around
the pulley, the forces due to the tensions
T
2
and
T
3
are in the same direction (up).
The
acceleration of the system will be down to
the right (
m
3
> m
1
+
m
2
), and each mass in
the system accelerates at the same rate (the
string does not stretch). Let this acceleration
rate be
a
and the tension over the pulley be
T
≡
T
2
=
T
3
.
For the lower lefthand mass
m
1
the accel
eration is up and
T
1
−
m
1
g
=
m
1
a .
(1)
For the upper lefthand mass
m
2
the acceler
ation is up and
T
−
T
1
−
m
2
g
=
m
2
a .
(2)
For the righthand mass
m
3
the acceleration
is down and
−
T
+
m
3
g
=
m
3
a .
(3)
Adding Eqs. (1), (2), and (3), we have
(
m
3
−
m
1
−
m
2
)
g
= (
m
1
+
m
2
+
m
3
)
a
(4)
a
=
m
3
−
m
1
−
m
2
m
1
+
m
2
+
m
3
g
(5)
=
8 kg
−
1 kg
−
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 Spring '08
 Turner
 Physics, Acceleration, Friction, Mass, Work, kg

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