Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 7 - Exercises and Problems: 62

Answer

$KE=1.6 \ J$

Work Step by Step

We call $\Delta x$ how far on the incline the block goes. Thus, we find using conservation of energy that: $ \frac{1}{2}kx^2 = mgh + \mu F_n \Delta x$ $ \frac{1}{2}kx^2 = mgh + \mu mgcos\theta \Delta x$ $ \frac{1}{2}(2000)(.1)^2 = (4.5)(\Delta x sin30) + (.5)(4.5)cos30 \Delta x$ $\Delta x = 2.38$ This is greater than 2, so the block reaches the top. We find its kinetic energy there: $ \frac{1}{2}(2000)(.1)^2 = (4.5)(\Delta x sin30) + (.5)(4.5)cos30 \Delta x+\frac{1}{2}mv^2 $ $ \frac{1}{2}(2000)(.1)^2 = (4.5)(2 sin30) + (.5)(4.5)cos30(2)+KE $ $KE=1.6 \ J$
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