#### Answer

(a) $W_g = -9.8\times 10^4~J$
(b) $W_T = 1.08\times 10^5~J$
(c) $KE_2 = 1.0\times 10^4~J$
(d) $v = 4.47~m/s$

#### Work Step by Step

(a) We can find the work that gravity does on the elevator.
$W_g = mg~d~cos(180^{\circ})$
$W_g = (1000~kg)(9.80~m/s^2)(10~m)~cos(180^{\circ})$
$W_g = -9.8\times 10^4~J$
(b) We can find the magnitude of the tension.
$\sum F = ma$
$T - mg = ma$
$T = m(g+a)$
$T = (1000~kg)(9.80~m/s^2+1.0~m/s^2)$
$T = 10,800~N$
We can find the work done by tension.
$W_T = T~d$
$W_T = (10,800~N)(10~m)$
$W_T = 1.08\times 10^5~J$
(c) We can use the work energy theorem to find the kinetic energy of the elevator as it reaches 10 m.
$KE_2 = KE_1 + W_g+W_T$
$KE_2 = 0 + -9.8\times 10^4~J+1.08\times 10^5~J$
$KE_2 = 1.0\times 10^4~J$
(d) We can find the speed of the elevator.
$KE = 1.0\times 10^4~J$
$\frac{1}{2}mv^2 = 1.0\times 10^4~J$
$v^2 = \frac{(2)(1.0\times 10^4~J)}{m}$
$v = \sqrt{\frac{(2)(1.0\times 10^4~J)}{m}}$
$v = \sqrt{\frac{(2)(1.0\times 10^4~J)}{1000~kg}}$
$v = 4.47~m/s$