hernandez (ejh742) – oldhomework 13 – Turner – (58120)
1
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001
10.0 points
The pulley system is in equilibrium, and the
pulleys are weightless and frictionless.
The
spring constant is 2 N
/
cm and the suspended
mass is 17 kg.
17 kg
2 N
/
cm
How much will the spring stretch?
The
acceleration of gravity is 9
.
8 m
/
s
2
.
Correct answer: 27
.
7667 cm.
Explanation:
Let :
k
1
= 2 N
/
cm
,
m
= 17 kg
,
and
g
= 9
.
8 m
/
s
2
.
m
1
k
1
T
1
T
1
T
1
T
2
The existence of a spring in a string defines
the tension in the string because the force
(tension) exerted by a spring is
T
=
F
=
k x .
At any
point in
the system
summationdisplay
F
up
=
summationdisplay
F
down
.
At pulley 1,
T
2
= 2
T
1
.
At the suspended mass,
T
2
+
T
1
=
m g
3
T
1
=
m g
3
k
1
x
1
=
m g
x
1
=
m g
3
k
1
=
(17 kg)
(
9
.
8 m
/
s
2
)
3 (2 N
/
cm)
=
27
.
7667 cm
.
002
(part 1 of 2) 10.0 points
The two blocks are connected by a light
string that passes over a frictionless pulley
with a negligible mass.
The block of mass
m
1
lies on a rough horizontal surface with a
constant coefficient of kinetic friction
μ
. This
block is connected to a spring with spring
constant
k
. The second block has a mass
m
2
.
The system is released from rest when the
spring is unstretched, and
m
2
falls a distance
h
before it reaches the lowest point.
Note:
When
m
2
is at the lowest point, its
velocity is zero.
m
1
m
2
k
m
1
m
2
h
h
μ
Consider the moment when
m
2
has de
scended by a distance
s
, where
s
is less than
h
.
At this moment the sum of the kinetic energy
for the two blocks
K
is given by
1.
K
=
m
2
g s

1
2
k s
2

μ
(
m
1
+
m
2
)
g s .
2.
K
=

m
2
g s
+
1
2
k s
2
+
μ
(
m
1
+
m
2
)
g s .
3.
K
=
m
2
g s

1
2
k s
2

μ m
1
g s .
correct
4.
K
= (
m
1
+
m
2
)
g s
+
1
2
k s
2

μ m
1
g s .
5.
K
=

(
m
1
+
m
2
)
g s
+
1
2
k s
2
+
μ m
1
g s .
6.
K
= (
m
1
+
m
2
)
g s

1
2
k s
2
+
μ m
1
g s .
7.
K
=

m
2
g s
+
1
2
k s
2
+
μ m
1
g s .
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hernandez (ejh742) – oldhomework 13 – Turner – (58120)
2
8.
K
= (
m
1
+
m
2
)
g s

1
2
k s
2

μ m
1
g s .
Explanation:
Basic Concepts:
WorkEnergy Theorem
Spring Potential Energy
Frictional
Force
according
to
the
Work
Energy Theorem
Solution:
W
ext
A
→
B
= (
K
B

K
A
) + (
U
g
B

U
g
A
)
+ (
U
sp
B

U
sp
A
) +
W
dis
A
→
B
For
the
present
case,
the
external
work
W
ext
A
→
B
= 0,
A
corresponds to the initial state
and
B
the state where
m
2
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 Spring '08
 Turner
 Physics, Force, Kinetic Energy, Mass, Work, Correct Answer, kg

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