#### Answer

The player's foot must be in contact with the ball over a distance of 0.168 meters.

#### Work Step by Step

$W = K_2-K_1$
$F~d = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$
$d = \frac{m(v_2^2-v_1^2)}{2F}$
$d = \frac{(0.420~kg)((6.00~m/s)^2-(2.00~m/s)^2)}{(2)(40.0~N)}$
$d = 0.168~m$
The player's foot must be in contact with the ball over a distance of 0.168 meters.