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hiarker (srh959) – homework 10 – Turner – (58220)
1
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001
10.0 points
In a certain system oF units, 38
.
5 thumbs = 1
meter, 1 pause = 43
.
4 s, and 1 glob = 17
.
1 kg.
The unit oF Force in this system is a perk (p).
How many kiloperks (kp) are in 42
.
6 N?
Correct answer: 180
.
657 kp.
Explanation:
The unit oF measure For Force in the new
system is 1 perk = 1
glob
·
thumb
pause
2
.
f
=
p
42
.
6
kg
·
m
s
2
P
·
1 glob
17
.
1 kg
·
38
.
5 thumbs
1 m
·
p
43
.
4 s
1 pause
P
2
·
1 kp
1000 p
=
180
.
657 kp
.
002
10.0 points
A person weighing 0
.
5 kN rides in an elevator
that has a downward acceleration oF 1
.
7 m
/
s
2
.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
What is the magnitude oF the Force oF the
elevator ±oor on the person?
Correct answer: 0
.
413265 kN.
Explanation:
Basic Concepts:
s
v
F
=
mva
v
W
=
mvg
Solution:
Since
W
=
m g
,
F
net
=
ma
=
W −
f
f
=
W −
ma
=
W
p
1
−
a
g
P
= (0
.
5 kN)
p
1
−
1
.
7 m
/
s
2
9
.
8 m
/
s
2
P
= 0
.
413265 kN
.
003
(part 1 oF 2) 10.0 points
A pulley is massless and Frictionless. 1 kg,
1 kg, and 11 kg masses are suspended as in
the fgure.
3
.
8 m
21
.
7 cm
ω
1 kg
1 kg
11 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: 16
.
5846 N.
Explanation:
Let :
R
= 21
.
7 cm
,
m
1
= 1 kg
,
m
2
= 1 kg
,
m
3
= 11 kg
,
and
h
= 3
.
8 m
.
Consider the Free body diagrams
1 kg
1 kg
11 kg
T
1
2
3
m
g
a
²or each mass in the system
v
F
net
=
mva .
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View Full Documenthiarker (srh959) – homework 10 – Turner – (58220)
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
=
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 Spring '09
 NUI
 Physics, Work

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