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

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Q. 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 $t$ seconds is

Work, Energy and Power

Solution:

Given $v = k\sqrt{x}$
or $\frac{dx}{dt} = k\sqrt{x}$ or $x^{\frac{1}{2}} dx = k dt$
Integrating both sides, we get
$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = kt + C$; assuming $x\left(0\right) = 0$
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,
$\Delta W =$ Increase in $KE$
$=\frac{1}{2}mv^{2} - \frac{1}{2}m\left(0\right)^{2} = \frac{1}{2}m\left[\frac{k^{2}t}{2}\right]^{2} = \frac{1}{8}mk^{4}t^{2}$