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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Similar questions

Q. 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?

An electric motor raises a load of 0.2 kg, at a constant speed, through a vertical distance of 3.0 m in 2 s. If the acceleration of free fall is 10 $m / s^{2}$ , the power in W developed by the motor in raising the load is :

(a) 0.3

(b) 1.2

(d) 6.0

If the power of a motor be 100kw, at what speed can it raise a load of 50,000 N?

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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?

The correct option is B 1.81 kWTotal mass = m = 800 kg Weight = 800 × 9.81 = 7848 N Total force = 7848 + 1200 = 9048 N Power = force × speed = 9048 × 0.2 = 1810 W = 1.81 kW

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