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
\[ \boxed{12.18 , \text{m/s}^2} \]