An object of mass $5kg$falls from rest through a vertical distance of $20m$and reaches a velocity of $10m/s$ the work done by the push of the air on the object is $\left( {g = 10m/{s^2}} \right)$:
A. $500J$
B. $ - 750J$
C. $ - 500J$
D. $750J$

Hint: Between energy and work done there is always a relation. A theory that relates the work done and energy is called the work energy theorem. The theory states that the total work done on a body will always equals to the kinetic energy of the body.

Formula used:
$ \Rightarrow W.D = \Delta K.E$
Where, $W.D$is the work done, and $\Delta K.E$ is the change in the kinetic energy.
$ \Rightarrow \Delta K.E = \dfrac{1}{2}m{v^2}$
Where $m$is the mass and $v$is the velocity.
$ \Rightarrow W.D = \left( {mg - F} \right)h$
Where, $m$ is the mass, $g$ is the acceleration due to gravity, $F$ is the force, and $h$ is the height.

Complete step by step solution:
Given an object of mass of $5kg$. It falls from the rest through $20m$ of distance. It reaches the velocity of $10m/s$.
Consider, the force due to the air is $F$.
The net force acting downwards is given as $mg - F$. To calculate the work done, the work-energy theorem is needed. The theorem that relates the work done on the object will always be equal to the change in the kinetic energy. That is,
$ \Rightarrow W.D = \Delta K.E$
Where, $W.D$is the work done, and $\Delta K.E$is the change in the kinetic energy.
Kinetic energy formula is given as,
$ \Rightarrow \Delta K.E = \dfrac{1}{2}m{v^2}$
Where,$m$ is the mass and $v$ is the velocity.
The work done on the object is given as,
$ \Rightarrow W.D = \left( {mg - F} \right)h$
Where, $m$is the mass, $g$is the acceleration due to gravity, $F$ is the force and $h$is the height.
We have,
$ \Rightarrow W.D = \Delta K.E$.
Substitute the kinetic energy and work done formula.
$ \Rightarrow \left( {mg - F} \right)h = \dfrac{1}{2}m{v^2}$
Substitute the values.
$ \Rightarrow \left( {5 \times 10 - F} \right) \times 20 = \dfrac{1}{2} \times 5 \times {10^2}$
Simplify the equation.
$ \Rightarrow \left( {50 - F} \right) \times 20 = \dfrac{1}{2} \times 5 \times 100$
$ \Rightarrow F = - 750J$
Therefore, the work done on the object is $ - 750J$.
Hence option $\left( B \right)$ is the correct answer.

Note:
The work-energy theorem follows the rules of the law of conservation of energy. According to that, we can only apply the theorem where we can transfer the energy from one form to another. The work done here donates all the work done by the forces acting on the bodies like friction, gravity, and other external forces.