## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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.