homework 16 – HETTIGER, CHRISTOF – Due: Feb 26 2008, 4:00 am
1
Question 1, chap 8, sect 1.
part 1 of 1
10 points
A 4
.
7 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
3
.
3 kg mass.
The acceleration of gravity is 9
.
8 m
/
s
2
.
6
.
4 m
ω
4
.
7 kg
3
.
3 kg
Use conservation of energy to determine the
final speed of the first mass after it has fallen
(starting from rest) 6
.
4 m.
Correct answer: 4
.
6853
m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
1
= 4
.
7 kg
,
m
2
= 3
.
3 kg
,
and
ℓ
= 6
.
4 m
.
Consider the free body diagrams
4
.
7 kg
3
.
3 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
4
.
7 kg

3
.
3 kg
4
.
7 kg + 3
.
3 kg
×
2 (9
.
8 m
/
s
2
)(6
.
4 m)
bracketrightbigg
1
/
2
= 4
.
6853 m
/
s
.
Question 2, chap 8, sect 1.
part 1 of 3
10 points
A block starts at rest and slides down a fric
tionless 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
.
506 g
4
.
2 m
2 m
x
9
.
81 m
/
s
2
v
What is the speed
v
of the ball when it
leaves the track?
Correct answer: 6
.
56993 m
/
s (tolerance
±
1
%).
Explanation:
Let :
x
= 4
.
19524 m
,
g
= 9
.
81 m
/
s
2
,
m
= 506 g
,
h
1
=

2
.
2 m
,
h
2
=

2 m
,
h
=
h
1
+
h
2
,
and
v
x
=
v .
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homework 16 – HETTIGER, CHRISTOF – Due: Feb 26 2008, 4:00 am
2
m
h
h
1
h
2
4
.
2 m
g
v
Basic Concepts:
Conservation of Me
chanical Energy
U
i
=
U
f
+
K
f
+
W .
(1)
since
v
i
= 0 m/s.
K
=
1
2
m v
2
(2)
U
g
=
m g h
(3)
W
=
μ m g ℓ .
(4)
Choosing the point where the block leaves the
track as the origin of the coordinate system,
Δ
x
=
v
x
Δ
t
(5)
h
2
=

1
2
g
Δ
t
2
(6)
since
a
x
i
= 0 m/s
2
and
v
y
i
= 0 m/s.
Solution:
1
2
m v
2
x
=

m g h
1
v
2
x
=

2
g
(
h

h
2
)
=

2
g h
1
v
x
=
radicalbig

2
g h
1
(4)
=
radicalBig

2 (9
.
81 m
/
s
2
) (

2
.
2 m)
= 6
.
56993 m
/
s
.
Question 3, chap 8, sect 1.
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 Spring '10
 Turner
 Physics, Energy, Force, Friction, Mass, Potential Energy, Work, Light

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