2
5
. (a) The net upward force is given by
()
N
FF
mMg mM
a
+−+
=+
where
m
= 0.250 kg is the mass of the cheese,
M
= 900 kg is the mass of the elevator cab,
F
is the force from the cable, and
3.00 N
N
F
=
is the normal force on the cheese.
On the
cheese alone, we have
2
3.00 (0.250)(9.80)
2.20 m/s
0.250
N
Fm
g
m
a
a
−
−=
¡
==
.
Thus the force from the cable is
4
(
)
1
.
08 10 N
N
M
a
gF
+−=
×
, and the work
done by the cable on the cab is
44
1
(1.80 10 )(2.40)
2.59 10 J.
WF
d
×
=
×
(b) If
92.61 kJ
W
=
and
2
10.5 m
d
=
, the magnitude of the normal force is
4
2
9.261 10
(0.250 900)(9.80)
2.45 N.
10.5
N
W
M
g
d
×
+
−
=
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3
. (a) This is a situation where Eq. 728 applies, so we have
Fx
=
1
2
kx
2
¡
(3.0 N)
x
=
1
2
(50 N/m)
x
2
which (other than the trivial root) gives
x
=
(3.0/25) m = 0.12 m.
(b) The work done by the applied force is
W
a
=
Fx
= (3.0 N)(0.12 m) = 0.36 J.
(c) Eq. 728 immediately gives
W
s
= –
W
a
= –0.36 J.
(d) With
K
f
= K
considered variable and
K
i
= 0, Eq. 727 gives
K = Fx
–
1
2
kx
2
.
We take
the derivative of
K
with respect to
x
and set the resulting expression equal to zero, in
order to find the position
x
c
which corresponds to a maximum value of
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 Spring '10
 Lee
 Physics, Force, Mass, 90°, KF, 12.0 m, 3.23 m, 41.7 J

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