A particle is released from a height s. At a | vedclass

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(D) Let the total energy of the particle be $E = mgs$. At the given instant,let the potential energy be $PE = x$. According to the problem,the kinetic

Vedclass · Accepted Answer

$\frac{s}{4}, \sqrt{\frac{3gs}{2}}$

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(D) Let the total energy of the particle be $E = mgs$. At the given instant,let the potential energy be $PE = x$. According to the problem,the kinetic energy is $KE = 3x$. By the law of conservation of energy,$KE + PE = E$. $3x + x = mgs \Rightarrow 4x = mgs \Rightarrow x = \frac{mgs}{4}$. Since $PE = mgh = x$,we have $mgh = \frac{mgs}{4}$,which gives the height $h = \frac{s}{4}$. Now,for the speed $v$,we use $KE = \frac{1}{2}mv^2 = 3x$. $\frac{1}{2}mv^2 = 3 \left( \frac{mgs}{4} \right) = \frac{3mgs}{4}$. $v^2 = \frac{3gs}{2} \Rightarrow v = \sqrt{\frac{3gs}{2}}$. Thus,the height is $\frac{s}{4}$ and the speed is $\sqrt{\frac{3gs}{2}}$.

$A$ particle is released from a height $s$. At a certain height,its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively:

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