A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the…

A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $v=\alpha \sqrt{x}$, where $\alpha$ is a constant. The total work done by all the forces applied on the particle during its displacement from $x=0$ to $x=\mathrm{d}$, will be :
  1. $\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}$
  2. $\frac{\mathrm{md}}{2 \alpha^2}$
  3. $2 \mathrm{~m} \alpha^2 \mathrm{~d}$
  4. $\frac{m \alpha^2 \mathrm{~d}}{2}$

Solution

$\begin{aligned} & \mathrm{v}=\alpha \sqrt{\mathrm{x}} \\ & \text { at } \mathrm{x}=0: \mathrm{v}=0 \\ & \& \text { at } \mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}} \\ & \text { W.D }=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}} \\ & \text { W.D }=\frac{1}{2} \mathrm{~m}(\alpha \sqrt{\mathrm{d}})^2-\frac{1}{2} \mathrm{~m}(0)^2 \\ & \Rightarrow \text { W.D }=\frac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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