Question

An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor). Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend (during the period of constant acceleration). Take the acceleration due to gravity to be 9.81 m/s2. Show your work.

Answer 1

Step-by-step explanation:

The lift is going down with acceleration

Initial speed u = 0

Final speed v = 6 m/s

distance s = 15.25 m

acceleration a = ?

v² = u² + 2 a s

6² = 0 + 2 x a x 15.25

a = 1.18 m /s²

Elevator is going down with acceleration .

mg - T = ma where T is tension in the cable .

722 x 9.8 - T = 722 x 1.18

7075.6 - T = 851.96

T = 6223.64 N .