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

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 63

Answer

(a) $v = \sqrt{2gh}$ (b) The speed at the bottom is 2.4 m/s in both cases.

Work Step by Step

(a) The acceleration down the ramp is $a = g~sin(\theta)$. We can find the distance $d$ the rock slides; $d = \frac{h}{sin(\theta)}$ We can find the speed at the bottom; $v^2 = v_0^2+2ad = 0 +2ad$ $v = \sqrt{2ad} = \sqrt{(2)(g~sin(\theta))(\frac{h}{sin(\theta)})}$ $v = \sqrt{2gh}$ (b) $v = \sqrt{2gh}$ $v = \sqrt{(2)(9.80~m/s^2)(0.30~m)}$ $v = 2.4~m/s$ Since the speed at the bottom does not depend on the angle, the speed at the bottom is 2.4 m/s in both cases.
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