Certain forces acting on a 20kg mass change its velocity from $5m{s^{ - 1}}$ to $2m{s^{ - 1}}$ Calculate the work done by the force.
A). -210 Joule
B). 210 Joule
C). -105 Joule
D). 420 Joule

Hint- Before solving this problem we will see what is the work done by a force and then we will use the formula of work done in terms of mass and velocity. Work is said to be done when an object moves along the direction of the applied force.

Formula used - $W = F \times S $
$W = \Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v^2}_i$.

Complete step-by-step solution -
The work is defined as a product of force and displacement.
$W = F \times S $
The mechanical energy used while applying a force is called work.
S.I unit of work done is joule.
$W = F \times S $ = (Force is measured in Newtons and the displacement in meters).
The Work-Energy Theorem
This theorem states that the net work done on a particle is equal to the change in its total kinetic energy. i.e. initial kinetic energy – final kinetic energy.
The work done on the particle is equal to the change in the kinetic energy of the particle.
\[W = \Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v^2}_i\]
Where vi and vf are the speeds of the particle before and after the application of force, and m is the particle’s mass.
Thus
\[ W = \Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v^2}_i \\
   = \dfrac{1}{2} \times 20 \times \left( {4 - 25} \right) \\
   = - 210J \\ \]
Hence, the correct option is “A”.

Note- In order to do the work two conditions are required to be fulfilled. The force must be applied to the object. The object must cover a certain distance. The force applied and the distance cover must not be perpendicular to each other. Otherwise the work done will be zero. Even after applying the force on an object if the object does not move then the work done is zero.