#### Answer

$v = 3.76~m/s$

#### Work Step by Step

We can find the work that gravity does on the block.
$W_g = mg~d~cos(180^{\circ})$
$W_g = (1.50~kg)(9.80~m/s^2)(2.00~m)~cos(180^{\circ})$
$W_g = -29.4~J$
We can find the work done by tension.
$W_T = T~d$
$W_T = (20~N)(2.00~m)$
$W_T = 40.0~J$
We can use the work energy theorem to find the kinetic energy of the block as it reaches 2.00 m.
$KE_f = KE_0 + W_g+W_T$
$KE_f = 0 - 29.4~J+40.0~J$
$KE_f = 10.6~J$
We can find the speed of the block.
$KE_f = 10.6~J$
$\frac{1}{2}mv^2 = 10.6~J$
$v^2 = \frac{(2)(10.6~J)}{m}$
$v = \sqrt{\frac{(2)(10.6~J)}{m}}$
$v = \sqrt{\frac{(2)(10.6~J)}{1.50~kg}}$
$v = 3.76~m/s$