Question

A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18km/h. If g = 10 m/s², the power of its engine is :

• A. 4 kW
• B. 50 kW
• C. 625 kW
• D. 25 kW

Answer 1

Answer: (A)

4 kW

Answer 2

• Given Information:

  • Mass of the car (m) = 1000 kg
  • Speed of the car (v) = 18 km/h
  • Acceleration due to gravity (g) = 10 m/s²
  • Slope of the incline = 2 in 25

• Convert Speed to SI Units: $v = 18 \text{ km/h} = 18 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s}$

• Interpret the Slope: A slope of "2 in 25" means that for every 25 units of distance traveled along the incline, the vertical height gained is 2 units. Therefore, the sine of the angle of inclination (θ) is: $\sin(\theta) = \frac{\text{Vertical Rise}}{\text{Distance along Incline}} = \frac{2}{25}$

• Calculate the Force Required: Since the car is moving at a "steady speed," there is no acceleration. This implies that the engine force pulling the car up the incline is equal to the component of gravity pulling the car down the incline. The component of gravitational force acting down the incline is $F_g = mg \sin(\theta)$. Engine Force ($F_{engine}$) = $F_g = mg \sin(\theta)$ $F_{engine} = 1000 \text{ kg} \times 10 \text{ m/s}^2 \times \frac{2}{25}$ $F_{engine} = 10000 \times \frac{2}{25} = 400 \times 2 = 800 \text{ N}$

• Calculate the Power of the Engine: Power (P) is the product of force and velocity. $P = F_{engine} \times v$ $P = 800 \text{ N} \times 5 \text{ m/s}$ $P = 4000 \text{ W}$

• Convert Power to Kilowatts: $P = 4000 \text{ W} = 4 \text{ kW}$

The power of the engine is 4 kW.