Question

The elevator and its passenger together have a mass of 1200.0 kg. While moving upwards, the elevator slows down from 10 m/s to 0 m/s in 2.5 seconds. How much tension (in N) must be in the cable during that time? (Use g=10N/kg)

Answer 1

The deceleration of the elevator

$$a=\frac{\Delta v}{\Delta t}=\frac{0-10}{2.5}=-4.0\:\rm m/s^2$$

The tension in the cable

$$T=m(g+a)\\ =1200(9.8-4.0)=7000\:\rm N$$