Tilt a surface by an angle $\theta$ (the Greek letter theta, said “THAY-ta”) above the horizontal and gravity's single downward pull splits into two perpendicular jobs. Along the slope it drives the slide, and there its share is $g\sin\theta$. Into the slope it presses the body against the surface, and there its share is $g\cos\theta$. Sine goes with the side OPPOSITE the angle, cosine with the side that touches it — swap them and every answer in this lesson is wrong.

So a body sliding freely down a smooth slope has $a = g\sin\theta$ — a equals g sine theta — where $a$ is the acceleration down the slope in $\mathrm{m/s^2}$, $g = 9.8\ \mathrm{m/s^2}$, and $\theta$ is the slope angle above horizontal. Look hard at what is missing: the mass. It cancels, so on a frictionless ramp a marble and a machine tool accelerate identically. Check the ends, too — at 0° the sine is zero and nothing moves; at 90° it is one and you have free fall.

Add friction and the press does the paying: the normal force is $N = mg\cos\theta$, $N$ the normal force in newtons, $m$ the mass in kilograms — and kinetic friction claims $\mu_k$ of it, where $\mu_k$ (mu-k, said “mew”) is the coefficient of kinetic friction, a naked number with no units at all. Subtract that from the driving share and the mass cancels a second time, leaving $a = g\left(\sin\theta - \mu_k\cos\theta\right)$. If $\mu_k$ ever reaches $\tan\theta$ the bracket hits zero and nothing slides at all — that angle has a name, the angle of repose, and it is why gravel piles have the shape they do.