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11. We choose +
x
as the direction of motion (so
G
a
and
G
F
are negativevalued).
(a) Newton’s second law readily yields
2
(85kg)( 2.0m/s )
F
=−
G
so that
2
1
.
71
0N
FF
==×
G
.
(b) From Eq. 216 (with
v
= 0) we have
()
2
22
0
2
37m/s
02
3
.
4
1
0
m
2
2.0m/s
va
x
x
=+Δ
¡
Δ=−
= ×
−
.
Alternatively, this can be worked using the workenergy theorem.
(c) Since
G
F
is opposite to the direction of motion (so the angle
φ
between
G
F
and
G
d
x
=Δ
is 180°) then Eq. 77 gives the work done as
4
5.8 10 J
WF
x
=− Δ =−
×
.
(d) In this case, Newton’s second law yields
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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

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