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 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\(^{-2}\)):
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When calculating power on an incline, always consider the component of gravitational force parallel to the incline.
Updated On: Mar 13, 2025
- \(2 { kW}\)
- \(3 { kW}\)
- \(3.6 { kW}\)
- \(1 { kW}\)
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The Correct Option is B
Solution and Explanation
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 \( P = Fv \):
\[
P = 600 \times 5 = 3000 { W} = 3 { kW}.
\]
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