#### Answer

(a) $v = 3.94~m/s$
(b) The magnitude of the acceleration is $11.8~m/s^2$ and the acceleration is directed away from the post.

#### Work Step by Step

(a) The kinetic energy of the block will be equal to the sum of the work done by the force and the spring.
$\frac{1}{2}mv^2 = F~x - \frac{1}{2}kx^2$
$v^2 = \frac{2F~x - kx^2}{m}$
$v = \sqrt{\frac{2F~x - kx^2}{m}}$
$v = \sqrt{\frac{(2)(54.0~N)(0.400~m) - (76.0~N/m)(0.400~m)^2}{2.00~kg}}$
$v = 3.94~m/s$
(b) We can use a force equation to find the acceleration.
$\sum F = ma$
$F-kx = ma$
$a = \frac{F-kx}{m}$
$a = \frac{54.0~N-(76.0~N/m)(0.400~m)}{2.00~kg}$
$a = 11.8~m/s^2$
The magnitude of the acceleration is $11.8~m/s^2$ and the acceleration is directed away from the post.