Question

An elevator in a tall building is allowed to reach a maximum speed of 3.0 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.4 m if the elevator has a mass of 1500 kg including occupants? Need step by step help please!

Answer 1

find the accelearation:

Vf^2=Vi^2+2ad

a= -1/2 Vi^2/d= in my head....-4.5/2.4

tension= mass(g+4.5/2.4)