Calculate time using work done and useful output power Steps:

• Work required to lift load: W = force × distance = 40 N × 0.5 m = 20 J.
• Useful output power: P_out = efficiency × input power = 0.25 × 20 W = 5 W.
• Time to perform work: t = W / P_out = 20 J / 5 W = 4 s. Why D is correct:
• Power is work per unit time, so t = W / (η P_in) directly gives 4 s using the efficiency formula for mechanical systems. Why the others are wrong:
• A: Requires P_out = 20 J / 0.04 s = 500 W, exceeding motor capability even without losses.
• B: From error of P_out = P_in / η = 80 W, inverting efficiency wrongly.
• C: Implies P_out = 40 W, perhaps from v = P_in / F = 0.5 m/s and t = h / v but ignoring efficiency. Final answer: D