7.33.
IDENTIFY:
Apply Eq.(7.16).
SET UP:
The sign of
x
F
indicates its direction.
EXECUTE:
4
33
4
(4.8 J m )
x
dU
F
x
x
dx
.
4
3
( 0.800 m)
(4.8 J m )( 0.80 m)
2.46 N.
x
F
The force
is in the
direction.
x
EVALUATE:
0
x
F
when
0
x
and
0
x
F
when
0
x
, so the force is always directed towards the origin.
7.36.
IDENTIFY:
Apply Eq.(7.18).
SET UP:
23
12
d
dx x
x
and
d
dy y
y
.
EXECUTE:
ˆˆ
UU
xy
F =
i
j
since
U
has no
z
dependence.
22
and
so
,
3
3
3
3
2
x
y
x
y
ij
F =
i +
j
+
.
EVALUATE:
x
F
and
x
have the same sign and
y
F
and
y
have the same sign. When
0
x
,
x
F
is in the
direction,
x
and so forth.
7.83.
2
ˆ
,
Fj
3
2.50 N/m
IDENTIFY:
F
is not constant so use Eq.(6.14) to calculate
W
.
F
must be evaluated along the path.
(a)
SET UP:
The path is sketched in Figure 7.83a.
d
dx
dy
l
i
j
2
d
xy dy
Fl
On the path,
so
3
d
y dy
Figure 7.83a
EXECUTE:
2
2
1
1
2
3
4
4
4
21
1
(
)
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 Fall '07
 Evrard
 Energy, Force, Kinetic Energy, Potential Energy, dl

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