Question

To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at $45°$ and the block is released from rest. The block takes a time $t$ in moving a distance $d$. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time $\frac{t}{2}$ in moving down the same distance $d$. The coefficient of friction is

• A. $\frac{3}{4}$
• B. $\frac{5}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A

$\frac{3}{4}$

Step 1: Given Data

Initial velocity $u=0$

Since the block is moving downwards, so displacement $S=-d$

The angle of inclination is $\theta =45°$

For the rough surface, the time taken $t_1=t$

For the smooth surface, the time taken $t_2=\frac{t}{2}$

Step 2: Formula used

According to Newton's second equation of motion,

$S=ut+\frac{1}{2}a{t}^2$

Step 3: Time taken for rough surface

Let the coefficient of friction be $\mu$.

Let the mass of the block be $m$.

Let the acceleration due to gravity be $g$.

From the figure, we can see that

Normal $N=mg\cos \theta$

Frictional force $f_s=\mu N=\mu mg\cos \theta$

$$\begin{gathered} \therefore mg\sin \theta -f_s=ma \\ \Rightarrow mg\sin \theta -\mu mg\cos \theta =ma \\ \Rightarrow a=g\sin \theta -\mu g\cos \theta \end{gathered}$$

According to Newton's second equation of motion,

$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ -d=0-\frac{1}{2}\left(g\sin \theta -\mu g\cos \theta \right){t_1}^2 \\ {t_1}^2=\frac{2d}{g\sin \theta -\mu g\cos \theta } \\ {t}_1=\sqrt{\frac{2d}{g\sin \theta -\mu g\cos \theta }}\left(1\right) \end{gathered}$$

Step 4: Time taken for a smooth surface

From the figure, we can see that

Acceleration $a=-g\sin \theta$

According to Newton's second equation of motion,

$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ -d=0+\frac{1}{2}\left(-g\sin \theta \right){t_2}^2 \\ {t_2}^2=\frac{2d}{g\sin \theta } \\ {t}_2=\sqrt{\frac{2d}{g\sin \theta }}\left(2\right) \end{gathered}$$

Step 5: Calculate the Coefficient of Friction

From the question we have,

$$\begin{gathered} \frac{t_1}{2}=t_2 \\ \Rightarrow \frac{\sqrt{\frac{2d}{g\sin \theta -\mu g\cos \theta }}}{2}=\sqrt{\frac{2d}{g\sin \theta }} \\ \Rightarrow \frac{1}{4}\times \frac{2d}{g\sin \theta -\mu g\cos \theta }=\frac{2d}{g\sin \theta } \\ \Rightarrow 4\left(\sin \theta -\mu \cos \theta \right)=\sin \theta \\ \Rightarrow 4\mu \cos \theta =3\sin \theta \\ \Rightarrow \mu =\frac{3}{4}\tan \theta \end{gathered}$$

Given that $\theta =45°\Rightarrow \tan 45°=1$

$\therefore \mu =\frac{3}{4}$

Hence, the correct answer is option (A).