Answer
(a) $\sum F = F_T - mg$
The maximum acceleration is $a = 2.93~m/s^2$.
(b) On the moon, the maximum acceleration is $a = 11.1~m/s^2$.
Work Step by Step
(a) $\sum F = F_T - mg$
Let's assume that $F_T = 28,000~N$. We can find the maximum acceleration of the elevator.
$F_T - mg = ma$
$a = \frac{F_T-mg}{m} = \frac{(28,000~N)-(2200~kg)(9.80~m/s^2)}{2200~kg}$
$a = 2.93~m/s^2$
The maximum acceleration is $a = 2.93~m/s^2$.
(b) $F_T - mg = ma$
$a = \frac{F_T-mg}{m} = \frac{(28,000~N)-(2200~kg)(1.62~m/s^2)}{2200~kg}$
$a = 11.1~m/s^2$
On the moon, the maximum acceleration is $a = 11.1~m/s^2$.
