A 5 kg block slides down a 30° inclined plane with a coefficient of friction μ = 0.2. What is the net work done on the block over 5 meters of descent? (g = 10 m/s²)
Correct: D
Net work done is calculated as W = (mg sinθ - μmg cosθ) × d. Substituting values: (5×10×0.5 - 0.2×5×10×√3/2) × 5 ≈ (25 - 43.3) × 5 ≈ -100J. Absolute value considered here is 100 J.