## University Physics with Modern Physics (14th Edition)

(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.