#### Answer

(a) The speed of the block after it leaves the spring is 3.11 m/s.
(b) The block travels a distance of 0.821 meters up the incline.

#### Work Step by Step

(a) The kinetic energy of the block will be equal in magnitude to the initial potential energy in the spring.
$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$
$v^2 = \frac{kx^2}{m}$
$v = \sqrt{ \frac{kx^2}{m}}$
$v = \sqrt{ \frac{(400~N/m)(0.220~m)^2}{2.00~kg}}$
$v = 3.11~m/s$
The speed of the block after it leaves the spring is 3.11 m/s.
(b) The increase in potential energy will be equal to the initial potential energy in the spring.
$mgh = \frac{1}{2}kx^2$
$mgd~sin(\theta) = \frac{1}{2}kx^2$
$d = \frac{kx^2}{2mg~sin(\theta)}$
$d = \frac{(400~N/m)(0.220~m)^2}{(2)(2.00~kg)(9.80~m/s^2)~sin(37.0^{\circ})}$
$d = 0.821~m$
The block travels a distance of 0.821 meters up the incline.