Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 10 - Energy - Exercises and Problems: 23

Answer

(a) The rock leaves the spring with a speed of 15 m/s (b) The contestants will go hungry.

Work Step by Step

(a) The rock's kinetic energy when it leaves the spring will be equal to the energy $U_s$ stored in the spring initially. We can find the speed of the rock. $KE = U_s$ $\frac{1}{2}mv^2=\frac{1}{2}kx^2$ $v^2=\frac{kx^2}{m}$ $v=\sqrt{\frac{kx^2}{m}}$ $v=\sqrt{\frac{(1000~N/m)(0.30~m)^2}{0.400~kg}}$ $v = 15~m/s$ The rock leaves the spring with a speed of 15 m/s. (b) The potential energy of the rock at maximum height will be equal to the rock's kinetic energy when it leaves the spring. We can find the maximum height $h$ reached by the rock. $PE = KE$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(15~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 11.5~m$ Since the rock's maximum height is less than the 15 meters required to reach the fruit, the contestants will go hungry.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.