A particle of mass $m$ moves on a straight line with its velocity increasing with distance according to the equation $\mathrm{v}=\alpha \sqrt{\mathrm{x}}$, 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:

(B) $\frac{\mathrm{md}}{2 \alpha^{2}}$

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Work, Power & Energy

A particle of mass

m

moves on a straight line with its velocity increasing with distance according to the equation

v = α x

, where

α

is a constant. The total work done by all the forces applied on the particle during its displacement from

x = 0

x = d

, will be:

(A) $\frac{m}{2 α ^{2} d}$
(B) $\frac{md}{2 α ^{2}}$
(C) $\frac{m α ^{2} d}{2}$ ✓ Correct
(D) $2 m α^{2} d$

Answer: (C)

Solution

v = α x

, so at

x = 0

v = 0

and at

x = d

v = α d

. By work–energy theorem,

W = \frac{1}{2} m v^{2} - 0 = \frac{1}{2} m (α d)^{2} = \frac{m α ^{2} d}{2}

. Option (3).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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