Certain force acting on a 20kg mass changes its velocity from $5m{{s}^{-1}}$ to $2m{{s}^{-1}}$. Calculate the work done by the force.

Hint: This problem can be solved by using work- energy theorem which gives the relation between work done and kinetic energy. It states that: the total work done on a body is equal to change in its velocity.
When force is applied on a body, from Newton’s second law of motion we know that the body is accelerated which means that there is a change in velocity with respect to the time this change in velocity is nothing but work done.

Complete step by step answer:
Consider a body of mass ‘m’ if u is the initial velocity of a body u and v is the final velocity of a body then from work – energy theorem, work done is given by
$W=\dfrac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)$…… (1)
Here, in this problem it is given that
u =$5m{{s}^{-1}}$
v =$2m{{s}^{-1}}$
m = 20kg
on substituting we get
$\begin{align}
  & W=\dfrac{1\times 20\times \left( {{2}^{2}}-{{5}^{2}} \right)}{2} \\
 & \\
 & W=\dfrac{20\times \left( -21 \right)}{2} \\
\end{align}$
$\therefore W=-210J$
Here -ve sign indicates that work is done by the system implying that retarding nature of applied force.
Additional information:
Work done is also defined as the dot product of force and displacement and is given by,
$\begin{align}
  & W=\vec{F}\bullet \vec{S} \\
 & \Rightarrow W=FS\cos \theta \\
\end{align}$
Which means that work is measured by the displacement of an object in the direction of force applied.
It is a scalar and derived physical quantity.
SI unit of work done is joule represented by a symbol J and one more SI unit is Nm.
Its dimensional formula is $\left[ M{{L}^{-1}}{{T}^{-2}} \right]$

Note: Work energy theorem is also stated as - the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
If work done is positive, then work is done on the system.
While writing the answer along with the numerical you must write a proper SI unit of that physical quantity.