Question

Starting from rest ,a body slides down a $45°$ inclined plane in twice the time takes to slide down the same distance in the absence of friction .The coefficient of friction between the body and inclined plane is

Answer 1

Step 1. Given data.

1. The inclination of the plane is $45°$
2. The body slide down in twice the time it takes to slide down the same distance in the absence of friction.

Step. 2 Formula used

1. $F=ma$
2. $x=ut+\frac{1}{2}a{t}^2$

Where $F$ is the force, $m$the mass of the body, $a$ the acceleration, $x$ the distance covered, $t$ the time, $u$ the initial velocity and $a$ the acceleration

Step 3. Find the time when there is no friction

Let the acceleration be $a_1$

From newton's laws of motion

$F=ma$

$\Rightarrow mg\sin \left(\theta \right)=m{a}_1$ [Where $g$ is the acceleration due to gravity]

$\Rightarrow a_1=g\sin \left(\theta \right)$

Let the body travel $x$ distance in time $t_1$ with the initial velocity $u=0m/s$

$$\begin{gathered} \therefore x=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow x=\frac{1}{2}{a}_1{t_1}^2 \\ \Rightarrow t_1=\sqrt{\frac{2x}{a_1}} \\ \Rightarrow t_1=\sqrt{\frac{2x}{g\sin \left(\theta \right)}} \end{gathered}$$

Step 4. Find the time when there is friction

Let the acceleration be $a_2$ and $\mu$ the coefficient of friction.

From newton laws of motion

$F=ma$

$\Rightarrow mg\sin \left(\theta \right)-\mu mg\cos \left(\theta \right)=m{a}_2$ [Where $g$ is the acceleration due to gravity]

$\Rightarrow a_2=g\sin \left(\theta \right)-\mu g\cos \left(\theta \right)$

Let the body travel $x$ distance in time $t_2$ with the initial velocity $u=0m/s$

$$\begin{gathered} \therefore x=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow x=\frac{1}{2}{a}_2{t_2}^2 \\ \Rightarrow t_2=\sqrt{\frac{2x}{a_2}} \\ \Rightarrow t_2=\sqrt{\frac{2x}{g\sin \left(\theta \right)-\mu g\cos \left(\theta \right)}} \end{gathered}$$

Step 5. Find the coefficient of friction

Given that the body slide down in twice the time takes to slide down the same distance in the absence of friction.

$$\begin{gathered} \therefore 2t_1=t_2 \\ \Rightarrow 2\sqrt{\frac{2x}{g\sin \left(\theta \right)}}=\sqrt{\frac{2x}{g\sin \left(\theta \right)-\mu g\cos \left(\theta \right)}} \\ \Rightarrow \frac{8x}{g\sin \left(\theta \right)}=\frac{2x}{g\sin \left(\theta \right)-\mu g\cos \left(\theta \right)} \\ \Rightarrow \frac{4}{\sin \left(\theta \right)}=\frac{1}{\sin \left(\theta \right)-\mu \cos \left(\theta \right)} \end{gathered}$$

$\Rightarrow 4\sqrt{2}=\frac{\sqrt{2}}{1-\mu }$ [Substitute the value of $\theta =45°$]

$\Rightarrow \mu =\frac{3}{4}=0.75$

Hence, the coefficient of the friction is $0.75$.