Question

A block is projected up to a rough plane of inclination $ 30°$. If the time of ascending is half the time for descending and the coefficient of friction is $ \mathrm{\mu } =\raisebox{1ex}{$ 3$}\!\left/ \!\raisebox{-1ex}{$5$}\right. \sqrt{\mathrm{n}} $. Then $ \mathrm{n}=$

2

Answer 1

Answer: ($n=3$)

Step 1: Given data

The inclination of the rough plane is $30°$

The coefficient of friction is $\mathrm{\mu }=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5\sqrt{\mathrm{n}}$}\right.$

Step 2: To calculate the value of n

Upward motion:

$\mathrm{Retardation}:{\mathrm{a}}_{\mathrm{A}}=\mathrm{gsin\theta }+\mathrm{\mu gcos\theta }$

Downward motion:

$$\begin{gathered} \mathrm{Accelaration}:{\mathrm{a}}_{\mathrm{D}}=\mathrm{gsin\theta }-\mathrm{\mu gcos\theta } \\ \mathrm{S}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{a}}_{\mathrm{A}}{{\mathrm{t}}_{\mathrm{D}}}^2\dots \dots \dots \dots \dots \left(1\right) \\ \mathrm{S}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{a}}_{\mathrm{D}}{{{\mathrm{t}}_{\mathrm{D}}}^2}_{}\cdots \cdots \cdots \cdots \cdots \cdots \left(2\right) \end{gathered}$$

From equation $\left(1\right)$ and $\left(2\right)$:

$$\begin{gathered} \frac{{{\mathrm{t}}_{\mathrm{A}}}^2}{{\mathrm{t}}_{\mathrm{D}}^2}=\frac{{\mathrm{a}}_{\mathrm{D}}}{{\mathrm{a}}_{\mathrm{A}}} \\ \frac{{\mathrm{t}}_{\mathrm{A}}^2}{{\mathrm{t}}_{\mathrm{D}}^2}=\frac{\mathrm{gsin\theta }-\mathrm{\mu gcos\theta }}{\mathrm{gsin\theta }+\mathrm{\mu gcos\theta }} \\ \frac{{{\mathrm{t}}^2}_{\mathrm{A}}}{{{\mathrm{t}}^2}_{\mathrm{D}}}=\sqrt{\frac{\mathrm{gsin\theta }-\mathrm{\mu gcos\theta }}{\mathrm{gsin\theta }+\mathrm{\mu gcos\theta }}} \\ \frac{1}{2}=\sqrt{\frac{1-\sqrt{3\mathrm{\mu }}}{1-\sqrt{3\mathrm{\mu }}}} \\ =1+\sqrt{3\mathrm{\mu }}=4-4\sqrt{3}\mathrm{\mu } \\ =5\sqrt{3\mathrm{\mu }}=3 \\ \mathbf{\mu }\mathbf{=}\mathbf{3}\mathbf{/}\mathbf{5}\sqrt{\mathbf{3}}\mathbf{}\mathbf{} \end{gathered}$$