Question

The relation between displacement x and time, t for a body of mass 2 kg moving under the action of a force is given by $x=\dfrac{{{t}^{3}}}{3}$, where x is in meters and t in seconds. Calculate the work done by the body in the first two seconds.

Answer 1

Hint: We are given that the mass of the body is 2 kg but in order to find out the force, we need to have the acceleration. We are given a relationship between the displacement and the time. We can differentiate this equation to find a relationship between the differentials and then we can integrate it to find out the velocity and then the acceleration. After the acceleration we can find the force.

Complete step by step answer:
Now the velocity can be calculated as,
$x=\dfrac{{{t}^{3}}}{3}$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{3{{t}^{2}}}{3}$
$\Rightarrow dx={{t}^{2}}dt$---(1)
$\Rightarrow v={{t}^{2}}$
Now again differentiating with respect to time on both sides,
$\Rightarrow \dfrac{dv}{dt}=2t$
$\Rightarrow a=2t$
Now force can be given as: $F=ma$
$\Rightarrow F=2\times 2t$
$\Rightarrow F=4t$---(1)
Now in two seconds, work can be found out as using eq (1)
$dW=Fdx$
$\Rightarrow \int{dW}=\int\limits_{0}^{2}{Fdx}$
$\Rightarrow W=\int\limits_{0}^{2}{4t\times {{t}^{2}}dt}$
$\Rightarrow W=4\int\limits_{0}^{2}{{{t}^{3}}dt}$
$\Rightarrow W=4[\dfrac{{{t}^{4}}}{4}]_{0}^{2}$
$\Rightarrow W=4[\dfrac{{{2}^{4}}-0}{4}]$
$\therefore W=16J$

So, the work done comes out to be 16 J.

Note: Work is a scalar quantity and it is given as the dot product of the force acting on the body and the displacement of the body during the course of time.Dot product or scalar product of two vectors is given by,
\[\overrightarrow{A}.\overrightarrow{B}=AB\cos \alpha \]
Where \[\alpha \] is the angle between the two vectors. Here the angle between the force and the displacement is equal to \[90{}^\circ \].Here we do not have the vectors and so we have used integration to arrive at the meaningful and sound answer.