JEE Main 2024 — Laws Of Motion Question with Solution

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.