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.
This preview has intentionally blurred sections. Sign up to view the full version.
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 )
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Nandi
 Physics, Energy, Kinetic Energy, Potential Energy, m/s

Click to edit the document details