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=
mg
/10, so
F
= 11
mg
/10. Since the force
G
F
and the displacement
G
d
are in the same
direction, the work done by
G
F
is
2
4
11
11 (72 kg)(9.8 m/s )(15 m)
1.164 10 J
10
10
F
mgd
WF
d
==
=
=
×
which (with respect to significant figures) should be quoted as 1.2
×
10
4
J.
(b) The force of gravity has magnitude
mg
and is opposite in direction to the
displacement. Thus, using Eq. 77, the work done by gravity is
24
(72 kg)(9.8 m/s )(15 m)
1.058 10 J
g
Wm
g
d
=−
×
which should be quoted as – 1.1
×
10
4
J.
(c) The total work done is
W
=×−×=
×
1164
.
10 J 1.058 10 J
1.06 10 J
44
3
. Since the
astronaut started from rest, the workkinetic energy theorem tells us that this (which we
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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, Acceleration, Force, Gravity

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