Answer
The particle will be turned back at the position $~~x = 12~cm$
Work Step by Step
We can find the total energy in the system:
$E = K+U$
$E = \frac{1}{2}m~v_{max}^2+0$
$E = (\frac{1}{2})(2.0~kg)~(0.85~m/s)^2$
$E = 0.7225~J$
We can find the value of $b$:
$U = bx^2$
$b = \frac{U}{x^2}$
$b = \frac{2.0~J}{(0.20~m)^2}$
$b = 50~J/m^2$
We can find the position $x$ where $U = 0.7225~J$:
$U = 0.7225~J$
$bx^2 = 0.7225~J$
$x^2 = \frac{0.7225~J}{b}$
$x = \sqrt{\frac{0.7225~J}{b}}$
$x = \sqrt{\frac{0.7225~J}{50~J/m^2}}$
$x = 0.12~m$
$x = 12~cm$
The particle will be turned back at the position $~~x = 12~cm$
