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

$W = K_2 - K_1$ $mg~d~cos(53.1^{\circ}) = \frac{1}{2}mv^2 - 0$ $v^2 = 2g~d~cos(53.1^{\circ})$ $v = \sqrt{2g~d~cos(53.1^{\circ})}$ $v = \sqrt{(2)(9.80~m/s^2)(1.35~m)~cos(53.1^{\circ})}$ $v = 3.99~m/s$ The final speed is 3.99 m/s.