Question

Starting from rest, the time taken by a body sliding down on a rough inclined plane at ${45}^{°}$ with the horizontal is twice the time taken to travel on a smooth plane of same inclination and same distance. Then, find the coefficient of kinetic friction.

• A. $0·25$
• B. $0·33$
• C. $0·50$
• D. $0·75$

Answer 1

The correct option is D

$0·75$

Step 1: Given data and assumptions

The angle of an inclined plane from the horizontal $\left(\theta \right)={45}^{°}$

Let us suppose the time taken to travel on a smooth plane is $t_1$.

Time taken to slide down the inclined plane is $t_2$.

It is given that,

$$\begin{gathered} t_2=2t_1 \\ \frac{t_2}{t_1}=2 \end{gathered}$$

Suppose the ratio is $n$. then-

$\frac{t_2}{t_1}=n=2$

Step 2: Formula used

The coefficient of dynamic friction will be given by-

$Coefficient\left(\mu \right)=\tan \theta \left(1-\frac{1}{n^2}\right)$

Step 3: Calculating the coefficient of kinetic friction

From the formula,

$Coefficient\left(\mu \right)=\tan \theta \left(1-\frac{1}{n^2}\right)$

Now, putting the given values,

$$\begin{gathered} \mu =\tan 45\times \left(1-\frac{1}{2^2}\right)\tan 45=1 \\ \mu =1\times \frac{3}{4} \\ \mu =\frac{3}{4} \\ \mu =0·75 \end{gathered}$$

Thus, the coefficient of kinetic friction is $0·75$.

Hence, option D is the correct answer.