An elevator and its load have a combined mass of 1700 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 13 m/s, is brought to rest with constant acceleration in a distance of 43 m.

Answer 1

The cable tension will increase to
T = M(g+a)
when the elevator is slowing down at deceleration rate a.

You can get a from the relation
V^2/2 = a X
where X is the stopping distance, and V is the initial velocity.

Answer 2

6089

Answer 3

169/2=a*43
84.5=a*43
a=84.5/43
a=1.965
T=M(g+a)
T=1700(-9.8+1.965)
T=1700*(-7.835)
T=-13,319.5