(i) The power required to overcome the resistance is given by:
\(P = \text{DF} \times v = 850 \times 36\)
\(P = 30600 \text{ W} = 30.6 \text{ kW}\)
(ii) The driving force (DF) is the sum of the resistance and the component of the weight along the hill:
\(\text{DF} = 1250g \times 0.1 + 850\)
The power is given by:
\(P = \text{DF} \times v\)
\(63000 = (1250 \times 9.8 \times 0.1 + 850) \times v\)
Solving for \(v\), we get:
\(v = 30 \text{ m s}^{-1}\)
(iii) Using the energy method:
Gain in kinetic energy (KE):
\(\frac{1}{2} \times 1250 \times (24^2 - 20^2) = 110000 \text{ J}\)
Loss in potential energy (PE):
\(1250 \times 9.8 \times 176 \times 0.1 = 220000 \text{ J}\)
Work done by the car's engine:
\(20000 \times 8 = 160000 \text{ J}\)
Total work done against resistance:
\(160000 + 220000 = 270000 \text{ J} = 270 \text{ kJ}\)