Question

A particle of mass $2kg$ travels along straight line with velocity $v=a\sqrt{x}$, where $x$ is constant. The work done by net force during the displacement of particle from $x=0$ to $x=4m$ is

• A. $a^2$
• B. $2a^2$
• C. $4a^2$
• D. $\sqrt{2}{a}^2$

Answer 1

The correct option is C $4a^2$

$v=a\sqrt{x}\Rightarrow \frac{dx}{dt}=a{x}^{1/2}\Rightarrow \int x^{1/2}dx=\int adt$
$2\sqrt{x}=at\Rightarrow x=\frac{a^2{t}^2}{4}$
$\Rightarrow v=\frac{a^{2}t}{2}$
$\Rightarrow a^{\prime }=\frac{a^2}{2}$
$F=m{a}^{\prime }=\frac{m{a}^2}{2}$
$Work=Force\times Displacement$
$\Rightarrow Work=\frac{m{a}^2}{2}\times 4=2m{a}^2$
As, $m=2$
$\therefore Work=4a^2$