Question

A particle of mass $2kg$ travels along a straight line with velocity $v=a\sqrt{X}$, where $a$ is a 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 $4a^2$

Given, $mass=2kg,velocity=a\sqrt{x}$

Since, $v=a\sqrt{x},v=a\sqrt{4}$

So, $V^2=u^2+2as\Rightarrow v^2-u^2=2as\Rightarrow v^2=2as\to a=\frac{v^2}{2s}$ Since, $u=0$

$a=\frac{2a^2}{2s}=\frac{4a^2}{4r^2}\Rightarrow a=\frac{a^2}{2}$

So, workdone $=f\times ds=2\times \frac{a^2}{2}\times 4=4a^2$