40. (a) The initial direction of motion is taken to be the +
x
direction (this is also the
direction of
G
E
). We use
vv
a
x
fi
22
2
−=∆
with
v
f
= 0 and
G
G
G
aFm e
Em
e
==
−
to solve for
distance
∆
x
:
∆
x
v
a
mv
eE
ie
i
=
−
=
−
−
=
−×
×
×
=×
−
−
−
31
19
2
911 10
2 160 10
712 10
.
.
.
kg 5.00 10 m s
C 1.00 10 N C
m.
6
2
3
c
hc
h
c
h
(b) Eq. 217 leads to
t
x
v
x
v
i
=
×
×
−
−
∆∆
avg
m
ms
s.
2
2 712 10
500 10
285 10
2
6
8
.
.
.
c
h
(c) Using
∆
v
2
= 2
a
∆
x
with the new value of
∆
x
, we find
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 Fall '10
 Rebello,NobelS
 Energy, Kinetic Energy, Mass, Trigraph, Joule, initial kinetic energy

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