Answer
The tension in the cable must be 17,600 N.
Work Step by Step
We can use kinematics to find the acceleration:
$a = \frac{v^2-v_0^2}{2y} = \frac{0-(3.5 ~m/s)^2}{(2)(2.6 ~m)} = -2.36 ~m/s^2$
The tension $F_T$ provides the force to decelerate the elevator at a rate of $2.36 ~m/s^2$:
$\sum ~F = ma$
$F_T - mg = ma$
$F_T = m(a+g)$
$F_T = (1450 ~kg)(2.36 ~m/s^2 + 9.80 ~m/s^2)$
$F_T = 17,600 ~N$
The tension in the cable must therefore be 17,600 N.
