Question

An elevator must lift 1000 kg a distance of 100 m at a velocity of 4 m/s. What is the average power the
elevator exerts during this trip?

Answer 1

To calculate the average power exerted by the elevator while lifting the 1000 kg mass, we need to determine the work done and the time taken to perform that work.

First, we can calculate the work done (W) to lift the object against gravity. The work done in lifting an object is given by the formula:

\[
W = m \cdot g \cdot h
\]

where:
- \( m = 1000 \, \text{kg} \) (mass)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 100 \, \text{m} \) (height)

Substituting in the values:

\[
W = 1000 \cdot 9.81 \cdot 100
\]
\[
W = 981000 \, \text{J}
\]

Next, we need to determine the time (t) it takes to lift the elevator. The velocity (v) is given as \( 4 \, \text{m/s} \), and the distance (d) is \( 100 \, \text{m} \). The time taken can be calculated using the formula:

\[
t = \frac{d}{v}
\]

Substituting the values:

\[
t = \frac{100}{4} = 25 \, \text{s}
\]

Now we can calculate the average power (P) using the formula:

\[
P = \frac{W}{t}
\]

Substituting in the values we calculated:

\[
P = \frac{981000 \, \text{J}}{25 \, \text{s}} = 39240 \, \text{W}
\]

Thus, the average power exerted by the elevator is:

\[
\boxed{39240} \, \text{W} \text{ or } 39.24 \, \text{kW}
\]