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 - General Problems - Page 108: 95

Answer

(a) v = 16 m/s (b) v = 13 m/s

Work Step by Step

If the slope is 1-in-4 then $\theta = sin^{-1}(\frac{1}{4})$. The angle of the road is $\theta = 14.5^{\circ}$ (a) We can use the force equation to find the acceleration. $ma = mg ~sin(\theta)$ $a = g ~sin(\theta) = (9.80 ~m/s^2)~sin(14.5^{\circ}) = 2.45 ~m/s^2$ We can use the acceleration to find the speed $v$ at the bottom. $v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$ $v = \sqrt{2ax} = \sqrt{(2)(2.45 ~m/s^2)(55 ~m)} = 16 ~m/s$ (b) We can use the force equation to find the acceleration. $ma = mg ~sin(\theta) - mg ~cos(\theta) ~\mu_s$ $a = g ~sin(\theta) - g ~cos(\theta) ~\mu_s$ $a = (9.80 ~m/s^2)~sin(14.5^{\circ}) - (9.80 ~m/s^2)~cos(14.5^{\circ}) \cdot (0.10))$ $a= 1.5 ~m/s^2$ We can use the acceleration to find the speed $v$ at the bottom. $v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$ $v = \sqrt{2ax} = \sqrt{(2)(1.5 ~m/s^2)(55 ~m)} = 13 ~m/s$
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