Motion on an Inclined Plane

From Force to Motion

Choose distance s positive down the ramp. The normal balance gives N = mg cos θ. Rest is possible when mg sin θ ≤ μ_sN. Static friction then matches the downhill pull; it is not automatically at its maximum.

Once the block slips downhill, f = μ_kN and Newton's second law becomes

a = g (sin θ − μ_k cos θ).

Integrating from the initial time t₀ fixes both constants of integration:

v(t) = v₀ + a(t − t₀)
s(t) = s₀ + v₀(t − t₀) + ½a(t − t₀)².

Here t₀ = s₀ = v₀ = 0. For a sliding release, the arrival time is √(2L/a) and the arrival speed is √(2aL). The mass cancels from the acceleration, but forces and energies still scale with mass.

Where the Energy Goes

With zero gravitational potential at the foot, K = ½mv², U = mg(L − s)sin θ, and the dissipated mechanical energy is Q = fs. Throughout each saved trajectory, K + U + Q = mgL sin θ = 98.1 J.

Frictionless motion converts all 98.1 J into kinetic energy. Sliding friction converts part into thermal/internal energy of the contact system. Static friction does no work on a block that does not move.

Assumptions

A nonrotating block on a fixed straight ramp; constant gravity; Coulomb friction with 0 ≤ μ_k ≤ μ_s; no air drag. The animation ends before any collision or change of surface. Force arrows share a scale; the block is drawn larger for visibility.