Question

A force acts on a $2\text{kg}$ object so that its position is given as a function of time as $x=3t^2+5$. What is the work done by this force in first $5$ seconds ?
$\left[\text{Mains-2019, Jan 9th, Shift 2}\right]$

• A. $\text{850 J}$
• B. $\text{950 J}$
• C. $\text{875 J}$
• D. $\text{900 J}$

Answer 1

The correct option is D $\text{900 J}$

Position, $x=3t^2+5$
$\therefore \text{Velocity},v=\frac{dx}{dt}$
$\Rightarrow v=\frac{d(3t^2+5)}{dt}$
$\Rightarrow v=6t+0$
$\begin{array}{cc}\text{At}t=0\text{sec}& v_i=0\text{m/s}\\ \text{At}t=5\text{sec}& v_f=30\text{m/s}\end{array}$
According to work-energy theorem, $W=\Delta KE.$
$W=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^{2}W=\frac{1}{2}\left(2\right)(30{)}^2-0=900\text{J}$