Work And Watts

1. Mar 4, 2008

BitterSuites

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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?

2. Mar 4, 2008

Staff: Mentor

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

3. Mar 4, 2008

BitterSuites

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

4. Mar 4, 2008

Staff: Mentor

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

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.

5. Mar 4, 2008

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?

6. Mar 4, 2008

Staff: Mentor

You are back on track.

7. Mar 4, 2008

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.

8. Mar 4, 2008

Staff: Mentor

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