Question

A truck of mass 1200 kg moves over an inclined plane raising 1 in 20, with a speed of 18 kmph. The power of the engine is 
(g = 10 m/s\(^{-2}\)):

• A. \(2 { kW}\)
• B. \(3 { kW}\)
• C. \(3.6 { kW}\)
• D. \(1 { kW}\)

Hint:

When calculating power on an incline, always consider the component of gravitational force parallel to the incline.

Answer 1

The Correct Option is B

Step 1: Calculate the velocity in m/s and the force needed to overcome gravity. The speed of the truck in m/s is \( v = \frac{18 \times 1000}{3600} = 5 { m/s} \). The incline ratio is 1:20, so the vertical height per meter horizontally is \( \frac{1}{20} \). Thus, the force due to gravity is: \[ F = mg \sin(\theta) = 1200 \times 10 \times \frac{1}{20} = 600 { N}. \] Step 2: Calculate the power required. Power is given by \( P = Fv \): \[ P = 600 \times 5 = 3000 { W} = 3 { kW}. \]