TS EAMCET · Physics · Work Power Energy
A particle of mass \(m \mathrm{~kg}\) moves along the \(X\)-axis with its velocity varying with the distance travelled as \(v=k x^\beta\), where \(k\) is a positive constant. The total work done by all the forces during displacement of the particle from \(x=0\) to \(x=d\) is close to
- A \(\frac{m k^2}{2}\)
- B \(\frac{m k^2}{2} d^{2 \beta}\)
- C \(\frac{m k^2}{2 \beta}\)
- D \(\frac{m k^2 d}{2 \beta}\)
Answer & Solution
Correct Answer
(B) \(\frac{m k^2}{2} d^{2 \beta}\)
Step-by-step Solution
Detailed explanation
Velocity, \(v=k x^\beta\) By work-energy theorem, Work done \(=\) Change in \(\mathrm{KE}\) \(=\) Final \(K E-\) Initial \(K E\)…
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