BLOCK ON INCLINE WITH FRICTION

On an inclined surface, gravity must be resolved into two perpendicular components: one along the slope (the driving force) and one perpendicular to it (balanced by the normal force). Converting the angle: θ = 35° × (π/180) ≈ 0.6109 rad. The along-slope gravity component is F‖ = mg sinθ = 15 × 9.807 × sin(35°) ≈ 84.4 N. This is the force trying to slide the block downhill. The perpendicular component determines the normal force: N = mg cosθ = 15 × 9.807 × cos(35°) ≈ 120.5 N. The ramp pushes back with exactly this magnitude — no perpendicular acceleration occurs. Kinetic friction resists the sliding: F_f = μk × N = 0.20 × 120.5 ≈ 24.1 N, directed up the slope. Net force along the slope: F_net = 84.4 − 24.1 ≈ 60.3 N downhill. Newton's second law gives a = F_net / m = 60.3 / 15 ≈ 4.02 m/s². Equivalently: a = g(sinθ − μk cosθ) = 9.807(0.574 − 0.20 × 0.819) ≈ 4.02 m/s². Note that mass cancels in the final formula — two blocks of different mass but the same shape and μk would accelerate identically down this slope. The most common exam error is using mg (full weight) instead of mg cosθ as the normal force, which overstates friction and understates the acceleration.