TS EAMCET · Physics · Work Power Energy
The work done in displacing a particle from \(y=a\) to \(y=2 a\) by a force \(-\frac{\mathrm{K}}{y^2}\) acting along \(y\)-axis is
- A \(-\frac{5 \mathrm{~K}}{8 a}\)
- B \(-\frac{14 \mathrm{~K}}{8 a^3}\)
- C \(-\frac{ \mathrm{~K}}{ a^2}\)
- D \(-\frac{ \mathrm{~K}}{2 a}\)
Answer & Solution
Correct Answer
(D) \(-\frac{ \mathrm{~K}}{2 a}\)
Step-by-step Solution
Detailed explanation
\(W = \int_{a}^{2a} F \, dy\) \(W = \int_{a}^{2a} \left(-\frac{K}{y^2}\right) \, dy\) \(W = -K \left[ -\frac{1}{y} \right]_{a}^{2a}\) \(W = K \left( \frac{1}{2a} - \frac{1}{a} \right)\) \(W = K \left( \frac{1 - 2}{2a} \right)\) \(W = -\frac{K}{2a}\)
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