homework 05 – KIM, JI – Due: Oct 3 2007, 4:00 am
1
Question 1, chap 8, sect 1.
part 1 of 1
10 points
A 8
.
2 kg mass is attached to a light cord
that passes over a massless, frictionless pulley.
The other end of the cord is attached to a
2
.
8 kg mass.
The acceleration of gravity is 9
.
8 m
/
s
2
.
6
.
1 m
ω
8
.
2 kg
2
.
8 kg
Use conservation of energy to determine the
final speed of the first mass after it has fallen
(starting from rest) 6
.
1 m.
Correct answer: 7
.
66114 m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
1
= 8
.
2 kg
,
m
2
= 2
.
8 kg
,
and
ℓ
= 6
.
1 m
.
Consider the free body diagrams
8
.
2 kg
2
.
8 kg
T
T
m
1
g
m
2
g
a
a
Let the figure represent the initial config
uration of the pulley system (before
m
1
falls
down).
From the conservation of energy
K
i
+
U
i
=
K
f
+
U
f
0 +
m
1
g ℓ
=
m
2
g ℓ
+
1
2
m
1
v
2
+
1
2
m
2
v
2
(
m
1
−
m
2
)
g ℓ
=
1
2
(
m
1
+
m
2
)
v
2
Therefore
v
=
radicalBigg
(
m
1
−
m
2
)
(
m
1
+
m
2
)
2
g ℓ
=
bracketleftbigg
8
.
2 kg
−
2
.
8 kg
8
.
2 kg + 2
.
8 kg
×
2 (9
.
8 m
/
s
2
)(6
.
1 m)
bracketrightbigg
1
/
2
= 7
.
66114 m
/
s
.
Question 2, chap 8, sect 1.
part 1 of 2
10 points
A 0
.
4 kg bead slides on a curved wire, start
ing from rest at point A as shown in the figure.
The segment from A to B is frictionless, and
the segment from B to C is rough. The point
A is at height 8
.
3 m and the point C is at
height 1
.
8 m with respect to point B.
The acceleration of gravity is 9
.
8 m
/
s
2
.
A
8.3 m
B
1.8 m
C
Find the speed of the bead at B.
Correct answer: 12
.
7546 m
/
s (tolerance
±
1
%).
Explanation:
Given :
m
= 0
.
4 kg
and
h
= 8
.
3 m
.
Choose the zero level for potential energy at
the level of B. Between A and B,
K
A
+
U
A
=
K
B
+
U
B
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homework 05 – KIM, JI – Due: Oct 3 2007, 4:00 am
2
0 +
m g H
=
1
w
m v
2
+ 0
v
=
radicalbig
2
g H
=
radicalBig
2(9
.
8 m
/
s
2
)(8
.
3 m)
= 12
.
7546 m
/
s
.
Question 3, chap 8, sect 1.
part 2 of 2
10 points
If the bead comes to rest at C, find the
change in mechanical energy due to friction
as it moves from B to C.
Correct answer:
−
25
.
48 J (tolerance
±
1 %).
Explanation:
Given :
h
= 1
.
8 m
Choose the starting point at B, the zero po
tential energy level at B, as before, and the
end point at C. Then the energy loss is equal
to the work done by the nonconservative
forces and is
W
nc
=
K
f
−
K
i
+
U
f
−
U
i
= 0
−
1
2
m v
2
B
+
m g h
−
0
=
(0
.
4 kg) (12
.
7546 m
/
s)
2
2
+ (0
.
4 kg) (9
.
8 m
/
s
2
) (1
.
8 m)
=
−
25
.
48 J
.
Question 4, chap 8, sect 1.
part 1 of 1
10 points
A block starts at rest and slides down a
frictionless track except for a small rough area
on a horizontal section of the track (as shown
in the figure below).
It leaves the track horizontally, flies through
the air, and subsequently strikes the ground.
The acceleration of gravity is 9
.
81 m
/
s
2
.
μ
=0
.
3
1
.
4 m
420 g
h
2
.
3 m
3
.
81 m
9
.
81 m
/
s
2
v
At what height
h
above the ground is the
block released?
Correct answer:
4
.
29784
m (tolerance
±
1
%).
Explanation:
Let :
x
= 3
.
81 m
,
g
= 9
.
81 m
/
s
2
,
m
= 420 g
,
μ
= 0
.
3
,
ℓ
= 1
.
4 m
,
h
2
=
−
2
.
3 m
,
h
=
h
1
−
h
2
,
and
v
x
=
v .
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 Fall '08
 Turner
 Physics, Friction, Mass, Potential Energy, Work, Light, Correct Answer

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