# 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

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