Question

An elevator weighs 2500 𝑙𝑏; a cable, which can sustain a maximum tension of 6000 𝑙𝑏, supports it. If the elevator is going down with a velocity of 9 𝑓𝑡/𝑠, find the minimum safe distance it can travel before coming to a stop.

Answer 1

Find the maximum acceleration that will create a 6000-lb-force on stop:

$$ma=-mg+T,\\ a=\frac Tm-g,\\ d=\frac{v^2}{2a}=\frac{v^2m}{2T-mg}=0.9\text{ ft}.$$