For the arrangement of forces in Problem 81, a $2.00 \mathrm{~kg}$ particle is released at $x=5.00 \mathrm{~m}$ with an initial velocity of $3.45 \mathrm{~m} / \mathrm{s}$ in the negative direction of the $x$ axis. (a) If the particle can reach $x=0 \mathrm{~m}$, what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive $x$ direction when it is released at $x=5.00 \mathrm{~m}$ at speed $3.45 \mathrm{~m} / \mathrm{s}$. (b) If the particle can reach $x=13.0 \mathrm{~m}$, what is its speed there, and if $\mathrm{it}$ cannot, what is its turning point?