JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 _______.
- A \(\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}\)
- B \(\frac{\mathrm{md}}{2 \alpha^2}\)
- C \(\frac{m \alpha^2 d}{2}\)
- D \(2 m \alpha^2 d\)
Answer & Solution
Correct Answer
(C) \(\frac{m \alpha^2 d}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{v}=\alpha \sqrt{\mathrm{x}}\) \(\text { at } \mathrm{x}=0: \mathrm{v}=0\) \(\&\text { at } \mathrm{x}=\mathrm{d} ; \mathrm{v}=\alpha \sqrt{\mathrm{d}}\) \(\text { W.D }=\mathrm{K}_1-\mathrm{K}_{\mathrm{i}}\)…
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