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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2024
  • Hard
  • MCQ

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 _______.

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Answer & Solution

Correct Answer

(C) \(\frac{m \alpha^2 d}{2}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{v}=\alpha \sqrt{\mathrm{x}}\)

\(\text { at } \mathrm{x}=0: \mathrm{v}=0\)

\(\&\text { at } \mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}}\)

\(\text { W.D }=\mathrm{K}_1-\mathrm{K}_{\mathrm{i}}\)

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

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

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