A 1.1 kg seagull takes off from a piling along the California coast and accelerates within 0.50 s to a flight speed of 7.5 m/s along a slightly inclined path.At the end of the 0.50 s period,it is 5.3 m above its point of takeoff.What power was developed by the bird's wings?

Question

Quizplus · Accepted Answer

The power developed by the bird's wings is equal to the rate at which it does work. The work done by the bird is equal to the change in its kinetic energy plus the change in its potential energy. 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, $v_i$ is its initial velocity, and $v_f$ is its final velocity. The change in potential energy is:$$\Delta U = mgh$$where g is the acceleration due to gravity and h is the height to which the bird has risen. The power developed by the bird's wings is:$$P = \frac{\Delta K + \Delta U}{\Delta t}$$where $\Delta t$ is the time interval over which the work is done. Substituting the given values into this equation, we get:$$P = \frac{\frac{1}{2}(1.1 	ext{ kg})(7.5 	ext{ m/s})^2 + (1.1 	ext{ kg})(9.8 	ext{ m/s}^2)(5.3 	ext{ m})}{0.50 	ext{ s}} = 180 	ext{ W}$$Therefore, the correct answer is A.

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