To calculate the average power output of the elevator, we need to apply the following formulas:

1. Work done (W) by the elevator is calculated using: $W = F \cdot d$ where $F$ is the force, and $d$ is the distance the elevator moves.

2. The force $F$ is equal to the weight of the elevator, which is: $F = m \cdot g$ where $m = 1300 \, \text{kg}$ is the mass of the elevator and passengers, and $g$ is the acceleration due to gravity ($g \approx 9.8 \, \text{m/s}^2$).

3. Substituting the values: $W = 1300 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 50 \, \text{m}$

4. Finally, power $P$ is the work done divided by the time $t$: $P = \frac{W}{t} = \frac{1300 \times 9.8 \times 50}{30}$

Now, let's calculate this step-by-step:

$W = 1300 \times 9.8 \times 50 = 637,000 \, \text{Joules}$

$P = \frac{637,000}{30} = 21,233.33 \, \text{W} \approx 2.1 \times 10^4 \, \text{W}$

Thus, the average power output is approximately $2.1 \times 10^4 \, \text{W}$, which corresponds to option (d).

Would you like further clarification or more details?

Here are 5 questions to extend your understanding:

1. How would the power change if the elevator's mass increased?
2. What happens to power if the elevator moved faster but over the same distance?
3. How is power related to the energy consumption of the elevator?
4. How would the result differ if the elevator was moving downward?
5. What is the significance of "constant speed" in this problem?

Tip: Remember, constant speed means no net acceleration, so the force required is only to counteract gravity.