A 1.1 kg seagull takes off from a piling along the California coast and accelerates in 0.50 s to a flight speed of 7.5 m/s. Assuming the seagull flies along a horizontal path (at the same height as its takeoff), what average power is delivered by the bird's wings during the takeoff?

Question

Quizplus · Accepted Answer

The average power delivered by the bird's wings during takeoff can be calculated using the formula for power, which is $P = \frac{\Delta W}{\Delta t}$, where $\Delta W$ is the work done and $\Delta t$ is the time over which the work is done. The work done to accelerate the bird can also be expressed as the change in kinetic energy, $\Delta KE = \frac{1}{2}mv^2$, where $m$ is the mass of the bird and $v$ is its final velocity. Substituting the given values, we get $\Delta KE = \frac{1}{2}(1.1 \, \text{kg})(7.5 \, \text{m/s})^2 = 30.9375 \, \text{J}$. Since the acceleration takes 0.50 s, the average power is $P = \frac{30.9375 \, \text{J}}{0.50 \, \text{s}} = 61.875 \, \text{W}$, which rounds to 62 W.

A 11 Kg Seagull Takes Off from a Piling Along the the California