#### Answer

$F_{max} = 8.0~N$

#### Work Step by Step

We can use the work-energy theorem to find the work done on the particle from x = 0 to x = 2 m.
$KE_0+W = KE_f$
$W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$
$W = \frac{1}{2}(0.50~kg)(6.0~m/s)^2 - \frac{1}{2}(0.50~kg)(2.0~m/s)^2$
$W = 8.0~J$
The work done on the particle is equal to the area under the force versus position graph. We can find the value of $F_{max}$.
$\frac{1}{2}F_{max}~x = W$
$F_{max} = \frac{2W}{x}$
$F_{max} = \frac{(2)(8.0~J)}{2~m}$
$F_{max} = 8.0~N$