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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2024 · 9 April · Morning Shift · Physics · Work, Power and Energy

A particle of mass m moves on a straight line with its velocity...

A particle of mass mm moves on a straight line with its velocity increasing with distance according to the equation v=αxv=\alpha \sqrt{x}, where α\alpha is a constant. The total work done by all the forces applied on the particle during its displacement from x=0x=0 to x=dx=\mathrm{d}, will be :

Options

  1. Am2α2 d\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}
  2. Bmd2α2\frac{\mathrm{md}}{2 \alpha^2}
  3. C2 mα2 d2 \mathrm{~m} \alpha^2 \mathrm{~d}
  4. Dmα2 d2\frac{m \alpha^2 \mathrm{~d}}{2}

Correct answer

D. mα2 d2\frac{m \alpha^2 \mathrm{~d}}{2}

Step-by-step solution

v=αx at x=0:v=0 at x=d;v=αd W.D =Kf−Ki W.D =12m(αd)2−12m(0)2⇒ W.D =mα2d2\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}
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