Question

The displacement x of a body of mass $1kg$ on smooth horizontal surface as a function of time $t$ is given by $x = \dfrac{{{t^3}}}{3}$ (where $x$ is in metres and $t$ is in seconds). Find the work done by the external agent for the first one second.

Answer 1

Hint: This question utilizes the concept of differentiation to find out acceleration through given equations. We double differentiate the equation concerning the position and time to find out the acceleration of the body and then find the force. Then we put in the values and find out the work done.

Formulae used:
\[W = \int {Fdx} \] Where \[\int {Fdx} \] is the integral of the Force $F$ multiplied by small displacement element $dx$
$\dfrac{{dx}}{{dt}} = v$ where $\dfrac{{dx}}{{dt}}$ is differentiation of position $x$ with respect to time $t$ and $v$ is velocity.
$\dfrac{{dv}}{{dt}} = a$ where $\dfrac{{dv}}{{dt}}$ is differentiation of velocity $v$ with respect to time $t$ and $a$ is acceleration.
$F = ma$ where $F$ is the force, $m$ is the mass and $a$ is the acceleration

Complete step by step answer:
According to the question
$m = 1kg$
From the given displacement – time relationship, we have
$x = \dfrac{{{t^3}}}{3}$
Differentiating both sides with respect to time, we get
$ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{{3{t^2}}}{3}$
$ \Rightarrow dx = {t^2}dt$ ------------(i)
We know that $\dfrac{{dx}}{{dt}} = v$
Thus, we get
$ \Rightarrow v = {t^2}$ ---------------(ii)
Further differentiating equation (ii) with respect to time, we get
$ \Rightarrow \dfrac{{dv}}{{dt}} = 2t$
We know that $\dfrac{{dv}}{{dt}} = a$
Thus, the equation becomes
$ \Rightarrow a = 2t$ ------------(iii)
Substituting the value of $a$ from equation (iii) in the equation $F = ma$ , we get
$ \Rightarrow F = m2t$
Now, substituting the value of mass, we get
 $ \Rightarrow F = 1 \times 2t$
$ \Rightarrow F = 2t$ ---------------(iv)
Now, work done can be calculated as
$ \Rightarrow W = \int\limits_{t = 0}^{t = 1} {Fdx} $ (since work done is asked for the first $1s$)
Substituting the values of $F$ and $dx$ from eq (iv) and (i) respectively, we have
\[
   \Rightarrow W = \int\limits_{t = 0}^{t = 1} {2t \times {t^2}dt} \\
   \Rightarrow W = 2\int\limits_{t = 0}^{t = 1} {{t^3}dt} \\
   \Rightarrow W = 2\left[ {\dfrac{{{t^4}}}{4}} \right]_0^1 \\
   \Rightarrow W = \dfrac{2}{4} = \dfrac{1}{2}Joule \\
 \]
Therefore, work done by the external agent for the first one second will be $\dfrac{1}{2}Joule$

Note: Work done is a scalar quantity and it is given as the dot product of the force acting on the body and the displacement produced by the force. Dot product or scalar product of two vectors is given by $\overrightarrow A \cdot \overrightarrow B = AB\cos \theta $ , Where $\theta $ is the angle between the two vectors. Here, we have not been given vectors, hence we use integration to find out the answer .