#### Answer

The net work done on the box is 390 J

#### Work Step by Step

We first find the final speed of the box:
$v = a~t = (2.0~m/s^2)(7.0~s)$
$v = 14~m/s$
Then, we find the net work done on the box:
$KE_1+Work = KE_2$
$0+Work = \frac{1}{2}mv^2$
$Work = \frac{1}{2}(4.0~kg)(14~m/s)^2$
$Work = 390~J$
The net work done on the box is 390 J