Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 10 - Interactions and Potential Energy - Exercises and Problems: 71

Answer

When the bungee cord reaches maximum elongation, the person is 10.9 meters above the water.

Work Step by Step

We can use conservation of energy to solve this question. The sum of the energy stored in the bungee cord and the final potential energy will be equal to the initial potential energy. Note that the bungee cord begins to stretch after the person falls 30 meters Let $x$ be the distance that the bungee cord stretches; $U_s+PE_f = PE_0$ $\frac{1}{2}kx^2+mg(100-30-x) = mg~(100)$ $\frac{1}{2}kx^2+ mg(-x-30) = 0$ $\frac{1}{2}(40~N/m)x^2+ (80~kg)(9.80~m/s^2)(-x-30) = 0$ $(20~N/m)~x^2-(784~N)~x-23,520~J = 0$ We can use the quadratic formula to find $x$. $x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $x = \frac{-(-784)\pm \sqrt{(-784)^2-(4)(20)(-23,520)}}{(2)(20)}$ $x = -19.9~m, 59.1~m$ Since the negative value is unphysical, the solution is $x = 59.1~m$. The height above the water is $100~m-30~m-59.1~m$ which is 10.9 meters. When the bungee cord reaches maximum elongation, the person is 10.9 meters above the water.
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