Physics, asked by kharevaibhavi621, 2 days ago

An elevator weighing 200 kg is to be lifted up atte constant velocity of 0.4 m/s. What would be the minimum horse power of the motor to be used?​

Answers

Answered by meshramkhushi746
8

Answer:

1.07hp

Explanation:

power=work done/time

=mgD/t

=200×10×0.4/1=8000w

=1.07hp

Answered by Anonymous
1

Given - Mass and velocity

Find - Minimum horsepower of the motor

Solution - The minimum horsepower of the motor to be used is 1 HP.

Required formulas for solving the question are as follows -

Power = Force*velocity

Force = mass*accelaration due to gravity.

Power = mass*accelaration due to gravity*velocity

Power = 200*10*0.4

Power = 800 Watt

Now, 1 Watt = 1/736 Horsepower

800 Watt = 1 Horsepower

Thus, the minimum horsepower of the motor to be used is 1 HP.

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