## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Let $d$ be the length of the ramp. We can find the height $h$ of the ramp. $\frac{h}{d} = sin(\theta)$ $h = d~sin(\theta)$ $h = (3.0~m)~sin(20^{\circ})$ $h = 1.03~m$ The puck's minimum kinetic energy at the bottom of the ramp should be equal to the potential energy at the top of the ramp. $KE = PE$ $\frac{1}{2}mv^2 = mgh$ $v^2 = 2gh$ $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(1.03~m)}$ $v = 4.5~m/s$ The puck needs a minimum speed of 4.5 m/s.