1. The problem statement, all variables and given/known data A typical escalator in the London Underground rises at an angle of 30° to the horizontal. It lifts people through a vertical height of 15 m in 1 minute. Assuming all the passengers stand still whilst on the escalator, 75 people can step on at the bottom and off at the top in each minute. Take the average mass of a passenger to be 75 kg. (a) Find the power needed to lift the passengers when the escalator is fully laden. For this calculation assume that any kinetic energy given to the passengers by the escalator is negligible. (b) The frictional force in the escalator system is 1.4 * 104 N when the escalator is fully laden. Calculate the power needed to overcome the friction. Hence find the power input for the motor driving the fully laden escalator, given that the motor is only 70 % efficient. (c) When the passengers walk up the moving escalator, is more or less power required by the motor to maintain the escalator at the same speed? Explain your answer. Answers: (a) 1.4 * 104 W, (b) 7.0 * 103 W, 3.0 * 104 W 2. The attempt at a solution (a) I. P = W / t = Fs / t = mgs / t mg = (75 * 75) * (10) [simplified g) = 56 250 kg Since the force (F = mg) is inclined at an angle 30°, therefore F = mg cos 30°, and the height h equals s / tan 30°. P = [56 250 * cos 30° * (15 / tan 30°)] / 60 seconds = 21 094 W -- wrong. II. P = mgs / t = (56 250 * 15) / 60 = 14 063 W -- looks correct. But why don't we use the 30°? (b) I. P = [(F - R) * s] / t 14 063 = (15F - 210 000) / 60 F = 70 252 N II. P = Rs / t = (14 000 * 15) / 60 = 3500 W In any case I am confused about (a) and not sure about (b) either. Any ideas please?