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
8
Answer:
1.07hp
Explanation:
power=work done/time
=mgD/t
=200×10×0.4/1=8000w
=1.07hp
Answered by
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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