Inclined plane

When you release a block on a tilted board, it does not fall straight down. It travels along the board and gains speed more slowly than a ball dropped from your hand. That is an inclined plane: a slope or ramp at angle α to the horizontal, where motion is easier to time than a vertical drop.

Weight mg acts straight down. On the board it is useful to split it into two parts: a parallel component mg sin α (down the slope) and a perpendicular component mg cos α (into the board). The normal force balances the perpendicular part, so N = mg cos α.

With no friction, only mg sin α remains along the board. Newton’s second law is F = m·a, so mg sin α = m·a. Mass cancels and you get:

a = g sin α

A heavier block and a lighter one (no drag, no friction) share the same acceleration. What matters is the angle and g, not the mass. It is the same idea as free fall, only the full g is cut by sin α.

First numeric intuition: at α = 30°, sin 30° = 1/2. With g = 9.81 m/s² you get a = 4.905 m/s², or about 4.91 m/s²: exactly half of g. As α approaches 0° the board is nearly level and a falls to zero. As α approaches 90° the board becomes vertical and a returns to g.

Once you have a, a slide of length L is ordinary uniformly accelerated motion along the board. Starting from rest (v₀ = 0), L = ½·a·t², so:

t = √(2L/a)

The speed at the bottom follows from v_k² = 2·a·L:

v_k = √(2·a·L)

Second numeric intuition: for α = 30°, L = 5 m and g = 9.81 m/s² you get a ≈ 4.91 m/s², t ≈ 1.43 s and v_k ≈ 7 m/s. That is a short warehouse ramp or a slide segment: under a second and a half and already about 7 m/s at the bottom. The same numbers appear in the example below: work them out by hand, then check in the frictionless-incline calculator.

With kinetic friction μ, the friction force T = μ·N = μ·mg cos α acts up the slope while sliding down. The net force along the board is mg sin α − μ·mg cos α. From F = m·a, mass cancels again:

a = g (sin α − μ cos α)

To slide at all you need sin α > μ cos α, that is tan α > μ. If at rest tan α is smaller than the static friction coefficient, the block may not start. Problem texts often give one μ and assume motion has already begun.

For α = 30° and μ = 0.1, cos 30° = √3/2 ≈ 0.866, so a = g(0.5 − 0.1·0.866) ≈ g·0.4134 ≈ 4.06 m/s². The same L takes longer, and v_k at the bottom is smaller. Good habit: ideal case first, then μ. The gap from 4.91 m/s² to 4.06 m/s² is already a clearly slower slide.

Which tool? Ideal board, no mention of friction: incline without friction. When μ is given: incline with friction. When you already know a and mass and want net force: F = m·a.

Units: α in degrees (DEG mode on the calculator), L in meters, g and a in m/s², t in seconds, v in m/s. Do not enter “30” while thinking in radians. Compare a with g: for a slide on a fixed board you always have a ≤ g. If a comes out larger than g, hunt a sin/cos mix-up or the angle mode.

Without friction you get the same v_k from energy: the loss Ep = mg·Δh becomes Ek = ½ m v², and Δh = L sin α, so v = √(2 g L sin α). That matches √(2 a L) with a = g sin α. With friction, friction work takes some of the energy, so you arrive slower. Energy explains “why slower”; kinematics with a gives the time. When the two routes disagree, you usually mixed L with height h.

A dramatized common mistake looks like this: someone sees the board angle, takes cos instead of sin, and writes a = g cos α. At 30° that yields a ≈ 8.50 m/s² instead of 4.91 m/s² and a “surprisingly fast” slide. Or they take L as height instead of board length, drop μ into the frictionless formula, or expect mass to change a in the no-drag model. A percent grade (for example 10%) is not a 10° angle: convert to α first.

Before you calculate, sketch the slope and mark mg, N, and T if present. Axis along the board, axis perpendicular. Write F = m·a along the slope, find a, then use the formulas for t and v_k. If a comes out negative for a “slide down”, check μ and α: in the model the body may not accelerate downhill at all.

The model leaves out strong air drag, rolling with rotational energy, and a board whose angle changes during the slide. For a typical slide and a quick ramp estimate, fixed α and fixed μ are enough. When the body rolls, part of the energy goes into spin and bottom v falls below the sliding result.

The incline joins F = m·a with simple kinematics and a bridge to energy. Once you can find a from the angle, the natural next step is the Ek and Ep balance in energy in motion: without friction, bottom speed depends on height; with friction, on path length and μ.

Before you start, check: (1) friction silent or μ given, (2) L is board length not height, (3) α in degrees, (4) a between zero and g, (5) whether the question wants a, t, v_k, or F = m·a. Then open the matching calculator and compare with the 30° and L = 5 m intuition.