The escalator is powered by a 7.5 kW electric motor. (i) State the equation linking efficiency, useful energy output and total energy input. (ii) The escalator lifts 30 people each minute. Each person has a mass of 78 kg and is lifted a height of 5m. Calculate the efficiency of the escalator.

(i) The equation linking efficiency, useful energy output and total energy input is:

Efficiency = (Useful energy output / Total energy input) x 100%

(ii) To calculate the efficiency of the escalator, we first need to calculate the useful energy output and the total energy input.

Useful energy output can be calculated using the formula for gravitational potential energy, which is mass x gravity x height.

For one person, the useful energy output is 78 kg x 9.81 m/s² x 5 m = 3823.9 Joules.

Since the escalator lifts 30 people each minute, the total useful energy output per minute is 30 x 3823.9 Joules = 114717 Joules.

The total energy input is the power of the motor times the time. Power is energy per unit time, so the energy input for one minute is 7.5 kW x 1 minute. However, we need to convert kW to W and minute to seconds to match the units of the useful energy output.

So, the total energy input is 7500 W x 60 s = 450000 Joules.

Now we can calculate the efficiency using the formula:

Efficiency = (Useful energy output / Total energy input) x 100%

Efficiency = (114717 Joules / 450000 Joules) x 100% = 25.49%

So, the efficiency of the escalator is approximately 25.49%.