#### Answer

At x = 2 m, the particle's velocity is 5.1 m/s
At x = 4 m, the particle's velocity is 4.0 m/s

#### Work Step by Step

The work done on the particle is equal to the area under the force versus position graph.
We can find the work done during the interval 0 - 2 m:
$W = \frac{1}{2}(10~N)(2~m)$
$W = 10~J$
We can use the work-energy theorem to find the particle's velocity at x = 2 m.
$KE_2 = KE_1+W$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W$
$v_2^2 = \frac{mv_1^2+2W}{m}$
$v_2 = \sqrt{\frac{mv_1^2+2W}{m}}$
$v_2 = \sqrt{\frac{(2.0~kg)(4.0~m/s)^2+(2)(10~J)}{2.0~kg}}$
$v_2 = 5.1~m/s$
The particle's velocity at x = 2 m is 5.1 m/s
We can find the work done during the interval 2 - 4 m:
$W = \frac{1}{2}(-10~N)(2~m)$
$W = -10~J$
The total work done on the particle during the interval 0 - 4 m is zero. Therefore the kinetic energy at x = 4 m is the same as the kinetic energy at x = 0. Therefore, the particle's velocity at x = 4 m is 4.0 m/s