Bozanich, Katherine – Final 1 – Due: May 10 2006, 5:00 pm – Inst: W E Drummond
1
This
printout
should
have
28
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
Answer 24 questions. Do NOT answer more
than 24.
001
(part 1 of 1) 10 points
A block of mass
m
is pushed a distance
D
up
an inclined plane by a horizontal force
F
. The
plane is inclined at an angle
θ
with respect to
the horizontal. The block starts from rest and
the coe
ffi
cient of kinetic friction is
μ
k
.
m
D
μ
k
F
θ
What is the final speed of the block?
1.
v
=
2
m
(
F
sin
θ

μ
k
N
)
D
2.
v
=
2
m
(
F
sin
θ
+
μ
k
N
)
D
3.
v
=
2
m
(
F
cos
θ
+
m g
sin
θ

μ
k
N
)
D
4.
v
=
2
m
(
F
cos
θ

m g
sin
θ
)
D
5.
v
=
2
m
(
F
cos
θ

m g
sin
θ

μ
k
N
)
D
correct
6.
v
=
2
m
(
F
cos
θ
+
m g
sin
θ
)
D
7.
v
=
2
m
(
F
cos
θ

μ
k
N
)
D
8.
v
=
2
m
(
F
cos
θ

m g
sin
θ
+
μ
k
N
)
D
Explanation:
The work done by gravity is
W
grav
=
m g D
cos(90
◦
+
θ
)
=

m g D
sin
θ
.
The work done by the force
F
is
W
F
=
F D
cos
θ
.
From the workenergy theorem we know that
W
net
=
Δ
K ,
W
F
+
W
grav
+
W
friction
=
1
2
m v
2
f
.
Thus
v
f
=
2
m
(
F
cos
θ

m g
sin
θ

μ
k
N
)
D .
002
(part 1 of 1) 10 points
Mass
m
1
moves in a circular path of radius
r
on a frictionless horizontal table.
It is at
tached to a string that passes through a fric
tionless hole in the center of the table.
A
second mass
m
2
is attached to the other end
of the string.
r
v
m
1
m
2
Determine an expression for
r
in terms of
m
1
,
m
2
, and the period
T
(the time for one
revolution).
1.
r
=
m
1
g
T
2
2
π
2
m
2
2.
r
=
m
2
g
T
2
4
π
2
m
1
correct
3.
r
=
m
2
g
T
2
2
π
2
m
1
4.
r
=
m
2
g
T
2
π
2
m
1
5.
r
=
m
1
g
T
2
4
π
2
m
2
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Bozanich, Katherine – Final 1 – Due: May 10 2006, 5:00 pm – Inst: W E Drummond
2
Explanation:
The linear velocity
v
can be expressed in
terms of the distance it travels each revolu
tion:
v
=
2
π
r
T
.
Consider the forces acting on each mass:
N
T
m
1
g
T
m
2
g
The tension
T
in the string provides the
centripetal force required to keep
m
1
moving
in a circular path. Applying
F
x
=
m a
x
to
m
1
, we obtain
T
=
m
1
a
c
=
m
1
v
2
r
and
F
y
=
m a
y
= 0 to
m
2
, so
m
2
g

T
= 0
m
2
g
=
m
1
v
2
r
m
2
g
=
m
1
4
π
2
r
T
2
r
=
m
2
g
T
2
4
π
2
m
1
.
003
(part 1 of 2) 10 points
Note:
P
atm
= 101300 Pa. The viscosity of the
fluid is negligible and the fluid is incompress
ible.
A liquid of density 1349 kg
/
m
3
flows with
speed 2
.
38 m
/
s into a pipe of diameter 0
.
12 m.
The diameter of the pipe decreases to 0
.
05 m
at its exit end.
The exit end of the pipe is
6
.
52 m lower than the entrance of the pipe,
and the pressure at the exit of the pipe is
1
.
2 atm.
The acceleration of gravity is 9
.
8 m
/
s
2
.
P
1
2
.
38 m
/
s
0
.
12 m
P
2
v
2
0
.
05 m
1
.
2 atm
6
.
52 m
What is the velocity
v
2
of the liquid flowing
out of the exit end of the pipe?
Correct answer: 13
.
7088 m
/
s.
Explanation:
Let :
v
1
= 2
.
38 m
/
s
,
d
1
= 0
.
12 m
,
and
d
2
= 0
.
05 m
.
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 Spring '10
 SHIH
 Force, Friction, Potential Energy, Cos, Katherine – Final, Bozanich

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