An escalator joins one floor with another one 8.20 m above. The escalator is 13.3 m long and moves along its length at 62.0 cm/s. (a) What power must its motor deliver if it is required to carry a maximum of 100 persons per minute, of average mass 75.0 kg? (b) An 83.5-kg man walks up the escalator in 9.50 s. How much work does the motor do on him? (c) If this man turned around at the middle and walked down the escalator so as to stay at the same level in space, would the motor do work on him? If so, what power does it deliver for this purpose? (d) Is there any (other?) way the man could walk along the escalator without consuming power from the motor?

							
According to conservation of energy 
Work done to carry persons = change in potential energy 
Work = mgh₁ - mgh₂ 
Initially they were at ground level h₁ = 0 
Hence, Work = -m(-g)h₂ = mgh₂ = 100 × 75× 9.8 × 8.2 = 602700J 
Power = Work/time = 602700/60 = 10045W = 10.045kW
						

							
(a) P = mgh=t, but m=t is the persons per minute times the average mass, so
P = (100 people/min)(75.0kg)(9.81m/s2)(8.20 m) = 1.01*10^4W:
(b) In 9.50 s the Escalator has moved (0.620m/s)(9.50 s) = 5.89 m; so the Escalator has \lifted"
the man through a distance of (5.89 m)(8.20m/13:3 m) = 3.63 m. The man did the rest himself.
The work done by the Escalator is then W = (83.5 kg)(9.81m/s2)(3.63 m) = 2970 J.
(c) Yes, because the point of contact is moving in a direction with at least some non-zero component
of the force. The power is
P = (83.5m/s2)(9.81m/s2)(0.620m/s)(8.20m/13:3 m) = 313W:
(d) If there is a force of contact between the man and the Escalator then the Escalator is doing
work on the man.