To solve the problem, we need to find the potential energy of the particle at the mean position given the force acting on it, the amplitude of oscillation, the mass of the particle, and the total mechanical energy.
### Step-by-Step Solution:
1. **Identify the Force Constant (k)**:
The force acting on the particle is given by \( F = -8x \). In simple harmonic motion (SHM), the force can also be expressed as \( F = -kx \). By comparing both equations, we find that:
\[
k = 8 \, \text{N/m}
\]
2. **Calculate Angular Frequency (\(\omega\))**:
The angular frequency \(\omega\) can be calculated using the formula:
\[
\omega = \sqrt{\frac{k}{m}}
\]
where \( m = 0.5 \, \text{kg} \). Substituting the values:
\[
\omega = \sqrt{\frac{8}{0.5}} = \sqrt{16} = 4 \, \text{rad/s}
\]
3. **Understand Total Mechanical Energy (E)**:
The total mechanical energy \( E \) in SHM is given as:
\[
E = \frac{1}{2} k A^2
\]
where \( A \) is the amplitude. Given that \( A = 2 \, \text{m} \):
\[
E = \frac{1}{2} \times 8 \times (2)^2 = \frac{1}{2} \times 8 \times 4 = 16 \, \text{J}
\]
However, we are given that the total mechanical energy is \( 20 \, \text{J} \). This indicates that the potential energy at the mean position is not zero.
4. **Calculate Kinetic Energy at Mean Position**:
At the mean position, the displacement \( x = 0 \), and the kinetic energy \( KE \) can be calculated as:
\[
KE = E - PE
\]
where \( PE \) is the potential energy at the mean position. We need to find \( PE \).
5. **Using the Energy Conservation**:
Since total mechanical energy \( E = KE + PE \), we can rearrange it to find \( PE \):
\[
PE = E - KE
\]
At the mean position, the kinetic energy is maximum. We can use the formula for kinetic energy:
\[
KE = \frac{1}{2} m v^2
\]
However, since we already know the total energy is 20 J, we can directly find \( PE \) as:
\[
PE = 20 \, \text{J} - KE
\]
6. **Calculate Potential Energy**:
Since we need to find \( KE \) at the mean position, we can use:
\[
KE = \frac{1}{2} k A^2 = \frac{1}{2} \times 8 \times 2^2 = 16 \, \text{J}
\]
Thus, substituting back into the equation for potential energy:
\[
PE = 20 \, \text{J} - 16 \, \text{J} = 4 \, \text{J}
\]
### Final Answer:
The potential energy of the particle at the mean position is \( \boxed{4 \, \text{J}} \).
