A car of mass 1200\text{ kg} ascends a straight road inclined at an angle \theta to the ho… — JAMB UTME

A car of mass $1200\text{ kg}$ ascends a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$ , at a steady speed of $15\text{ m s}^{-1}$ . If the total frictional resistance to motion is $400\text{ N}$ , what is the useful mechanical power output of the engine in kilowatts? (Take $g = 10\text{ m s}^{-2}$ )

Answer: 24 kW

Answer

24 kW

To maintain a constant ascending speed, the engine must supply a force equal to the sum of the component of weight parallel to the incline (

mg\sin\theta = 1200\text{ N}

) and the frictional force (

400\text{ N}

), resulting in a total force of

1600\text{ N}

. Multiplying this total force by the constant speed of

15\text{ m s}^{-1}

yields a power output of

24000\text{ W}

, which corresponds to

24\text{ kW}

Step-by-Step Solution

Calculate the gravitational force component acting down the slope

1200 N

The component of the car's weight parallel to the incline opposes upward motion:

F_g = m g \sin\theta = 1200 \text{ kg} \times 10 \text{ m s}^{-2} \times 0.1 = 1200 \text{ N}

Determine the total tractive force required from the car engine

1600 N

Because the velocity is constant, the net force is zero; hence, the engine force must balance both the gravitational slope component and the frictional resistance:

F_{\text{engine}} = F_g + F_{\text{friction}} = 1200 \text{ N} + 400 \text{ N} = 1600 \text{ N}

Calculate the mechanical power delivered by the engine

24 kW

Power is the product of tractive force and constant speed:

P = F_{\text{engine}} \times v = 1600 \text{ N} \times 15 \text{ m s}^{-1} = 24000 \text{ W} = 24 \text{ kW}

Key Concept

Power required to maintain motion against opposing forces on an inclined plane

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