Power equals horizontal force component times velocity

Steps:

• Horizontal component of pulling force: $F_x = 12 \cos 30^\circ = 12 \times \frac{\sqrt{3}}{2} \approx 10.4$ N.
• Since velocity is horizontal and constant, power $P = F_x \times v$.
• Substitute values: $P = 10.4 \times 2.0 = 20.8$ W.
• This is the rate of work done by the pulling force.

Why B is correct:

• Matches formula $P = F \cos \theta \times v$, where only the motion-parallel force component contributes to power.

Why the others are wrong:

• A uses vertical component ($12 \sin 30^\circ \times 2 = 12$ W), which does no work on horizontal motion.
• C uses full force magnitude ($12 \times 2 = 24$ W), ignoring angle.
• D uses weight ( W), irrelevant to pulling power.

Final answer: B