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A particle of mass $m \mathrm{~kg}$ moves along the $X$-axis with its velocity varying with the distance travelled as $v=k x^\beta$, where $k$ is a positive constant. The total work done by all the forces during displacement of the particle from $x=0$ to $x=d$ is close to
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Verified Answer
The correct answer is:
$\frac{m k^2}{2} d^{2 \beta}$
Velocity, $v=k x^\beta$
By work-energy theorem,
Work done $=$ Change in $\mathrm{KE}$
$=$ Final $K E-$ Initial $K E$
$$
\begin{aligned}
& =\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2 \\
& =\frac{1}{2} m\left(k d^\beta\right)^2-\frac{1}{2} m(k \times 0)^2=\frac{m k^2}{2} \cdot d^{2 \beta}
\end{aligned}
$$
By work-energy theorem,
Work done $=$ Change in $\mathrm{KE}$
$=$ Final $K E-$ Initial $K E$
$$
\begin{aligned}
& =\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2 \\
& =\frac{1}{2} m\left(k d^\beta\right)^2-\frac{1}{2} m(k \times 0)^2=\frac{m k^2}{2} \cdot d^{2 \beta}
\end{aligned}
$$
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