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Determine the horse power required to drive a lift in raising a load of 2000 Kgf at a speed of 2m/sec ,if the efficiency is 70%.




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Load (F)= 2000 kgf = 2000×9.8 N = 19600N = 19.6 kN
Speed = 2m/s
Power = Force × velocity = 19.6×2 = 39.2 kW
But efficiency = 70%
Let power required = P

 \frac{39.2}{P}  \frac{70}{100}

⇒P =  \frac{100*39.2}{70}

⇒P = 56 kW
1kW = 1.341hp
Thus power required = 56×1.341 = 75.1hp

0 0 0
they r 75.2 hp
76.2 hp ,86.2 hp ,96.2 hp
it's just how much accuracy you are looking at.... actually i had got 75.09722408
the answer will be 75.2
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