zhou (cz3574) – HW6 – mackie – (10611)
1
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printout
should
have
16
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A 2
.
58 kg block initially at rest is pulled to the
right along a horizontal surface by a constant,
horizontal force of 18 N.
The coefficient of
kinetic friction is 0
.
0972.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Find the speed of the block after it has
moved 3
.
09 m.
Correct answer: 6
.
10159 m
/
s.
Explanation:
We have to use the equation
Δ
K
=

f s
(1)
,
to calculate the change in the kinetic energy,
Δ
K
.
The net force exerted on the block is
the sum of the applied 18 N force and the
frictional force.
Since the frictional force is
in the direction opposite the displacement,
it must be subtracted. The magnitude of the
frictional force is
f
=
μ
N
=
μ m g .
Therefore
the net force acting on the block is
F
net
=
F

μ m g
= 18 N

(0
.
0972) (2
.
58 kg) (9
.
8 m
/
s
2
)
= 15
.
5424 N
.
Multiplying this constant force by the dis
placement, and using equation (1), we obtain
Δ
K
=
F
net
s
= (15
.
5424 N) (3
.
09 m)
= 48
.
026 J
=
1
2
m v
2
,
since the initial velocity is zero. Therefore,
v
f
=
radicalbigg
2 Δ
K
m
=
radicalBigg
2 (48
.
026 J)
2
.
58 kg
=
6
.
10159 m
/
s
.
002
10.0 points
You leave your 125 W portable color TV on
for 3 hours each day and you do not pay
attention to the cost of electricity.
If the dorm (or your parents) charged you
for your electricity use
and the cost was
$0
.
3
/
kW
·
h, what would be your monthly
(30 day) bill?
Correct answer: 3
.
375 dollars.
Explanation:
Let :
P
= 125 W
and
t
= 3 h
/
day
,
The energy consumed in each day is
W
=
P t
= (125 W) (3 h
/
day)
·
kW
1000 W
= 0
.
375 kW
·
h
/
day
.
In 30 days, you would use
(30 day) (0
.
375 kW
·
h
/
day) = 11
.
25 kW
·
h
,
which would cost you
(11
.
25 kW
·
h) ($0
.
3
/
kW
·
h) =
$3
.
375
.
003
10.0 points
A car of weight 3150 N operating at a rate of
154 kW develops a maximum speed of 39 m
/
s
on a level, horizontal road.
Assuming that the resistive force (due to
friction and air resistance) remains constant,
what is the car’s maximum speed on an incline
of 1 in 20;
i.e.
, if
θ
is the angle of the incline
with the horizontal, sin
θ
= 1
/
20 ?
Correct answer: 37
.
5041 m
/
s.
Explanation:
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zhou (cz3574) – HW6 – mackie – (10611)
2
If
f
is the resisting force on a horizontal
road, then the power
P
is
P
=
f v
horizontal
,
so that
f
=
P
v
h
=
(1
.
54
×
10
5
W)
(39 m
/
s)
= 3948
.
72 N
.
On the incline, the resisting force is
F
=
f
+
m g
sin
θ
=
f
+
W
20
=
P
v
h
+
W
20
.
And,
F v
=
P ,
so
v
=
P
F
=
P
P
v
h
+
W
20
=
(1
.
54
×
10
5
W)
(1
.
54
×
10
5
W)
(39 m
/
s)
+
(3150 N)
20
=
37
.
5041 m
/
s
.
004 (part 1 of 2) 10.0 points
A frictionless roller coaster is given an ini
tial velocity of
v
0
at height
h
= 13 m
.
The
radius of curvature of the track at point
A
is
R
= 25 m.
The acceleration of gravity is 9
.
8 m
/
s
2
.
R
2
3
h
h
h
′
v
0
A
B
Find the maximum value of
v
0
so that the
roller coaster stays on the track at
A
solely
because of gravity.
Correct answer: 12
.
6517 m
/
s.
Explanation:
let :
h
= 13 m
and
R
= 25 m
.
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 Spring '10
 Hiyishi
 Energy, Force, Friction, Potential Energy, Correct Answer, Zhou

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