Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - Problems - Page 104: 57

Answer

(a) $a = 3.67 ~m/s^2$ (b) v = 9.39 m/s

Work Step by Step

(a) We can use the force equation to find the acceleration. $ma = \sum ~F$ $ma = mg ~sin(\theta)$ $a = g ~sin(\theta) = (9.80 ~m/s^2)\cdot ~sin(22.0^{\circ})$ $a = 3.67 ~m/s^2$ (b) $v^2 = v_0^2 + 2ax = 0 + 2ax$ $v = \sqrt{2ax} = \sqrt{(2)(3.67 ~m/s^2)(12.0 ~m)} = 9.39 ~m/s$
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