A car weighing 1000 kg is going up an incline with a slope of 2 in 25 at steady speed of 18kmph.if g=10m per second square , the power engine is?

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Solution

Dear Student,

$s l o p e i s \tan (θ) = \frac{2}{25} = \sin (θ) (a p r o x i m a t e s o F = 1000 \times 10 \times \frac{2}{25} a n d p o w e r = F v = 10000 \times \frac{2}{25} \times 18 \times \frac{5}{18} = 4000 W$
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