fisica1-youn-e-freedman-exercicios-resolvidos

6.102: In both cases, a given amount of fuel represents a given amount of work W
0
that
the engine does in moving the plane forward against the resisting force. In terms of the
range R and the (presumed) constant speed v,
⎛ 2 β ⎞
W0 = RF = R⎜α
v + .
2
⎟
⎝ v ⎠
In terms of the time of flight T , R = vt,
so
⎛ 3 β ⎞
W0 = vTF = T⎜α
v + ⎟.
⎝ v ⎠
a) Rather than solve for R as a function of v, differentiate the first of these relations with
dW
dR dF
respect to v, setting 0 to obtain F R = 0.
For the maximum range, = 0,
dv0
=
dv
+
dv
dF = 0. Performing the differentiation, dF
= 2α v − 2β
v
3 = 0 , which is solved for
dv
1 4
dv
5 2 2
⎛ β ⎞ ⎛ 3.5×
10 N ⋅ m s ⎞
v = ⎜ ⎟ =
⎜
= 32.9 m s = 118 km h.
2 2
0.30 N s m
⎟
⎝ α ⎠ ⎝ ⋅ ⎠
d
b) Similarly, the maximum time is found by setting ( Fv)
= 0;
performing the
2 2
differentiation, 3α v − β v = 0,
which is solved for
⎛ β ⎞
v = ⎜ ⎟
⎝ 3α
⎠
1 4
5 2
⎛ 3.5×
10 N ⋅ m s
=
⎜
2 2
⎝ 3(0.30 N ⋅ s m
1 4
2
dv
1 4
⎞
⎟
⎠
= 25 m s = 90 km
h.
dR
dv
so
6.103: a) The walk will take one-fifth of an hour, 12 min. From the graph, the oxygen
3
consumption rate appears to be about 12 cm kg ⋅ min, and so the total energy is
(12 cm
3
kg ⋅ min)
(70 kg) (12 min) (20J
3
5
cm ) = 2.0×
10 J.
b) The run will take 6 min. Using an estimation of the rate from the graph of about
3
5
33 cm kg ⋅ min gives an energy consumption of about 2.8× 10 J. c) The run takes 4
min, and with an estimated rate of about
3
50 cm
kg ⋅ min, the energy used is about
5
2.8× 10 J. d) Walking is the most efficient way to go. In general, the point where the
slope of the line from the origin to the point on the graph is the smallest is the most
efficient speed; about 5 km h.