- #1

BitterSuites

- 38

- 0

## 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?