An elevator and its load weigh a total of 166 kg. Find the tension T in the supporting cable when the elevator, originally moving downwards at `20m//s` is brought to rest with constant acceleration in a distance of 50 m.

Explanation

As the lift, moving downwards, stops after travelling a distance of 50 m from `v^(2)=u^(2)+2as`
we get, `0^(2)=20^(2)+2axx50`
i.e., `a=-4 m//s^(2)`
i.e., lift is accelerated up with an acceleration `4m//s^(2)`
`W=m(g+a)=1600(10+4)=22400 "newton"`