Calculating Power for Escalator in Dept Store

[pooface]

Homework Statement

An escalator in a department store is designed to carry 300 passengers per minute from the ground floor to the mezzanine, 10m vertically higher. The design assumes an average weight per person of 70 kg. Allowing 30 percent loss for friction, calculate the power output which the driving motor must have.

Homework Equations

KE = 0.5mv^2

The Attempt at a Solution

So calculating the velocity of the escalator
since it travels 10 meters in 60 seconds
10m/60 = 0.16667 m/s

mass is 300 passengers * 70kg = 21000 kg

work=0.5mv^2 = 291.167 J

power = 291.167J/60 s = 4.86 W

Is this good so far? 4.86 W is pretty low.

 

[anap40]

You also need to account for the potential energy(massxgravityxheight) that each person gains from moving up 10m

 

[pooface]

so in this case it would be

work=0.5mv^2 + mgh?

= 2.06 MJ

power = 34.34 KW

so i should add 30% to this for the motor to compensate for 30% frictional loss?

power the motor should output = 44.642 KW?

thanks.

 

[anap40]

That looks correct to me.