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(A) The particle is released from rest at the origin $(x = 0)$,so the initial kinetic energy $K_i = 0$ and initial potential energy $U_i = U(0) = 0^2

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$2$

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(A) The particle is released from rest at the origin $(x = 0)$,so the initial kinetic energy $K_i = 0$ and initial potential energy $U_i = U(0) = 0^2 - 3(0) = 0 \ J$.By the law of conservation of mechanical energy,the total energy $E = K + U$ remains constant.$E_i = K_i + U_i = 0 + 0 = 0 \ J$.At $x = 2$,the potential energy is $U_f = U(2) = 2^2 - 3(2) = 4 - 6 = -2 \ J$.Since $E_i = E_f$,we have $0 = K_f + U_f$.$0 = K_f - 2 \ J$.Therefore,$K_f = 2 \ J$.

$A$ particle is released from rest at the origin. It moves under the influence of a potential field $U = x^2 - 3x$. The kinetic energy at $x = 2$ is ............... $J$.

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