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7.1
On August 10, 1972, a large meteorite skipped across the atmosphere above
the western United States and western Canada, much like a stone skipped across
water. The accompanying fireball was so bright that it could be seen in the daytime
sky and was brighter than the usual meteorite trail. The meteorite’s mass was about
4x10
6
kg
; its speed was about 15 km/s. Had it entered the atmosphere vertically,
it would have hit Earth’s surface with about the same speed. (a) Calculate the
meteorite’s loss of kinetic energy (in joules) that would have been associated with
the vertical impact. (b) Express the energy as a multiple of the explosive energy of
1 megaton of TNT, which is 4.2x10
15
J. (c) The energy associated with the atomic
bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how
many Hiroshima bombs would the meteorite impact have been equivalent?
262
1
4
fi
11
m
K
K
K
0J
mv
(4.0x10 kg)(15000
)
4.5x10 J
22
s
Δ=
−
=
−
=
−
=
−
a)
b)
14
15
1megaton TNT
K
( 4.5x10 J)
0.107 megaton TNT
4.2x10 J
−
=
−
c)
1Hiroshima bomb
0.107megaton TNT
107 kiloton TNT
8.24 Hiroshima bomb
13 kiloton TNT
⎛⎞
==
⎜⎟
⎝⎠
(the problem asks for the energy as
a multiple
of the explosive energy which
is really a dimensionless quantity)
(negative sign expresses loss
of kinetic energy)
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A 3.0 kg body is at rest on a frictionless horizontal air track when a constant
horizontal force
acting in the positive direction of an
x
axis along the track is
applied to the body. A stroboscopic graph of the position of the body as it slides
to the right is shown in the Fig.
The force
is applied to the body at t = 0 , and the
graph records the position of the body at 0.50 s intervals. How much work is done
on the body by the applied force
between t = 0
and t = 2.0 s ?
F
G
F
G
F
G
2
oo
1
xx
v
t
a
t
2
=+
+
Using,
We find,
2
2x
a
t
=
Then,
22
2mx
2(3kg)(0.8m)
F
ma
1.2N
t(
2
s
)
==
=
=
Now,
f
o
x
0.8m
x0
W
F dx
(1.2N)
dx
=⋅
=
∫∫
G
G
[ ]
0.8m
0
W
(1.2N) x
(1.2N)(0.8m
0)
=
=−
W0
.
9
6
J
=
Alternatively, from the acceleration
vv2
a
(
x
x
)
−
2
2
2
4x
v2
a
x
t
2
2
f
2
12
x
Km
v
m
2t
22
f
2mx
2(3kg)(0.8)
K0
.
9
6
J
t(
2
)
==
=
Then,
fi
W
K
K
K
0.96 J
0
0.96 J
=Δ =
−
=
− =
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View Full Document 7.13
The figure shows three forces applied to a trunk that moves leftward by
3.00 m over a frictionless floor. The force magnitudes are
,
,
and
, and the indicated angle is
θ
= 60
o
. During the displacement,
(a)
what is the net work done on the trunk by the three forces and
(b) does the kinetic energy of the trunk increase or decrease?
1
F5
.
0
0
N
=
2
F9
.
0
0
N
=
3
F3
.
0
0
N
=
We could calculate the net force of the 3 forces
and then use the formal expression,
f
o
x
x
WF
d
x
=⋅
∫
G
G
But let’s be smart about it and reduce our effort. Firstly, because the forces are
constant the net force comes out of the integral and the integral reduces to the
displacement, so that,
d
G
G
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This note was uploaded on 04/22/2008 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
 Spring '08
 Field
 Physics

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