The elevator of a large building is to raise a net mass of 600 kg at a constant speed of 12 m/s using an electric motor. What is the minimum power rating of the motor?

Explanation:

Given that.

Mass=600kg

v=12m/s

Since it is moving up then gravity is acting on it

take g=9.8m/s²

Then,

Power=workdone/time

Work done=force ×distance

Power=force×distance/time

Then, velocity =distance/time

So,

Power=force ×velocity.

The force to raise the object up is the weight of the object

Force=weight=mass ×gravity

Power=mass×gravity ×velocity.

Power=600×9.8×12

Power, =70560watts

Power=70.56KW

it is a negative acceleration

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft