#### Answer

(a) When $x = 1.0$:
$U = 20~J$
When $x = 2.0$:
$U = 40~J$
When $x = 3.0$:
$U = 60~J$
When $x = 4.0$:
$U = 70~J$
(b) $x = 2.56~m$

#### Work Step by Step

(a) We can write an expression for the potential energy:
$U(x) = -\int_{x_0}^{x_f} F_x~dx$
$U(x)$ is the negative of the area under the $F$ versus $x$ graph.
We can find $U$ when $x = 1.0$:
$U = -(-20~N)(1.0~m) = 20~J$
We can find $U$ when $x = 2.0$:
$U = -(-20~N)(2.0~m) = 40~J$
We can find $U$ when $x = 3.0$:
$U = -(-20~N)(3.0~m) = 60~J$
We can find $U$ when $x = 4.0$:
$U = -[(-20~N)(3.0~m)+(0.5)(-20~N)(1.0~m)] = 70~J$
(b) We can find the initial kinetic energy:
$K_0 = \frac{1}{2}mv^2$
$K_0 = \frac{1}{2}(0.100~kg)(25~m/s)^2$
$K_0 = 31.25~J$
We can use conservation of energy to find $x$:
$U_f+K_f = U_0+K_0$
$(20~N)(x)+0 = 20~J+31.25~J$
$x = \frac{20~J+31.25~J}{20~N}$
$x = 2.56~m$