An elevator can carry a maximum load of $$1800$$ kg (elevator$$ $$+$$ $$passengers) is moving up with a constant speed of $$2$$ $$ms-1$$. The frictional force opposing the motion is $$4000$$ N. What is minimum power delivered by the motor to the elevator?

Question

An elevator can carry a maximum load of $$1800$$ kg (elevator$$ $$+$$ $$passengers) is moving up with a constant speed of $$2$$ $$ms-1$$. The frictional force opposing the motion is $$4000$$ N. What is minimum power delivered by the motor to the elevator?

Infinity Learn · Accepted Answer

Here, m $$=$$ $$1800$$ kgFrictional force, f $$=$$ $$4000$$ NUniform speed, v $$=$$ $$2$$ $$ms-1Downward$$ force on elevator isF $$=$$ mg $$+$$ f     $$=$$ (1800$$ kg × $$10$$ $$ms-2)+4000$$ N $$=$$ $$22000$$ NThe motor must supply enough power to balance this force. Hence,P $$=$$ Fv $$=$$ $$(22000$$ $$N)(2$$ $$ms-1)  $$=$$ $$44000$$ W $$=$$ $$44$$×$$103$$ W $$=$$ $$44$$ kW

An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ${ms}^{- 1}$ . The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms-1. The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ${ms}^{- 1}$ . The frictional force opposing the motion is 4000 N. What is minimum power delivered by the motor to the elevator?