Power from driving force at constant velocity

Steps:

• Gravitational component down slope: $mg \sin 6^\circ = 1500 \times 9.8 \times 0.1045 \approx 1536$ N.
• Driving force from engine: $F = 2000 - 1536 = 464$ N (balances excess resistance).
• Power: $P = F \times v = 464 \times 30 = 13920$ W $= 14$ kW.

Why A is correct:

• Power equals driving force times velocity, where driving force equals resistive force minus downhill gravitational component for zero net force.

Why the others are wrong:

• B ignores gravitational assistance, using full resistance for force.
• C uses full weight $mg$ instead of $mg \sin \theta$.
• D approximates $\sin 6^\circ$ poorly or uses $g = 10$ m/s² incorrectly.

Final answer: A