Question

A block starts moving up an inclined plane of inclination $30°$ with an initial velocity of ${\mathrm{v}}_0$. It comes back to its initial position with velocity $\frac{{\mathrm{v}}_0}{2}$.The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{\mathrm{I}}{1000}$ . The nearest integer to $\mathrm{I}$ is ____.

Answer 1

Step 1: Given data and the diagram

Initial velocity of block when it is moving up an inclined plane$={\mathrm{v}}_0$

Angle of inclination$={30}^0$

When it comes back to its initial position its velocity$=\frac{{\mathrm{v}}_0}{2}$

The coefficient of kinetic friction, $\mathrm{\mu }=\frac{\mathrm{I}}{1000}$

Let distance travelled by block in going upwards$=\mathrm{l}$

The diagram of a block is given below:

Step 2: Calculating the value of $\mathrm{I}$

Apply the work energy theorem on the block, we get,

${\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{f}}+{\mathrm{W}}_{\mathrm{N}}={\left(\mathrm{K}.\mathrm{E}\right)}_{\mathrm{f}}-{\left(\mathrm{K}.\mathrm{E}\right)}_{\mathrm{i}}$

Here, we have,

Work done due to gravity, ${\mathrm{W}}_{\mathrm{g}}=0$

Work done due to normal force, ${\mathrm{W}}_{\mathrm{N}}=0$ (Because normal force is perpendicular to displacement)

Work done due to friction$={\mathrm{W}}_{\mathrm{f}}$

Final kinetic energy$={\left(\mathrm{K}.\mathrm{E}\right)}_{\mathrm{f}}$

Initial kinetic energy$={\left(\mathrm{K}.\mathrm{E}\right)}_{\mathrm{i}}$

Therefore,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{f}}=\frac{1}{2}\mathrm{m}{\left(\frac{{\mathrm{v}}_0}{2}\right)}^2-\frac{1}{2}{\mathrm{mv}}_0^2 \\ \Rightarrow -\mathrm{\mu mgcos}{\left(30\right)}^0\left(2\mathrm{l}\right)=\frac{1}{2}{\mathrm{mv}}_0^2\left(\frac{1}{4}-1\right) \\ \Rightarrow \mathrm{\mu gcos}{\left(30\right)}^0\left(2\mathrm{l}\right)=\frac{3{\mathrm{v}}_0^2}{8} \\ \Rightarrow \mathrm{\mu gcos}{\left(30\right)}^0\left(\frac{{\mathrm{v}}_0^2}{\mathrm{gsin}{\left(30\right)}^0+\mathrm{\mu gcos}{\left(30\right)}^0}\right)=\frac{3}{8}{\mathrm{v}}_0^2 \\ \Rightarrow \frac{\mathrm{\mu }\sqrt{3}}{2\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\mathrm{\mu }\right)}=\frac{3}{8} \\ \Rightarrow 8\sqrt{3}\mathrm{\mu }=3\sqrt{3}\mathrm{\mu }+3 \\ \Rightarrow 5\sqrt{3}\mathrm{\mu }=3 \\ \Rightarrow \mathrm{\mu }=\frac{\sqrt{3}}{5} \end{gathered}$$

Given that $\mathrm{\mu }=\frac{\mathrm{I}}{1000}$

Now comparing both the values of coefficient of kinetic friction we get,

$$\begin{gathered} \frac{\mathrm{I}}{1000}=\frac{\sqrt{3}}{5} \\ \Rightarrow \mathrm{I}=200\times \sqrt{3} \\ \Rightarrow \mathrm{I}=346 \end{gathered}$$

Therefore the value of $\mathrm{I}$ is equal to $346$