Answer
The tension in the cable that supports the elevator is $22,700~N$
Work Step by Step
We can find the net force on the elevator:
$\sum F = ma = (2010~kg)(1.50~m/s^2) = 3015~N$
We can find the tension $F_T$ in the cable that supports the elevator:
$F_T-mg = \sum F$
$F_T = \sum F+mg$
$F_T = 3015~N+(2010~kg)(9.8~m/s^2)$
$F_T = 22,700~N$
The tension in the cable that supports the elevator is $22,700~N$
