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()
2
3
12
1
22
1
0
2
2(2.80 kg)(9.80 m/s )
0.200 kg/s
0.653 m/s .
(
)
(2.80 kg 1.30 kg)
dm
m g
dm
da
da
dt
dm dt
m
m
dt
==
−
=
−
−
=
++
(b) At
3.00 s,
t
=
11
0
1
(
/
)
1.30 kg ( 0.200 kg/s)(3.00 s)
0.700 kg,
m
m
dm dt t
=+
=
+
−
=
and
the rate of change of acceleration is
2
3
1
1
2
2(2.80 kg)(9.80 m/s )
0.200 kg/s
0.896 m/s .
(
)
(2.80 kg 0.700 kg)
dm
m g
dm
da
da
dt
dm dt
m
m
dt
−
=
−
−
=
(c) The acceleration reaches its maximum value when
0
1
0
(
/
)
1.30 kg ( 0.200 kg/s) ,
m
m
dm dt t
t
+
=
+
−
or
6.50 s.
t
=
63. The freebody diagrams for
1
m
and
2
m
are shown in the
figures below. The only forces on the blocks are the upward
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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, Force

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