Question

Under the action of a force, a $2\text{kg}$ body moves such that its position $x-$ axis as a function of time $t$ is given by $x=\frac{t^2}{3}$, where $x$ is in metre and $t$ in second. The work done by the force in first two seconds is

• A. $1600\text{J}$
• B. $160\text{J}$
• C. $16\text{J}$
• D. $\frac{16}{9}\text{J}$

Answer 1

The correct option is D $\frac{16}{9}\text{J}$

Here the position of the particle is given by the equation,
$x=\frac{t^2}{3}$,
As we know that velocity is the time derivative of position.
$\Rightarrow v=\frac{dx}{dt}=\frac{2t}{3}$
Applying the Work energy theorem
Final velocity $v_{t=2}=\frac{4}{3}\text{m/s}$ and initial velocity is $0$
Work done in first two seconds,
$W=\Delta K.E.=\frac{1}{2}\left(2\right)[{\left(\frac{4}{3}\right)}^2-0]=\frac{16}{9}\text{J}$

Hence option D is the correct answer