A heavy box of mass $50 \mathrm{~kg}$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

<p>To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by:</p> <p>$F_{\text{friction}} = \mu_{k} \cdot N$</p> <p>where:</p> <ul> <li>$F_{\text{friction}}$ is the force of friction,</li> <li>$\mu_{k}$ is the coefficient of kinetic friction,</li> <li>$N$ is the normal force exerted by the surface onto the box.</li> </ul> <p>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 $9.8 \, \text{m/s}^2$.</p> <p>Substituting the given values:</p> <ul> <li>$m = 50 \, \text{kg}$,</li> <li>$\mu_k = 0.3$,</li> <li>$g = 9.8 \, \text{m/s}^2$.</li> </ul> <p>First, calculate the weight of the box, which is the normal force:</p> <p>$N = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$</p> <p>Then, use this value to find the force of kinetic friction:</p> <p>$F_{\text{friction}} = \mu_{k} \cdot N = 0.3 \times 490 \, \text{N} = 147 \, \text{N}$</p> <p>Thus, the force of kinetic friction acting on the box is <strong>147 N</strong>. The correct answer is <strong>Option B</strong>.</p>