Power input

The construction lift lifted a load weighing 300 kg to a height of 12 m with a uniform movement. How long has it taken for the load to be lifted if the efficiency is 75% and the engine has a power input of 5 kW?

Correct answer:

t =  9.6 s

Step-by-step explanation:

P=5 kW W=5 1000  W=5000 W ξ=75%=10075=0.75 V=ξ P=0.75 5000=3750 W  m=300 kg h=12 m  g=10 m/s2  F=m g=300 10=3000 N  W=F h=3000 12=36000 J  t=W/V=36000/3750=9.6 s

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