Answer to Question #252910 in Mechanics | Relativity for rah

Question

The walkway system has an efficiency of 78%. Calculate the power input to the system when

15 passengers of average mass 72 kg are standing on the walkway.

Answer 1

The weight of the passengers is

$$W = mg$$

where $m = 1080kg$ is they mass, and $g = 9.8N/kg$ is the gravitational acceleration. The component of the weight that acts along the walkway is:

$$W_a = W\sin \theta$$

where $\theta = 30\degree$.

By definition, the power is:

$$P = Wv$$

where $v = 0.51m/s$ is the speed of motion. Thus, obtain:

$$P = mgv\sin\theta\\ P = 1080\cdot 9.8\cdot 0.51\cdot \sin 30\degree \approx 2698.92W$$

Then, given the efficiency of the walkway, the power input has to be equal to the required power input (2698.92 W) divided by the efficiency (78% = 0.78):

$\text{efficiency} = \frac{\text{Power delivered}}{\text{Power input}} \\ \implies \text{Power input} = \frac{\text{Power delivered}}{\text{efficiency}}= \frac{2698.92 \text{ W}}{0.78} \\ \implies \text{Power input} = 3460.15 \text{ W}$

Answer. 3460.15 W.