There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (µk) between the object and the rough surface is close to

To find the coefficient of kinetic friction (${\mu }_k$) between the object and the rough inclined surface, we can analyze the motion of the body in both cases: sliding down a perfectly smooth inclined plane, and sliding down a rough inclined plane.

Case 1: Motion on the smooth inclined plane
When the body slides down a smooth incline of length $L$ and angle of inclination $\theta =45°$, the only force acting along the incline is the component of gravity. The acceleration of the body ($a_s$) is given by:
$a_s=g\sin \left(\theta \right)$
Since the initial velocity ($u$) is $0$, we can use the equation of motion $s=ut+\frac{1}{2}a{t}^2$ to find the time taken ($t_s$) to cover length $L$:
$L=\frac{1}{2}{a}_s{t}_s^2$
$t_s=\sqrt{\frac{2L}{a_s}}=\sqrt{\frac{2L}{g\sin \left(\theta \right)}}$

Case 2: Motion on the rough inclined plane
When the body slides down the rough incline, it experiences a friction force acting up the incline, oppose to its motion. The frictional force is $f_k={\mu }_{k}N={\mu }_{k}mg\cos \left(\theta \right)$.
The net force acting down the incline is:
$F_{\mathrm{net}}=mg\sin \left(\theta \right)-{\mu }_{k}mg\cos \left(\theta \right)$
Therefore, the acceleration on the rough surface ($a_r$) is:
$a_r=g\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)$
Similarly, the time taken ($t_r$) to slide down the same distance $L$ starting from rest is:
$t_r=\sqrt{\frac{2L}{a_r}}=\sqrt{\frac{2L}{g\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)}}$

Relating the two times
We are given that the body takes 2 times as much time to slide down the rough surface than on the smooth surface:
$t_r=2t_s$
Squaring both sides gives:
$t_r^2=4t_s^2$
Substituting the expressions for the times:
$\frac{2L}{g\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)}=4·\frac{2L}{g\sin \left(\theta \right)}$
Simplifying the equation by canceling common terms $2L$ and $g$ on both sides:
$\frac{1}{\sin \left(\theta \right)-{\mu }_k\cos \left(\theta \right)}=\frac{4}{\sin \left(\theta \right)}$
Cross-multiplying yields:
$\sin \left(\theta \right)=4\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)$
Since $\theta =45°$, we have $\sin (45°)=\cos (45°)=\frac{1}{\sqrt{2}}$. Dividing both sides of the equation by $\cos \left(\theta \right)$ (or directly using the value $\frac{1}{\sqrt{2}}$):
$1=4(1-{\mu }_k)$
$1=4-4{\mu }_k$
Rearranging the terms to solve for ${\mu }_k$:
$4{\mu }_k=4-1$
$4{\mu }_k=3$
${\mu }_k=\frac{3}{4}=0.75$

Thus, the coefficient of kinetic friction is 0.75.