In the given figure, the block of mass $$m$$ is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is

Let the ground be chosen as the reference level for zero potential energy.

At point $$A$$: The block is dropped from rest, so its initial velocity $$v_A = 0$$.

$$\text{Kinetic Energy } (K_A) = 0$$ and $$\text{Potential Energy } (U_A) = mgy$$

At point $$B$$: The block has fallen by a vertical distance $$y_0$$, meaning its height above the ground is $$(y - y_0)$$.

$$\text{Potential Energy } (U_B) = mg(y - y_0)$$

By conservation of total mechanical energy: $$K_A + U_A = K_B + U_B$$

$$0 + mgy = K_B + mg(y - y_0)$$

$$K_B = mgy - mgy + mgy_0 = mgy_0$$