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

Physics · Work, Power & Energy

JEE Main 2024 — 9 April, Shift 1 — Question 44

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

Option A:
$\frac{m}{2 \alpha^{2} d}$
Option B:
$\frac{m d}{2 \alpha^{2}}$
Option C:
$\frac{m \alpha^{2} d}{2}$
Correct
Option D:
$2 m \alpha^{2} d$

Answer: C

Step-by-step solution

$\quad v=\alpha \sqrt{\mathrm{x}}$

at $\mathrm{x}=0: \mathrm{v}=0$

$\&$ at $\mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}}$

W.D $=\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}$

W.D $=\frac{1}{2} m(\alpha \sqrt{\mathrm{~d}})^{2}-\frac{1}{2} m(0)^{2}$

$\Rightarrow$ W.D $=\frac{m \alpha^{2} \mathrm{~d}}{2}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Work, Power & Energy
Topic: Kinetic Energy and Work-Energy Theorem