Physics: Principles with Applications (7th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 166: 50

Answer

$v_2 = 20.6 ~m/s$

Work Step by Step

We can use work and energy to solve this question. Let $W_f$ be the work done by friction as the car moves from point 1 to point 2. Then, $E_2 = E_1 + W_f$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mgy -0.23mgd$ $v_2^2 = v_1^2 + 2gy-2(0.23gd)$ $v_2 = \sqrt{v_1^2 + 2gy-2(0.23gd)}$ $v_2 = \sqrt{(1.30 ~m/s)^2 + (2)(9.80 ~m/s^2)(32 ~m) - (2)(0.23)(9.80 ~m/s^2)(45.0 ~m)}$ $v_2 = 20.6 ~m/s$
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