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Q.
A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at a steady speed of 18 kmph. If $g=10 \, ms^{- 2}$ , the power of its engine is
NTA AbhyasNTA Abhyas 2022
Solution:
Power of engine can also be expressed as, $P=\overset{ \rightarrow }{F}\cdot \overset{ \rightarrow }{v}$
Component of force along the inclined plane, $F=mgsin\theta =1000\times 10\times \frac{2}{25}=800N$
Component of velocity along the plane is,
$v=18kmph=18kmph\times \frac{1000}{3600}\frac{m / s}{kmph}=5m/s$
Power $P=Fvcos\phi,$ $\phi$ is the angle between $\overset{ \rightarrow }{F}and\overset{ \rightarrow }{v}$
$Power=800\times 5\left(\because \phi = 0 ^\circ \right)=4kW$