There are two inclined surfaces of equal length (L) and same angle of inclination 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 surfa

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Solution

Q.
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 ( $μ_{k}$ ) between the object and the rough surface is close to [2025]

1 0.5

2 0.75

3 0.25

4 0.40

Ans.
(2)

Given: Angle of inclination, $θ = 45 °$

$t_{rough} = 2 t_{smooth}$

Length of both inclined surfaces is equal.

For smooth surface, initial velocity is zero, i.e., $u = 0,$ the distance $' S'$ is given by

$S = \frac{1}{2} a_{smooth} t_{smooth}^{2}$

Acceleration of smooth plane is $a_{smooth} = g \sin θ$

Using $θ = 45 °$ , we get

$a_{smooth} = g \sin 45 ° = \frac{g}{\sqrt{2}}$ or $\frac{g \sqrt{2}}{2}$

$\Rightarrow S = \frac{1}{2} g \frac{\sqrt{2}}{2} t_{smooth}^{2} \Rightarrow t_{smooth} = \sqrt{\frac{4 S}{g \sqrt{2}}}$

For rough surface :

$S = \frac{1}{2} a_{rough} t_{rough}^{2} (∵ u = 0)$

$a_{rough} = g \sin θ - μ_{k} g \cos θ$

Using $θ = 45 °$ , we get $a_{rough} = \frac{g \sqrt{2}}{2} - μ_{k} g \frac{\sqrt{2}}{2}$

$∴$ $S = \frac{1}{2} (g \frac{\sqrt{2}}{2} - μ_{k} g \frac{\sqrt{2}}{2}) t_{rough}^{2}$

or $t_{rough} = \sqrt{\frac{2 S}{g \frac{\sqrt{2}}{2} (1 - μ_{k})}} = \sqrt{\frac{4 S}{g \sqrt{2} (1 - μ_{k})}}$

As $t_{rough} = 2 t_{smooth}$

$\sqrt{\frac{4 S}{g \sqrt{2} (1 - μ_{k})}} = 2 \sqrt{\frac{4 S}{g \sqrt{2}}}$

Squaring on both sides, we get

$\Rightarrow \frac{4 S}{g \sqrt{2} (1 - μ_{k})} = 4 \cdot \frac{4 S}{g \sqrt{2}} \Rightarrow 4 = \frac{1}{1 - μ_{k}}$

$\Rightarrow 4 (1 - μ_{k}) = 1 or 1 = 4 - 4 μ_{k}$

$μ_{k} = \frac{3}{4} = 0.75$

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