Question

A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as $v=k\sqrt{x}$,where k is a positive constant . The work done by all the forces acting on the particle, during the first t second is

• A. $\frac{m{k}^4}{t^2}$
• B. $\frac{m{k}^4{t}^2}{4}$
• C. $\frac{m{k}^4{t}^2}{8}$
• D. $\frac{m{k}^4{t}^2}{16}$

Answer 1

The correct option is C

$\frac{m{k}^4{t}^2}{8}$

GIven $v=k\sqrt{x}$ or $\frac{dx}{dt}=k\sqrt{x}$ or $x^{-\frac{1}{2}}dx=kdt$

Integrating both sides, we get

$\frac{x^{\frac{1}{2}}}{\frac{1}{2}}=kt+C$; Assuming x(0) = 0

Therefore, C = 0

$2\sqrt{x}=kt\Rightarrow x=\frac{k^2{t}^2}{4}$ or $v=\frac{k^{2}t}{2}$

Therefore, work done,

$△W$ = Increase in KE

= $\frac{1}{2}m{v}^2-\frac{1}{2}m(0{)}^2=\frac{1}{2}m{\left[\frac{k^{2}t}{2}\right]}^2=\frac{m{k}^4{t}^2}{8}$