Q. A particle is released from a height $s$. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
A) $\frac{s}{4},\frac{3gs}{2}$
B) $\frac{s}{4},\frac{\sqrt{3gs}}{2}$
C) $\frac{s}{2},\frac{\sqrt{3gs}}{2}$
D) $\frac{s}{4},\frac{\sqrt{3gs}}{2}$
Explanation
(d) Velocity at $B$ when dropped from $A$
where $AC = s$
$v^{2} = u^{2} + 2g(s - x)$ ….(i)
$v^{2} = 2g(s - x)$ ….(ii)
Potential energy at $B = mgx$
$\therefore$ Kinetic energy $= \ 3 \times$ potential energy
$\frac{1}{2}m \times 2g(s - x) = 3 \times mgx$
or $(s - x) = 3x$
or $s = 4x$ or $x = \frac{s}{4}$
From Eq. (i)
$v^{2} = 2g(s - x)$
$= 2g\left( s - \frac{s}{4} \right)$
$= \frac{2g \times 3s}{4} = \frac{3gs}{2}$
$\therefore\ x = \frac{s}{4}$ and $v = \sqrt{\frac{3gs}{2}}$
[image:6_image104.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.