To determine the acceleration of the box, we first need to calculate the frictional force acting on it. The frictional force (\(F_f\)) can be calculated using the formula:
\[
F_f = \mu \times F_n
\]
where:
- \( \mu \) is the coefficient of friction (0.2 in this case).
- \( F_n \) is the normal force, which for a horizontal surface is equal to the weight of the box (347 N).
Calculating the frictional force:
\[
F_f = 0.2 \times 347 \, \text{N} = 69.4 \, \text{N}
\]
Next, we need to find the net force (\(F_{\text{net}}\)) acting on the box. The net force is the difference between the pushing force and the frictional force:
\[
F_{\text{net}} = F_{\text{push}} - F_f
\]
Given the pushing force is 500 N:
\[
F_{\text{net}} = 500 \, \text{N} - 69.4 \, \text{N} = 430.6 \, \text{N}
\]
Now, we can use Newton's second law to find the acceleration (\(a\)) of the box:
\[
F_{\text{net}} = m \cdot a
\]
First, we need to calculate the mass (\(m\)) of the box using its weight:
\[
F_n = m \cdot g \implies m = \frac{F_n}{g}
\]
Where \(g \approx 9.81 \, \text{m/s}^2\):
\[
m = \frac{347 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 35.34 \, \text{kg}
\]
Now substituting into \(F_{\text{net}} = m \cdot a\):
\[
430.6 \, \text{N} = 35.34 \, \text{kg} \cdot a
\]
Solving for \(a\):
\[
a = \frac{430.6 \, \text{N}}{35.34 \, \text{kg}} \approx 12.18 \, \text{m/s}^2
\]
Thus, the acceleration of the box is approximately
\[
\boxed{12.18 \, \text{m/s}^2}
\]