Question

A particle of mass m moves with a variable velocity $v_1$ 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 seconds 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

Answer: (C)

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

Given,

$v=k\sqrt{x}$

By integrating both side,

$2\sqrt{x}=kt+x$

At $t=0,x=0,c=0$

$2\sqrt{x}=kt$

$x=\frac{k^2{t}^2}{4}$

Velocity, $v=\frac{dx}{dt}=\frac{k^{2}t}{2}$

Acceleration, $a=\frac{dv}{dt}=\frac{k^2}{2}$

Work done, $W=\int F.dx$

$W=\int ma.dx$

$W=\int \frac{m{k}^2}{2}.dx$

$W=\frac{m{k}^2}{2}x=\frac{m{k}^2}{2}.\frac{k^2{t}^2}{4}$

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

The correct option is C.