Answer
a) $101.6\;\rm W$
b) $406\;\rm W$
c) $\rm 349\; cal$
Work Step by Step
a) We know that the drag force at small speeds is given by
$$F_D=\frac{1}{2}C\rho Av^2$$
where $C$ is the drag coefficient, $\rho$ is the density of the air, $A$ is the cross-sectional area of the cyclist with his bicycle, and $v$ is his speed.
Our cyclist is applying a force just equal to the magnitude of this drag force to move at a constant speed of 7.3 m/s.
Thus,
$$F_{applied}=F_D$$
Thus,
$$F_{applied}=\frac{1}{2}C\rho Av^2$$
We know that power is given by
$$P=Fv$$
Hence, the power output of the cyclist is given by
$$P=F_{applied}v=\frac{1}{2}C\rho Av^3 $$
Plugging the known;
$$P =\frac{1}{2}(0.9)(1.29) (0.45) (7.3)^3 =\color{red}{\bf101.6}\;\rm W$$
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b) Since the output power is only one-fourth of the total energy consumed by the body.
So, the cyclist’s metabolic power while cycling at 7.3 m/s is 4 times his output power.
Thus,
$$P_{metabolic }=4P=4\times 101.6$$
$$P_{metabolic } =\color{red}{\bf 406}\;\rm W$$
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c) Recall that energy is given by work divided by time.
$$P=\dfrac{E}{\Delta t}$$
Thus,
Hence,
$$E=P\Delta t$$
And for our case,
$$E=P_{metabolic }\Delta t=406\times 60\times 60\;\rm J$$
Thus, the energy in calories, that is burned during his trip at 7.3 m/s, is given by
$$E=406\times 60\times 60\;\rm J\times \dfrac{1\;cal}{4190}$$
$$E=\color{red}{\bf 349}\;\rm cal$$
