Question

A given object takes n times the time to slide down 45° rough inclined plane as it takes the time to slide down an identical perfectly smooth 45° inclined plane. The coefficient of kinetic friction between the object and the surface of inclined plane is :  

(1) \(1-\frac{1}{n^{2}}\)

(2) \(1-n^{2}\)

(3) \(\sqrt{1-\frac{1}{\mathrm{n}^{2}}}\)

(4) \(\sqrt{1-n^{2}}\)  

Answer 1

Correct option is :  (1) \(1-\frac{1}{n^{2}}\) 

Case-1 : No friction

\( \begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta \end{aligned} \) 

\( \begin{aligned} \ell=\frac{1}{2}(\mathrm{~g} \sin \theta) \mathrm{t}_{1}^{2} \\ \end{aligned} \) 

\( \begin{aligned} \mathrm{t}_{1}=\sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \end{aligned} \) 

Case-2 : With friction

\( \begin{aligned} & \mathrm{a}=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta \end{aligned}\) 

\( \begin{aligned} \ell=\frac{1}{2}(\mathrm{~g} \sin \theta-\mu \mathrm{g} \cos \theta) \mathrm{t}_{2}^{2} \end{aligned} \) 

\( \begin{aligned} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta-\mu g \cos \theta}}=\mathrm{n} \sqrt{\frac{2 \ell}{\mathrm{g} \sin \theta}} \end{aligned} \) 

\( \begin{aligned} \mu=1-\frac{1}{\mathrm{n}^{2}} \end{aligned} \)