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 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 202: 59


The tension in the string was 13.1 N

Work Step by Step

The ball reaches a height 400 cm above the point where it was when the string was cut. We can find the ball's speed $v_0$ at the point where it started moving upward. $v^2-v_0^2 = 2gy$ $0-v_0^2 = 2gy$ $v_0 = \sqrt{-2gy}$ $v_0 = \sqrt{-(2)(-9.80~m/s^2)(4.0~m)}$ $v_0 = 8.85~m/s$ The tension in the string an instant before the string broke is equal to the centripetal force. $T = \frac{mv_0^2}{r}$ $T = \frac{(0.100~kg)(8.85~m/s)^2}{0.60~m}$ $T = 13.1~N$ The tension in the string was 13.1 N
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.