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  4. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: (1 HP = 746 W, g = 10 ms-2 )

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Q. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : ($1 \,HP = 746 \,W, \,g = 10\, ms^{-2}$ )

JEE MainJEE Main 2020Work, Energy and Power

Solution:

$4000 × V + mg × V = P$
$\frac{60\times746}{4000+20000}=V$
$V=1.86\,m/s. \approx 1.9\,m/s.$