Answer
(a) The original kinetic energy of the player is 1060 J.
(b) The average power required to stop the player is 1060 W.
Work Step by Step
(a) $KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(85 ~kg)(5.0 ~m/s)^2$
$KE = 1060 ~J$
The original kinetic energy of the player is 1060 J.
(b) $P = \frac{KE}{t}$
$P = \frac{1060 ~J}{1 ~s}$
$P = 1060 ~W$
The average power required to stop the player is 1060 W.