homework 04 – JACOBS, AMBER – Due: Feb 19 2008, 1:00 am
1
Question 1, chap 6, sect 2.
part 1 of 2
10 points
A block is released from rest on an inclined
plane and moves 3
.
4 m during the next 5
.
2 s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
12 kg
μ
k
34
◦
What is the magnitude of the acceleration
of the block?
Correct answer: 0
.
251479 m
/
s
2
(tolerance
±
1 %).
Explanation:
Given :
m
= 12 kg
,
ℓ
= 3
.
4 m
,
θ
= 34
◦
,
and
t
= 5
.
2 s
.
Consider the free body diagram for the
block
m g
sin
θ
N
=
m g
cos
θ
μ
N
a
m g
The acceleration can be obtained through
kinematics. Since
v
0
= 0,
ℓ
=
v
0
t
+
1
2
a t
2
=
1
2
a t
2
a
=
2
ℓ
t
2
(1)
=
2 (3
.
4 m)
(5
.
2 s)
2
=
0
.
251479 m
/
s
2
.
Question 2, chap 6, sect 2.
part 2 of 2
10 points
What is the coefficient of kinetic friction
μ
k
for the incline?
Correct answer: 0
.
643556
(tolerance
±
1 %).
Explanation:
Applying Newton’s Second Law of Motion
summationdisplay
F
i
=
m a
and Eq. 1, the sine component
of the weight acts down the plane and friction
acts up the plane. The block slides down the
plane, so
m a
=
m g
sin
θ

μ
k
m g
cos
θ
2
ℓ
t
2
=
g
parenleftBig
sin
θ

μ
k
cos
θ
parenrightBig
2
ℓ
=
g t
2
parenleftBig
sin
θ

μ
k
cos
θ
parenrightBig
μ
k
=
g t
2
sin
θ

2
ℓ
g t
2
cos
θ
(2)
= tan
θ

2
ℓ
g t
2
cos
θ
= tan 34
◦

2 (3
.
4 m)
(9
.
8 m
/
s
2
) (5
.
2 s)
2
cos 34
◦
=
0
.
643556
.
Question 3, chap 6, sect 5.
part 1 of 2
10 points
Given:
The coefficient of static friction be
tween the person and the wall is 0
.
43
,
the
mass of the person is 59 kg
,
the radius of the
cylinder is 8
.
3 m
,
and
g
= 9
.
8 m
/
s
2
.
A barrel of fun consists of a large vertical
cylinder that spins about the vertical axis.
When it spins fast enough, any person inside
will be held up against the wall.
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homework 04 – JACOBS, AMBER – Due: Feb 19 2008, 1:00 am
2
ω
8
.
3 m
Find
ω
c
, the critical angular speed below
which a person will slide down the wall of the
cylinder.
Correct answer: 1
.
65707
rad
/
s (tolerance
±
1 %).
Explanation:
Let :
m
= 59 kg
,
R
= 8
.
3 m
,
and
μ
= 0
.
43
.
The normal force on the person by the wall
provides the centripetal acceleration
N
=
m ω
2
R .
The maximum force of static friction is given
by
f
=
μ
N
.
In equilibrium we have
f
=
m g .
Hence
m g
=
μ m ω
2
c
R .
or solving for
ω
c
ω
c
=
radicalbigg
g
μ R
=
radicalBigg
9
.
8 m
/
s
2
(0
.
43) (8
.
3 m)
=
1
.
65707 rad
/
s
.
Question 4, chap 6, sect 5.
part 2 of 2
10 points
Compare the following two cases where for
each case the angular speed is greater than
the critical value as defined in Part 1.
For
the first case, denote its angular speed and
frictional force by
ω
1
and
f
1
, respectively. For
the second case, its angular speed is
ω
2
= 2
ω
1
.
Express
f
2
, the frictional force for the sec
ond case, in terms of
f
1
.
1.
f
2
=
f
1
correct
2.
f
2
= 3
f
1
3.
f
2
=
f
1
3
.
5
4.
f
2
=
f
1
2
5.
f
2
= 2
f
1
6.
f
2
= 4
f
1
7.
f
2
= 1
.
5
f
1
8.
f
2
=
f
1
3
9.
f
2
=
f
1
4
10.
f
2
= 2
.
5
f
1
Explanation:
Since for each case the angular speed is
above the critical angular speed, the person
will remain in equilibrium in the vertical di
rection. Hence
f
2
=
f
1
=
m g .
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 Spring '07
 KOPP
 Physics, Acceleration, Force, Friction, Gravity, Work, Correct Answer, µk

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