There are two inclined surfaces of equal length (L) and same angle of inclination 45º 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 (𝜇_k) between the object and the rough surface is close to

The correct option is 0.75.

Let us analyze the motion of the body sliding down the two different inclined planes step-by-step.

1. Case 1: Perfectly Smooth Inclined Plane
When the surface is perfectly smooth, the only force acting along the incline is the component of gravity. For an angle of inclination $\theta =45°$, the acceleration $a_1$ of the body down the incline is given by:
$a_1=g\sin \left(\theta \right)$
Since the body starts from rest (initial velocity $u=0$), the distance $L$ traveled in time $t_1$ is:
$L=\frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}g\sin \left(\theta \right)t_1^2$

2. Case 2: Rough Inclined Plane
When the surface is rough, kinetic friction opposes the motion. The normal force acting on the body is $N=mg\cos \left(\theta \right)$, so the force of kinetic friction is:
$f_k={\mu }_{k}N={\mu }_{k}mg\cos \left(\theta \right)$
The net acceleration $a_2$ down the rough incline is:
$a_2=g\sin \left(\theta \right)-{\mu }_{k}g\cos \left(\theta \right)$
The distance $L$ traveled in time $t_2$ is:
$L=\frac{1}{2}{a}_2{t}_2^2=\frac{1}{2}g\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)t_2^2$

3. Relating the Two Cases
Since the length of both inclined planes is the same ($L$), we can equate the two expressions for $L$:
$\frac{1}{2}g\sin \left(\theta \right)t_1^2=\frac{1}{2}g\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)t_2^2$
Dividing both sides by $\frac{1}{2}g$ gives:
$\sin \left(\theta \right)t_1^2=\left(\sin \right(\theta )-{\mu }_k\cos (\theta \left)\right)t_2^2$

4. Applying the Given Conditions
We are given that the body takes 2 times as much time to slide down the rough surface than on the smooth surface:
$t_2=2t_1$
We also know the angle of inclination is:
$\theta =45°$
This means:
$\sin (45°)=\cos (45°)=\frac{1}{\sqrt{2}}$
Substituting these values into the equation:
$\frac{1}{\sqrt{2}}{t}_1^2=(\frac{1}{\sqrt{2}}-{\mu }_k\frac{1}{\sqrt{2}}){\left(2t_1\right)}^2$
We can divide both sides by $\frac{1}{\sqrt{2}}{t}_1^2$ to simplify:
$1=(1-{\mu }_k)\times 4$

5. Solving for the Kinetic Friction Coefficient
Now, we solve for ${\mu }_k$:
$1=4-4{\mu }_k$
$4{\mu }_k=4-1$
$4{\mu }_k=3$
${\mu }_k=\frac{3}{4}=0.75$

Thus, the coefficient of kinetic friction between the object and the rough surface is 0.75.