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

Hint: 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 $ .