A hoist operated by an electric motor has a mass of 500 kg. It raises a load of 300 kg vertically at a steady speed of 0.2 m/s. Frictional resistance can be taken to be constant at 1200 N. What is the power required?

1.51 kW

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1.81 kW

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2.01 kW

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1.61 kW

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Solution

The correct option is B 1.81 kW
Total mass = m = 800 kg
Weight = 800 $\times$ 9.81
= 7848 N

Total force = 7848 + 1200
= 9048 N

Power = force $\times$ speed
= 9048 $\times$ 0.2
= 1810 W
= 1.81 kW

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