A small block of mass 200 g is situated at the top of a...

By the principle of conservation of mechanical energy, the potential energy of the block at the top is converted entirely into kinetic energy when it reaches the ground.

$mgh = \dfrac{1}{2}mv^2$

$v = \sqrt{2gh}$

Given that the height of the incline is $h = 3.2 \text{ m}$ and $g = 10 \text{ m/s}^2$, we substitute these values into the equation:

$v = \sqrt{2 \times 10 \times 3.2}$

$v = \sqrt{64} = 8.0 \text{ m/s}$

The mass of the block and the length of the incline are extraneous information.