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 107: 87

Answer

The speed will be 23 m/s at the bottom.

Work Step by Step

We can use a force equation to find the acceleration. $ma = \sum F$ $ma = mg ~sin(\theta) - mg ~cos(\theta) \cdot \mu$ $a = g ~sin(\theta) - g ~cos(\theta) \cdot \mu$ $a = (9.80 ~m/s^2) ~sin(45^{\circ}) - (9.80 ~m/s^2) ~cos(45^{\circ}) \cdot (0.12)$ $a = 6.1 ~m/s^2$ $v_0 = (6.0 ~km/h)(\frac{1000 ~m}{1 ~km})(\frac{1 ~h}{3600 ~s}) = 1.67 ~m/s$ We can use kinematics to find the speed at the bottom. $v^2 = v_0^2 + 2ax$ $v = \sqrt{v_0^2 + 2ax}$ $v = \sqrt{(1.67)^2 + (2)(6.1 ~m/s^2)(45.0 ~m)}$ $v = 23~m/s$ The speed will be 23 m/s at the bottom.
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