A

The potential energy function is given as $U(x) = \frac{1}{2} k x^2$. For this problem, we have $k = 0.5 \, \text{N m}^{-1}$. Thus, we can write the potential energy as:
Step 1: Substitute the value of k into the potential energy function:
$$U(x) = \frac{1}{2} \times 0.5 \times x^2 = 0.25 x^2$$
Step 2: Calculate the potential energy at $x = 2m$ and $x = -2m$:
$$U(2) = 0.25 (2)^2 = 0.25 \times 4 = 1 \, \text{J} $$
$$U(-2) = 0.25 (-2)^2 = 0.25 \times 4 = 1 \, \text{J} $$
Step 3: The total mechanical energy E of the particle is the sum of kinetic energy (K) and potential energy (U):
$$ E = K + U $$
When the particle reaches $x = ±2m$, the potential energy $U(x)$ is equal to the total energy $1 ext{ J}$. This means that the kinetic energy $K$ will be zero at these points because:
$$ K = E - U \implies K = 1 - 1 = 0 \text{ J} $$
Hence, at $x = ±2m$, the particle has zero kinetic energy, causing it to 'turn back' or reverse its motion. Therefore, the particle will stop and change direction upon reaching these points.