KU
mgh
kx
AACC
+=+
+=
+
0
1
2
0
2
which yields
h
kx
mg
==
×
=
2
2
2
2
212
98
1.35 10 N m 0.055 m
kg
m s
0.174 m
4
c
hb
g
b
g c
h
.
.
Therefore,
A
°
=
°
=
x
h
sin 30
0.174 m
0.35 m .
(b) From this result, we find
A
=−
=
0.35 0.055
0.29 m , which means that
Δ
y
A
sin
m
θ
015
.
in sliding from point
A
to point
B
. Thus, Eq. 818 gives
0
1
2
0
ΔΔ
Δ
K
U
mv
mg h
B
2
which yields
vg
h
B
−
=
29
8
0
1
5
Δ
..
.
b
gb
g
1.7 m s
31. We refer to its starting point as
A
, the point where it first comes into contact with the
spring as
B
, and the point where the spring is compressed 
x
 = 0.055 m as
C
. Point
C
is
our reference point for computing gravitational potential energy. Elastic potential energy
(of the spring) is zero when the spring is relaxed. Information given in the second
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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, Energy, Potential Energy

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