STOPPING DISTANCE ON ROUGH FLOOR

The puck on ice feels only one horizontal force: kinetic friction backward. On a flat surface, the normal force equals the weight m·g, so the friction force is μk·m·g. Applying Newton's second law, the deceleration is a = μk·g = 0.5 × 9.807 = 4.90 m/s². Mass cancels — a heavier puck decelerates at the same rate as a lighter one (a fact traceable directly to Amontons, and equivalent to Galileo's inclined-plane result). With uniform deceleration, the puck goes from v₀ = 20 m/s to rest in t_stop = v₀/a = 20/4.90 = 4.08 s. The stopping distance follows from the constant-acceleration kinematic formula: d = v₀t − ½at² evaluated at t = t_stop, or more neatly from energy: ½mv₀² = μk·mg·d → d = v₀²/(2μk·g) = 400/(2 × 4.90) = 40.79 m. The energy route is illuminating: all the kinetic energy goes into frictional heat. μk = 0.5 means the friction force is half the weight — quite high for ice, which normally has μk ≈ 0.01–0.03. Real ice is slippery precisely because the low μk makes the stopping distance enormous. At μk = 0.02 the same puck would travel over a kilometre before stopping, which is why hockey rinks need boards.