Solving an Elevator Power Problem: Find the Average Power in kW

[BitterSuites]

Homework Statement

A 523 kg elevator starts from rest. It moves upward for 3.79s with a constant acceleration until it reaches its cruising speed of 1.78 m/s.

The acceleration of gravity is 9.8 m/s^2.

Find the average power delivered by the elevator motor during the period of this acceleration. Answer in units of kW.

Homework Equations

w = m(a+g)
p = w/t

The Attempt at a Solution

m = 523
a = 1.78
t = 3.79

w=m(a+g)=523(1.78+9.8)=6056.34

p=w/t=6056.34/3.79=1597.98W

1597.98W becomes 1.59798kW

I was incredibly confident in this answer, but it is wrong. Where did I make my silly error?

 

[Doc Al]

Homework Equations

w = m(a+g)

m(a+g) is the force exerted, not the work. What's missing?

 

[BitterSuites]

Hmm. I guess it is missing d, as in W=Fd. Am I capable of calculating d?

 

[Doc Al]

Hmm. I guess it is missing d, as in W=Fd.

Right.

Am I capable of calculating d?

One easy way to find distance is to use average speed X time. What's the average speed?

The Attempt at a Solution

m = 523
a = 1.78
t = 3.79

Your value for the acceleration is incorrect. That's the final speed after 3.79 seconds. Use the change in speed and the time to calculate the acceleration.

 

[BitterSuites]

Ok. So a = 1.78/3.79 = .469657
avg v = (V + Vo)/2 = (1.78 + 0)/2 = .89

So, d = avg v * t = .89 * 3.79

Am I at least on the right side of the highway?

 

[Doc Al]

You are back on track.

 

[BitterSuites]

So F = m(a+g) = 523 (.469657 + 9.8) = 5371.03

d = .89 * 3.79 = 3.3731

W = Fd = 5371.03 * 3.3731 = 18117W becomes 18.117 kW

Where am I still going wrong? This was not correct.

 

[Doc Al]

You calculated the work (in Joules). Now find the power (in Watts).