Question

A bird flying in the sky has
(A) KE only
(B) PE only
(C) Neither KE nor PE
(D) Both KE and PE

Answer 1

The kinetic energy is a form of energy possessed by a body due to the virtue of its motion. Potential energy is a form of energy possessed by a body due to the virtue of its position or height.
Formula used:
In this solution we will be using the following formulae;
$KE = \dfrac{1}{2}m{v^2}$ where $KE$ is the kinetic energy, $m$ is the mass of the body and $v$ is the speed of its motion.
$\Rightarrow PE = mgh$ where $PE$ is the potential energy, $g$ is the acceleration due to gravity and $h$ is height.

Complete step by step solution:
Fundamentally, the kinetic energy of a body can be defined as the form of energy that a body possesses due to the virtue of its motion. It can also be defined as the maximum amount of energy that can be extracted from decreasing the speed of a body in motion to zero. It is also the minimum amount of “opposing” energy required to bring a body in motion to rest.
Potential energy (PE), fundamentally can be defined as a form of energy possessed by a body by the reason of its position or height.It is also the minimum energy require to raise a body from a reference point to a particular height above that reference point. On a planet this reference point is usually taken as the surface, hence a body at a height above the surface possesses a potential energy relative to the surface.
Based in the definitions above, we can conclude that a bird flying (motion) in the sky (height above ground) possesses both Kinetic and potential energy
Hence, the correct option is D.

Note:
Alternatively, one can reason mathematically as follows: The kinetic energy of a body of mass $m$ is given by
$\Rightarrow KE = \dfrac{1}{2}m{v^2}$ . A flying bird must possess a speed of flight $v$ , hence $v \ne 0$ which implies that
$KE \ne 0$ .Hence bird possesses a kinetic energy.
Similarly, $PE = mgh$ where $h$ is height and $g$ is the acceleration due to gravity.
Bird in the sky implies that $h \ne 0$ hence $PE \ne 0$ .