# 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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by Monty1992

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by Monty1992

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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

⇒ =

⇒P =

⇒P = 56 kW

1kW = 1.341hp

Thus power required = 56×1.341 =**75.1hp**

Speed = 2m/s

Power = Force × velocity = 19.6×2 = 39.2 kW

But efficiency = 70%

Let power required = P

⇒ =

⇒P =

⇒P = 56 kW

1kW = 1.341hp

Thus power required = 56×1.341 =