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.
