A 1.1 kg seagull takes off from a piling along the California coast and accelerates within 0.50s to a flight speed of 7.5m/s along a horizontal path.What average power was developed by the bird's wings as it accelerated?

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Quizplus · Accepted Answer

The average power is the work done per unit time. The work done by the bird's wings is equal to the change in kinetic energy of the bird. The change in kinetic energy is:$$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$where:* $m$ is the mass of the bird* $v_f$ is the final velocity of the bird* $v_i$ is the initial velocity of the birdThe average power is then:$$P = \frac{\Delta K}{\Delta t} = \frac{\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}{\Delta t}$$Substituting the given values, we get:$$P = \frac{\frac{1}{2}(1.1	ext{ kg})(7.5	ext{ m/s})^2 - 0}{(0.50	ext{ s})}$$$$P = 62	ext{ W}$$

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