AP EAMCET · PHYSICS · Work Power Energy
A particle is released freely from a height \(H\). At a certain height, its kinetic energy is two times of its potential energy. Then, the height and the speed of the particle at that instant are respectively
( \(g=\) acceleration due to gravity)
- A \(\frac{H}{3}, \sqrt{\frac{2 g H}{3}}\)
- B \(\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}\)
- C \(\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}\)
- D \(\frac{H}{3}, \sqrt{2 g H}\)
Answer & Solution
Correct Answer
(B) \(\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}\)
Step-by-step Solution
Detailed explanation
If particle falls by a distance \(x\), then \[ \begin{aligned} \mathrm{KE}=\frac{1}{2} m v^2 & =\frac{1}{2} m(2 g x)=m g x \\ \mathrm{PE} & =m g(H-x) \end{aligned} \] As, \(\mathrm{KE}=2(\mathrm{PE}) \Rightarrow m g x=2 m g(H-x) \Rightarrow x=\frac{2 H}{3}\) So, height of…
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