# An Elevator and work

1. Oct 26, 2007

### KMjuniormint5

[SOLVED] An Elevator and work

1. The problem statement, all variables and given/known data
The loaded cab of an elevator has a mass of 3.0 103 kg and moves 204 m up the shaft in 23 s at constant speed. At what average rate does the force from the cable do work on the cab? Answer in kW.

2. Relevant equations
Power is the dot product of the force and velocity
Fnet = ma

3. The attempt at a solution
I went about this problem by since Power is the dot product of velocity and the force.
The force would be Tension - W = ma and since ma = 0. tension = W = mg so there is your force m*g

then i take 204m/23s to get the velocity of 68 m/s so. . .

mg*68 to give you the power in watts but since they want the velocity in kW, you divide by 1000 to get to that desided amount. . .

2. Oct 26, 2007

### PhanthomJay

Yes. But check your math in calculating velocity.

3. Oct 26, 2007

### KMjuniormint5

haha too funny. . .wow do not know where that number came from

4. Oct 26, 2007

### KMjuniormint5

ok so I take my force (mg) and multiply it by 8.87 to give me Power . . but is power what I am searching for?

5. Oct 27, 2007

### PhanthomJay

Yes. Power is the rate at which work is done (P=W/t). In this problem the work done by the cable force is the product of its force times distance(mgd), so the power is mgd/t, which is the same as mgv.

6. Oct 27, 2007

### KMjuniormint5

I do not know what i was doing. . some little mistake. . thanks phanthomJay