Question

A particle is released from a height h. At a certain height its kinetic energy is two-fifth of its potential energy, the speed of the particle at that instant is

Answer 1

Total Mechanical Energy = K.E + P.E = mgh
Given,

$KE=\frac{2}{5}PE$

Total Mechanical Energy = K.E + P.E = K.E + 5/2P.E = 7/2K.E.

mgh = 7/2K.E.

K.E.= 2mgh/7

Now we know that,

$KE=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{\frac{2KE}{m}}$

$v=\sqrt{\frac{4mgh}{7m}}$.