hernandez (ejh742) – homework 16 – Turner – (58120)
1
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001
(part 1 of 3) 10.0 points
A block of mass 0
.
5 kg is pushed against a hor
izontal spring of negligible mass, compressing
the spring a distance of Δ
x
as shown in the fig
ure. The spring constant is 319 N
/
m. When
released, the block travels along a frictionless,
horizontal surface to point
B
, the bottom of
a vertical circular track of radius 0
.
4 m, and
continues to move up the track.
The speed
of the block at the bottom of the track is
16 m
/
s, and the block experiences an aver
age frictional force of 5 N while sliding up the
track.
The acceleration of gravity is 9
.
8 m
/
s
2
.
m
k
R
v
B
v
T
T
B
∆
x
What is Δ
x
?
Correct answer: 0
.
633446 m.
Explanation:
From conservation of energy, the initial po
tential energy of the spring is equal to the
kinetic energy of the block at
B
. Therefore,
we write
1
2
k
(Δ
x
)
2
=
1
2
m v
2
B
Δ
x
=
radicalBigg
m v
2
B
k
=
radicalBigg
(0
.
5 kg) (16 m
/
s)
2
(319 N
/
m)
=
0
.
633446 m
.
002
(part 2 of 3) 10.0 points
What is the speed of the block at the top of
the track?
Correct answer: 14
.
6693 m
/
s.
Explanation:
The change in the total energy of the block
as it moves from
B
to
T
is equal to the work
done by the frictional force
Δ
E
=
W
f
E
T

E
B
=
W
f
.
The total energy at
B
is
E
B
=
1
2
m v
2
B
=
1
2
(0
.
5 kg) (16 m
/
s)
2
= 64 J
.
The work done by the frictional force is
W
f
=

f π R
=

(5 N) (
π
) (0
.
4 m)
=

6
.
28319 J
.
Therefore, the total energy at
T
is
E
T
=
E
B
+
W
f
= 64 J + (

6
.
28319 J)
= 57
.
7168 J
.
We can find now the speed of the block at
T
from
1
2
m v
T
2
=
E
T

m g h
T
.
Since
v
T
2
=
2
E
T
m

2
g h
T
,
=
2 (57
.
7168 J)
0
.
5 kg

2(9
.
8 m
/
s
2
) (0
.
8 m)
= 215
.
187 m
2
/
s
2
,
then
v
T
=
radicalBig
215
.
187 m
2
/
s
2
=
14
.
6693 m
/
s
.
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hernandez (ejh742) – homework 16 – Turner – (58120)
2
003
(part 3 of 3) 10.0 points
What is the centripetal acceleration of the
block at the top of the track?
Correct answer: 537
.
968 m
/
s
2
.
Explanation:
The centripetal acceleration at
T
is
a
c
=
v
2
T
R
=
(14
.
6693 m
/
s)
2
0
.
4 m
=
537
.
968 m
/
s
2
.
004
10.0 points
A compact car “consumes” 37
.
5 mi
/
gal when
traveling at 41
.
3 mi
/
h.
Its fuel efficiency is
16
.
4%.
(That is, 16
.
4% of the available fuel
energy is delivered to the wheels.)
How much power is delivered to the wheels
at a speed of 41
.
3 mi
/
h?
The mechanical
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 Spring '08
 Turner
 Physics, Energy, Kinetic Energy, Mass, Potential Energy, Work, Correct Answer, kg

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