In the given figure, the block of mass m is dropped from the point ' A ' . The expression for…

Here, gain in kinetic energy$=$loss in potential energy

Now loss in potential energy of particle in falling from height $y$ to $y_0$ is $mg\left(y-y_0\right)$.

So, $\Delta KE=$$K-0=-\left[mg\left(y-y_0\right)-mgy\right]$

Thus, kinetic energy at point $B$ is $K=mg{y}_0$