# Chapter 9 - Work and Kinetic Energy - Exercises and Problems - Page 229: 32

The player was running at a speed of 5.9 m/s

#### Work Step by Step

The amount of thermal energy is equal to the player's initial kinetic energy; $KE = E_{th}$ $\frac{1}{2}mv^2 = 950~J$ $v^2 = \frac{(2)(950~J)}{m}$ $v = \sqrt{\frac{(2)(950~J)}{m}}$ $v = \sqrt{\frac{(2)(950~J)}{55~kg}}$ $v = 5.9~m/s$ The player was running at a speed of 5.9 m/s.

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