Question Bank • QB_FP_142 NAT • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: I.C. engine systems – fuel, cooling, lubrication, ignition, electrical, intake and exhaust

Question Statement:

A pair of bullocks each weighing $450\text{ kg}$ develops an average pull equal to $10\%$ of their combined body weight while plowing at a walking speed of $3.6\text{ km/h}$ ($1.0\text{ m/s}$). Calculate the total power developed by the bullocks in $\text{kW}$ (take $g = 9.81\text{ m/s}^2$).

Answer Key:

0.88

Step-by-Step Solution & Explanation

Total weight of bullocks: $W = 2 \times 450 = 900\text{ kg}$ Pull exerted: $P = 0.10 \times 900\text{ kg} \times 9.81\text{ m/s}^2 = 882.9\text{ N}$ Forward speed $v = 1.0\text{ m/s}$. Power developed: $\text{Power} = \frac{P \times v}{1000} = \frac{882.9 \times 1.0}{1000} \approx 0.883\text{ kW}$ Acceptable range: $0.86 - 0.90\text{ kW}$.
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