Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 6 - Exercises and Problems - Page 111: 85

Answer

$12\ kJ$

Work Step by Step

a) We call $x_0$ the initial stretch length. (The book calls this $L_0$, but this is the same idea.) We know that work is equal to the integral of the force with respect to x. Thus: $W = \int_{x_0}^{2x_0}(-kx-bx^2-cx^3-dx^4)dx$ $W=.5kx_0^2+.33bx_0^3+.25cx_0^4+.2dx_0^5 $ b) Plugging in the given values, we find that the work done is $12\ kJ$.
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