# Elevator Problem

1. Apr 15, 2004

### KingNothing

Hey all..here is another stumper (for me).

"A 1000 kg elevator has a maximum load of 800 kg. If constant friction retards the elevator at 4000 N, to lift at 2 m/s, what does the motor's hp(horsepower) have to be?"

Now, I was trying to do a lot of things with this one. I thought it was an obstacle that we had a speed and not a distance. I tried getting around this by simply acknowledging that since we went to go 2 m/s, and since a watt is 1 J/s, that I could jsut sort of throw the /s off each of them and leave it at 2 m. Prolly totally off.

Any Help?

2. Apr 15, 2004

### Chen

Use energies. From point A to a higher point B, the equation looks like this:

$$W_{motor} + W_{friction} = \Delta E_m = \Delta E_p + \Delta E_k$$

$$Pt - f\Delta h = mg\Delta h + (0)$$

Where P is the power of the motor, t is the time the eleavtor takes to travel the distance from A to B, f is the force of friction and Δh is the distance from A to B. We need to get rid of the time there, so we express it as t = Δh/v, where v is the elevator's constant speed:

$$P\frac{\Delta h}{v} + f\Delta h = mg\Delta h$$

Now rearrange the equation, cancel Δh and you get:

$$P = v(mg - f)$$

And voila. I get P = 36.6hp.

Last edited: Apr 15, 2004
3. Apr 15, 2004

### KingNothing

I love you Chen.