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10.5.
Model:
Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic
and potential energy does not change as the car falls.
Visualize:
Solve:
(a)
The kinetic energy of the car is
2
2
5
C
C C
1
1
(1500 kg)(30 m/s)
6.75 10 J
2
2
K
m v
=
=
=
×
(b)
Let us relabel
K
C
as
K
f
and place our coordinate system at
f
0
y
=
m so that the car’s potential energy
U
gf
is
zero, its velocity is
v
f
, and its kinetic energy is
K
f
. At position
y
i
,
i
i
0 m/s or
0 J,
v
K
=
=
and the only energy the
car has is
gi
i
.
U
mgy
=
Since the sum
K
+
U
g
is unchanged by motion,
f
gf
i
gi
K
U
K
U
+
=
+
. This means
f
f
i
i
f
i
i
5
f
i
i
2
0
(
)
(6.75 10 J
0 J)
45.9 m
(1500 kg)(9.8 m/s )
K
mgy
K
mgy
K
K
mgy
K
K
y
mg
+
=
+
⇒
+
=
+

×

⇒
=
=
=
(c)
From part (b),
(
29
2
2
2
2
f
i
f
i
f
i
i
1
1
(
)
2
2
2
mv
mv
v
v
K
K
y
mg
mg
g



=
=
=
Free fall does
not
depend upon the mass.
10.11.
Model:
In the absence of frictional and airdrag effects, the sum of the kinetic and gravitational
potential energy does not change as the pendulum swings from one side to the other.
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View Full DocumentVisualize:
The figure shows the pendulum’s beforeandafter pictorial representation for the two situations described in
parts (a) and (b).
Solve:
(a)
The quantity
K
+
U
g
is the same at the lowest point of the trajectory as it was at the highest point.
Thus,
1
g1
0
g0
means
K
U
K
U
+
=
+
2
2
2
2
1
1
0
0
1
1
0
0
2
2
1
0
1
0
1
1
2
2
2
2
2 (0 m)
(0 m/s)
2
2
mv
mgy
mv
mgy
v
gy
v
gy
v
g
gy
v
gy
+
=
+
⇒
+
=
+
⇒
+
=
+
⇒
=
From the pictorial representation, we find that
0
cos30 .
y
L
L
=

°
Thus,
2
1
2
(1 cos30 )
2(9.8 m/s )(0.75 m)(1 cos30 )
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 Spring '08
 Nandi
 Physics, Energy, Kinetic Energy, Potential Energy

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