To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:

$F_{\text{friction}} = \mu_{k} \cdot N$

where:

• $F_{\text{friction}}$ is the force of friction,
• $\mu_{k}$ is the coefficient of kinetic friction,
• $N$ is the normal force exerted by the surface onto the box.

Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$, where $m$ is the mass of the box and $g$ is the acceleration due to gravity. For most calculations, $g$ is approximated as $$9.8 \, \text{m/s}^2$$.

Substituting the given values:

• $m = 50 \, \text{kg}$,
• $\mu_k = 0.3$,
• $g = 9.8 \, \text{m/s}^2$.

First, calculate the weight of the box, which is the normal force:

$N = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$

Then, use this value to find the force of kinetic friction:

$F_{\text{friction}} = \mu_{k} \cdot N = 0.3 \times 490 \, \text{N} = 147 \, \text{N}$

Thus, the force of kinetic friction acting on the box is 147 N. The correct answer is Option B.