Particle Moving in an Inclined Plane

For a block of mass $m$ sliding on a frictionless incline at angle $\theta$ to the horizontal, two forces act: the weight $mg$ (vertically down) and the normal force $N$ (perpendicular to the incline surface). Choosing axes along and perpendicular to the incline (rather than horizontal/vertical) makes the algebra far simpler, since the motion is purely along the incline.

Resolving $mg$ into these tilted axes: the component perpendicular to the surface is $mg\cos\theta$, and the component along the surface (down the slope) is $mg\sin\theta$ (the angle between $mg$ and the perpendicular to the incline equals the incline angle $\theta$ itself, by the geometry of the incline).

Perpendicular direction (no acceleration here): $N-mg\cos\theta=0\Rightarrow N=mg\cos\theta$.

Along the incline (the block accelerates down the slope): $mg\sin\theta=ma\Rightarrow a=g\sin\theta$.

Notably, this acceleration depends only on the angle $\theta$, not on the mass $m$ or the length of the incline — heavier and lighter blocks slide down at the same rate on the same frictionless incline. If $\theta=90°$ (a vertical drop), $a=g$, as expected.

Using the kinematic relation $v^2=u^2+2as$ with $u=0$ (starting from rest) and $s$ = the incline's length, the speed at the bottom is

$v=\sqrt{2gs\sin\theta}.$ …