Answer
The elevator's speed at $~t = 4.00~s~$ is $~2.27~m/s$
Work Step by Step
We can find the downward acceleration of the elevator:
$\sum F = ma$
$mg-F_T = ma$
$a = \frac{mg-F_T}{m}$
$a = \frac{(832~kg)(9.80~m/s^2)-7730~N}{832~kg}$
$a = 0.51~m/s^2$
We can find the velocity at $t= 0$:
$\Delta y = v_0~t+\frac{1}{2}at^2$
$v_0 = \frac{\Delta y-\frac{1}{2}at^2}{t}$
$v_0 = \frac{5.00~m-(\frac{1}{2})(0.51~m/s^2)(4.00~s)^2}{4.00~s}$
$v_0 = 0.23~m/s$
We can find the velocity at $t = 4.00~s$:
$v_f = v_0+at$
$v_f = 0.23~m/s+(0.51~m/s^2)(4.00~s)$
$v_f = 2.27~m/s$
The elevator's speed at $~t = 4.00~s~$ is $~2.27~m/s$
