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(B) The total mechanical energy $E$ of the particle is conserved. Since the particle is released from rest at $x = a$,its kinetic energy $K = 0$ at $x

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moves to a point $x = e$,and then moves to the left

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(B) The total mechanical energy $E$ of the particle is conserved. Since the particle is released from rest at $x = a$,its kinetic energy $K = 0$ at $x = a$,so its total energy is $E = U(a)$.As the particle moves,$E = K + U(x) = U(a)$. Since $K \ge 0$,the particle can only move in regions where $U(x) \le U(a)$.Looking at the graph,$U(x) = U(a)$ at $x = a$ and $x = e$. Between $x = a$ and $x = e$,$U(x) \le U(a)$.Therefore,the particle will move from $x = a$ towards the right,pass through $x = b$,$x = c$,and $x = d$,and reach $x = e$. At $x = e$,its kinetic energy becomes zero $(K = E - U(e) = 0)$,so it stops and then reverses its direction to move back to the left.

$A$ particle is released from rest at the point $x = a$ and moves along the $x$ axis subject to the potential energy function $U(x)$ shown. The particle:

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