Question

The position $x$ of a particle moving along $x-$axis at time $\left(t\right)$ is given by the equation $t=\sqrt{x}+2$, where $x$ is in meters and $t$ in second. Find the work done by the force in first four seconds.

• A. Zero
• B. $2J$
• C. $4J$
• D. $8J$

Answer 1

The correct option is A Zero

$t=\sqrt{x}+2$
$x=(t-2{)}^2$
On differentiating,
$V_{\left(t\right)}=2(t-2)$
$V_{\left(0\right)}=-4m/s\to K{E}_{(}0)=\frac{1}{2}\times m{V}_{\left(0\right)}^2=8m$
$V_{\left(4\right)}=4m/s\to K{E}_{(}4)=\frac{1}{2}\times m{V}_{\left(4\right)}^2=8m$
$\Delta KE=0$
$\therefore Workdone=0$ by work energy theorem