Question

Find the final velocity of the block when it has reached bottom of the incline after releasing from rest.

$a = gsin\theta$ $v^2 = u^2 + 2as$ $v^2 = 2gsin\theta s$

Answer 1

v = \sqrt{2gh}

Answer 2

For a block starting from rest and sliding down a frictionless incline, the loss in gravitational potential energy equals its gain in kinetic energy. That is,

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

Solving for $v$:

$$v = \sqrt{2gh}$$

Loss in gravitational potential energy ($mgh$) converts to kinetic energy ($\frac{1}{2}mv^2$), giving $v = \sqrt{2gh}$.