Your answer is partially correct.

The figure here shows a plot of potential energy $U$ versus position $x$ of a $0\mathrm{.}892$ kg particle that can travel only along an $x$ axis. $($Nonconservative forces are not involved.$)$ Three values are $U_A=15\mathrm{.}0J,U_B=35\mathrm{.}0J$ and $U_C=45\mathrm{.}0J.$ The particle is released at $x=4\mathrm{.}50m$ with an initial speed of $7\mathrm{.}76\frac{m}{s},$ headed in the negative $x$ direction. $($a$)$ If the particle can reach $x=1\mathrm{.}00m,$ what is its speed there, and if it cannot, what is its turning point? What are the $($b$)$ magnitude and $($c$)$ direction of the force on the particle as it begins to move to the left of $x=4\mathrm{.}00m?$ Suppose, instead, the particle is headed in the positive $x$ direction when it is released at $x=4\mathrm{.}50m$ at speed $7\mathrm{.}76\frac{m}{s}.($d$)$ If the particle can reach $x=7\mathrm{.}00m,$ what is its speed there, and if it cannot, what is its turning point? What are the $($e$)$ magnitude and $($ f $)$ direction of the force on the particle as it begins to move to the right of $x=5\mathrm{.}00m?$

$($a$)$ Number $$ Unit $$

$($b$)$ Number $$ Unit $$

$($c$)$