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A ball of mass 2 kg is thrown up with a velocity of 20 m s⁻¹. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s⁻²).
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(i) Upward motion: gravity (down) opposes displacement (up) → negative work. Downward motion: gravity (down) is along displacement (down) → positive work.
(ii) Initial KE \(= \tfrac{1}{2}\times 2 \times 20^2 = 400\ \text{J}\). At the top, PE \(= mgh = 2 \times 10 \times 19.4 = 388\ \text{J}\), KE = 0.
Energy lost to air resistance \(= 400 - 388 = 12\ \text{J}\), so the work done by air resistance \(= \mathbf{-12\ J}\).
