QJEEM24Ph5d02: JEE Main 2024 Physics - Class 11 JEEM24Ph5d02 Solution: A given object takes times the time to slide down rough…

To determine the coefficient of kinetic friction, let's analyze the time taken by the object to slide down each plane and use the equations of motion for both scenarios.

First, consider the perfectly smooth $45^{\circ}$ inclined plane (no friction). The acceleration of the object on this plane can be calculated using the component of gravitational force parallel to the incline. Since there is no friction, the only force acting down the plane is the component of the gravitational force:

$a_{\text{smooth}} = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}$

Let the time taken to slide down this smooth plane be $t_{\text{smooth}}$ . The distance $d$ covered by the object can be expressed using the equation of motion:

$d = \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 = \frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2$

Now, consider the rough $45^{\circ}$ inclined plane. The acceleration down the rough plane can be found by considering both the component of gravitational force and the kinetic friction force. Here, the friction force is $f_k = \mu_k N$ , where $N = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}$ . Thus the frictional force is:

$f_k = \mu_k \cdot \frac{mg}{\sqrt{2}}$

The net force acting on the object down the plane would be:

$F_{\text{net}} = mg \sin 45^{\circ} - f_k = \frac{mg}{\sqrt{2}} - \mu_k \cdot \frac{mg}{\sqrt{2}}$

So the net acceleration $a_{\text{rough}}$ is:

$a_{\text{rough}} = \frac{F_{\text{net}}}{m} = \frac{g}{\sqrt{2}} - \mu_k \cdot \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}} (1 - \mu_k)$

The time taken to slide down the rough plane is given as $nt_{\text{smooth}}$ . So, the distance $d$ can be written as:

$d = \frac{1}{2} a_{\text{rough}} (nt_{\text{smooth}})^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2$

Setting the distances equal for both cases, we get:

$\frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2$

Simplifying, we find:

$t_{\text{smooth}}^2 = (1 - \mu_k) n^2 t_{\text{smooth}}^2$

Therefore:

$1 = (1 - \mu_k) n^2$

Solving for $\mu_k$ , we get:

$\mu_k = 1 - \frac{1}{n^2}$

Thus, the coefficient of kinetic friction between the object and the surface of inclined plane is:

Option A: $1 - \frac{1}{\mathrm{n}^2}$