ACCELERATION ON ROUGH RAMP

Decompose the weight into two components: m·g·cos θ perpendicular to the slope (this becomes the normal force) and m·g·sin θ along the slope (this drives the block downhill). With m = 5 kg and θ = 30°, F_N = 5 × 9.807 × cos 30° = 42.46 N. Kinetic friction opposes motion, acting up the slope: F_friction = μk × F_N = 0.25 × 42.46 = 10.62 N. The net force driving the block downhill is the difference between the gravitational component and friction: F_net = m·g·sin 30° − 10.62 = 24.52 − 10.62 = 13.90 N. Applying Newton's second law, a = F_net / m = 13.90 / 5 = 2.78 m/s². Notice that mass cancels in the final formula: a = g·(sin θ − μk·cos θ). This means two blocks of different masses released simultaneously on the same ramp accelerate identically — Galileo's result survives even with friction, as long as μk does not depend on mass (Amontons' first law). The frictionless case gives g·sin 30° ≈ 4.90 m/s²; friction cuts the acceleration nearly in half here. A larger μk would reduce it further; at μk = tan 30° ≈ 0.577 the block would stop accelerating entirely — the critical angle where kinetic friction exactly balances gravity.