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Numerically, we have, with
m
= (5.29 N)/(9.80 m/s
2
)=0.54 kg,
2
2
(20.0 m/s)
19.4 m/s
2(9.80 m/s )(1 0.265/5.29)
h
==
+
.
(b) We notice that the force of the air is downward on the trip up and upward on the trip
down, since it is opposite to the direction of motion. Over the entire trip the increase in
thermal energy is
Δ
E
th
= 2
fh
. The final kinetic energy is
Km
v
f
=
1
2
2
, where
v
is the
speed of the stone just before it hits the ground. The final potential energy is
U
f
= 0. Thus,
using Eq. 831 (with
W
= 0), we find
1
2
2
1
2
2
0
2
mv
fh
mv
+=
.
We substitute the expression found for
h
to obtain
2
22
0
0
2
11
2(
1
/ ) 2
2
fv
mv
mv
gf
w
=−
+
which leads to
2
2
2
00
0
0
2
1
(1
/
)
/
)
fv
fv
f
wf
vv
v
v
v
mg
f w
w
f w
w
f
w
f
§·
−
=
−
=
¨¸
++
+
+
©¹
where
w
was substituted for
mg
and some algebraic manipulations were carried out.
Therefore,
0
5.29 N 0.265 N
(20.0 m/s)
19.0 m/s
5.29 N 0.265 N
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Resistance, Energy, Thermal Energy

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