91Ó°ÊÓ

The mass of the bird is $m_b=0.07kg$.

The minimum power output is localid="1663755763088" $p_{min}=10\mathrm{watt}/\mathrm{kg}$.

The minimum power output is $p_{max}=25\mathrm{watt}/\mathrm{kg}$.

The number of flaps per second is N = 10 beat/s.

The relationship between the power and work is given by,

$\mathit{P}\mathbf{=}\frac{\mathbf{W}}{\mathbf{t}}$

Here, P, is power, W is work done, and t is time.

Find the minimum power as follows.

$$\begin{gathered} P_{min}=P_{min}×m_b \\ =\left(10watt/kg\right)\left(0.07kg\right) \\ =0.7watt \end{gathered}$$

Find the maximum power as follows.

$$\begin{gathered} P_{max}=P_{max}×m_b \\ =\left(25watt/kg\right)\left(0.07kg\right) \\ =1.75watt \end{gathered}$$

Find the work done corresponding to the minimum power as follows.

$$\begin{gathered} W_{min}=\frac{0.7J/s}{10beat/s} \\ =0.07J/beat \end{gathered}$$

Find the work done corresponding to the maximum power as follows.

$$\begin{gathered} W_{max}=\frac{1.75J/s}{10beat/s} \\ =0.175J/beat \end{gathered}$$

Therefore, the required work done are 0.07 J/beat, and 0.175 J/beat.

Power output of the athlete is P = 1.4 kW or P = 1400 W.

The Mass of athlete is m = 70 kg.

Steady power output of the athlete is P = 500 W.

Power output per kilogram of the body mass is given by,

$$\begin{gathered} \frac{\mathrm{P}}{\mathrm{M}}=\frac{500\mathrm{W}}{70\mathrm{kg}} \\ ≈7\mathrm{W}/\mathrm{g} \end{gathered}$$

Since, the power output in order to fly ranges from 10 to 25 Watts, and the value of is P/m less when compared to the range of flying.

Now, the athlete can expend$\frac{1400\mathrm{W}}{70\mathrm{kg}}=20\mathrm{W}/\mathrm{kg}$ for short interval of time. Hence, it is not possible for a human powered aircraft to fly for extended periods by flapping its wings.