To find the coefficient of kinetic friction (\( \mu_k \)), we can use the formula:
\[ f_k = \mu_k \cdot N \]
where:
- \( f_k \) is the frictional force,
- \( \mu_k \) is the coefficient of kinetic friction,
- \( N \) is the normal force.
For a box sliding on a horizontal surface, the normal force \( N \) is equal to the weight of the box. In this case, the weight of the box is given as 45 N:
\[ N = 45 , \text{N} \]
The given frictional force is 13 N:
\[ f_k = 13 , \text{N} \]
Now, we can rearrange the formula to solve for \( \mu_k \):
\[ \mu_k = \frac{f_k}{N} = \frac{13 , \text{N}}{45 , \text{N}} \approx 0.289 \]
Therefore, the coefficient of kinetic friction is approximately 0.289.