Question

A force of $5\,N$ is applied on a $20\,kg$ mass at rest. The work done in the third second is:
(A) $\dfrac{{25}}{8}\,J$
(B) $\dfrac{{25}}{4}\,J$
(C) $12\,J$
(D) $25\,J$

Answer 1

Analyzing the question is the most important step. Note that the question asks to find the work done in the third second, not the work done after three seconds. As work done depends on the displacement, so by subtracting the displacement for two seconds from the displacement for three seconds, the displacement of the third second can be found.

Complete answer:
When a body having a mass, $m$ is subjected to an external disturbance known as force, $F$ , it gets accelerated. This acceleration characterizes the rate at which the state of the system changes. It can be formulated as:
$F = ma$
Where $F =$ force
$a =$ acceleration
$m =$ mass of the object
Mass can be defined as a measure of the tendency to oppose any change in the translational state also known as inertia.
The change in the position of an object is called displacement. It is often referred to as the shortest distance between the initial and final position of an object in a given interval of time. It gives both magnitude and direction, therefore it is a vector quantity.
Now, work done by a force on an object can be defined as the product of the force in the direction of displacement and the magnitude of displacement. It's S.I. unit is Joule and often represented as:
$W = F.\,s$
$\Rightarrow W = Fs\cos \theta$
Where
$W =$ work done
$F =$ the force acting on the object
$s =$ displacement
$\theta =$ the angle between force and displacement
In the question, the following are the given quantities:
Mass, $m = 20\,kg$
Force, $F = 5N$
Initial velocity, $u = 0$ as the body is at rest
$F = ma$
$\Rightarrow 5 = 20 \times a$
$\Rightarrow a = \dfrac{5}{{20}}$
$\Rightarrow a = \dfrac{1}{4}\,m/{s^2}$
To find the work done, displacement $s$ should be known. So, to find the displacement, the second equation of motion is used.
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $t =$ time taken
Displacement covered in two seconds can be given by:
$t = 2\,\operatorname{s}$
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow s = 0 \times t + \dfrac{1}{2} \times \dfrac{1}{4} \times {2^2}$
$\Rightarrow s = 0 + \dfrac{4}{8}$
$\Rightarrow s = \dfrac{4}{8}\,\,m$
Displacement covered in three seconds can be given by:
$t = 3\,\operatorname{s}$
$s = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow s = 0 \times t + \dfrac{1}{2} \times \dfrac{1}{4} \times {3^2}$
$\Rightarrow s = 0 + \dfrac{9}{8}$
$\Rightarrow s = \dfrac{9}{8}\,\,m$
Displacement covered in third second = Displacement covered in three seconds - Displacement covered in two seconds
${S_3}\, = \dfrac{9}{8} - \dfrac{4}{8}$
$\Rightarrow {S_3}\, = \dfrac{5}{8}\,\,m$
Work done in the third second is:
The direction of force is in the direction of displacement and hence angle between them is ${0^o}$
$W = F.\,s\,\left( {\because \cos \,{0^o} = 1} \right)$
$\Rightarrow W = 5 \times \dfrac{5}{8}$
$\Rightarrow W = \dfrac{{25}}{8}\,J$
Therefore, option A is the correct answer.

Note:
If $\theta = {0^o}$ then the force is in the direction of the displacement and work done is simply the product of the magnitude of force and displacement. If $\theta = {90^o}$ , then the force is perpendicular to the direction of displacement. In this case, work done is zero. Also work is a scalar quantity that only has magnitude and no direction.