University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 7 - Potential Energy and Energy Conservation - Problems - Exercises - Page 232: 7.47

Answer

The bungee cord should stretch a distance of 0.602 meters.

Work Step by Step

After falling a distance of 41.0 meters, the potential energy stored in the bungee cord will be equal in magnitude to the person's change in potential energy. Note that the bungee cord will be stretched a distance of 11.0 meters. $\frac{1}{2}kx^2 = mgh$ $k = \frac{2mgh}{x^2}$ $k = \frac{(2)(95.0~kg)(9.80~m/s^2)(41.0~m)}{(11.0~m)^2}$ $k = 631~N/m$ We can find the distance the bungee cord stretches when we pull on it with a force of 380.0 N $kx = F$ $x = \frac{F}{k} = \frac{380.0~N}{631~N/m}$ $x = 0.602~m$ The bungee cord should stretch a distance of 0.602 meters.
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