Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems: 9

Answer

The puck's minimum speed is 4.5 m/s

Work Step by Step

Let $d$ be the length of the ramp. We can find the height $h$ of the ramp as; $\frac{h}{d} = sin(\theta)$ $h = d~sin(\theta)$ $h = (3.0~m)~sin(20^{\circ})$ $h = 1.03~m$ 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's minimum speed is 4.5 m/s
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