Question 209: Laws of Motion

Question

There are two inclined surfaces of equal length $(L)$ and same angle of inclination $45^{\circ}$ 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 $\left(\mu_k\right)$ between the object and the rough surface is close to

Accepted Answer

The time taken to slide down the rough surface is twice that of the smooth surface: $t_{	ext{rough}} = 2t_{	ext{smooth}}$.
The acceleration on the smooth surface is given by:
$ a_{	ext{smooth}} = g \sin 	heta $
The time to slide down an inclined plane is inversely proportional to the square root of the acceleration:
$ t \propto \frac{1}{\sqrt{a}} \Rightarrow t_{	ext{smooth}} \propto \frac{1}{\sqrt{g \sin 	heta}} $
The acceleration on the rough surface is:
$ a_{	ext{rough}} = g \sin 	heta - \mu_k g \cos 	heta $
Relating the times on both surfaces, we have:
$ \frac{t_{	ext{rough}}}{t_{	ext{smooth}}} = \frac{\sqrt{\sin 	heta}}{\sqrt{\sin 	heta - \mu_k \cos 	heta}} = 2 $
Squaring both sides results in:
$ \frac{\sin 	heta}{\sin 	heta - \mu_k \cos 	heta} = 4 $
Substituting $	heta = 45^{\circ}$ where $\sin 	heta = \cos 	heta = \frac{1}{\sqrt{2}}$, we find:
$ \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \mu_k 	imes \frac{1}{\sqrt{2}}} = 4 $
Simplifying further:
$ 1 - \mu_k = \frac{1}{4} $
Solving for $\mu_k$:
$ \mu_k = \frac{3}{4} = 0.75 $
Thus, the coefficient of kinetic friction is $0.75$.

Answer

0.5

Answer

0.75

Answer

0.25

Answer

0.40

Laws of Motion

Choose the correct answer: