Question

A bus of mass 3 metric tons is pulled at a speed of 18 kmph on a smooth incline of inclination of $\theta ={37}^o$. Power of the engine is, $\left(g=10m{s}^{-2}\right)$.

• A. $15kW$
• B. $90kW$
• C. $30kW$
• D. $150kW$

Answer 1

The correct option is B $90kW$

man of engine bus $=3\times {10}^3$ kg $[1ton={10}^{3}kg]$

speed of bus $=18\times \frac{5}{18}$

$v=5m/s$ [convert kmph to m/s]

From the adjacent figure, the bus engine should apply a constant force of $F=mg\sin \theta$ so that it moves upwards

at a constant speed. $(Net\overrightarrow{a}=0)$

So, power applied by engine:

$P=Fv$

$=mgv\sin \theta$

Putting values,

$P=3\times {10}^3\times 10\times 5\times \sin \left(370\right)$

$=15\times {10}^4\times \frac{3}{5}=9\times {10}^4\Rightarrow P=90$ KW