Question

A particle of mass $m$ starts moving from origin along $x$-axis and its velocity varies with position $(x)$ as $v=k \sqrt{x}$. The work done by force acting on it during first "t''$ seconds is

• A. $\frac{m k^{4} t^{2}}{4}$
• B. $\frac{m k^{2} t}{2}$
• C. $\frac{m k^{4} t^{2}}{8}$
• D. $\frac{m k^{2} t^{2}}{4}$

Answer 1

Answer: (C)

$v=k \sqrt{x}$
Square both sides
$v^{2}=k^{2} x$ ...(1)
$v^{2}=(0)^{2}+2 a x$ ...(2)
Compare (1) and (2)
$2 a=k^{2}$
$a=\frac{k^{2}}{2}$
Displacement $x=\frac{1}{2} a t^{2}$
$=\frac{1}{2} \frac{k^{2}}{2} t^{2}$
$W=F x$
$=ma\, x$
$=\frac{m k^{2}}{2} \cdot \frac{1}{2} \frac{k^{2}}{2} t^{2}$
$=\frac{m k^{4} t^{2}}{8}$