A 0.50 kg ball is dropped from rest from a height of 20 m and reaches the ground at 18 m s−1. Taking g = 10 m s−2, find the work done against air resistance.
- Loss of PE = mgh = 0.50 × 10 × 20 = 100 J.
- Gain in KE = ½mv2 = ½ × 0.50 × 182 = 0.25 × 324 = 81 J.
- Work done against air resistance = 100 − 81 = 19 J.
Answer: 19 J, which becomes heat in the air and the ball.