P16_032.PDF
(
64 KB
)
Pobierz
Chapter 16 - 16.32
32. (a) The energy at the turning point is all potential energy:
E
=
2
kx
2
m
where
E
=1
.
00 J and
x
m
=
0
.
100 m. Thus,
k
=
2
E
x
2
m
= 200 N
/
m
.
(b) The energy as the block passes through the equilibrium position (with speed
v
m
=1
.
20 m/s)is
purely kinetic:
E
=
1
2
mv
2
m
=
⇒
m
=
2
E
v
2
m
=1
.
39 kg
.
(c) Eq. 16-12 (divided by 2
π
) yields
k
m
=1
.
91 Hz
.
f
=
1
2
π
Plik z chomika:
kf.mtsw
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