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(C) The potential energy of the particle is given by $U(x) = 2 - 20x + 5x^2 	ext{ J}$.At the starting position $x_i = -3 	ext{ m}$,the initial poten

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$7$

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(C) The potential energy of the particle is given by $U(x) = 2 - 20x + 5x^2 	ext{ J}$.At the starting position $x_i = -3 	ext{ m}$,the initial potential energy is:$U_i = 2 - 20(-3) + 5(-3)^2 = 2 + 60 + 45 = 107 	ext{ J}$.Since the particle is released from rest,its total mechanical energy $E$ is equal to its initial potential energy:$E = U_i = 107 	ext{ J}$.The particle will reach its maximum $x$ position when its kinetic energy becomes zero,meaning its potential energy at that point $x_f$ must equal the total mechanical energy $E$:$U(x_f) = E$$2 - 20x_f + 5x_f^2 = 107$$5x_f^2 - 20x_f - 105 = 0$Dividing by $5$:$x_f^2 - 4x_f - 21 = 0$Factoring the quadratic equation:$(x_f - 7)(x_f + 3) = 0$This gives two solutions: $x_f = 7 	ext{ m}$ or $x_f = -3 	ext{ m}$.Since the particle starts at $x = -3 	ext{ m}$ and moves to reach a maximum $x$,the maximum value is $x = 7 	ext{ m}$.

$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

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