A force f is acting on a block of mass m. Coe | vedclass

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Question

$\mathrm{d}=\frac{1}{2}(\mathrm{g} \sin 	heta) \mathrm{t}_{1}$

$\mathrm{d}=\frac{\mathrm{1}}{2}(\mathrm{g} \sin 	heta-\mu \mathrm{g} \cos 	heta)

Vedclass · Accepted Answer

$\cot \frac{	heta }{2} \ge \mu $

Vedclass · Answer

$\mathrm{d}=\frac{1}{2}(\mathrm{g} \sin 	heta) \mathrm{t}_{1}$

$\mathrm{d}=\frac{\mathrm{1}}{2}(\mathrm{g} \sin 	heta-\mu \mathrm{g} \cos 	heta) \mathrm{t}_{2}$

$t_{1}=\sqrt{\frac{2 d}{g \sin 	heta}}$

$\mathrm{t}_{2}=\sqrt{\frac{2 \mathrm{d}}{\mathrm{g} \sin 	heta-\mu \mathrm{g} \cos 	heta}}$

According to question, $\mathrm{t}_{2}=\mathrm{nt}_{1}$ $\sqrt[n]{\frac{2 d}{g \sin 	heta}}=\sqrt{\frac{2 d}{g \sin 	heta-\mu g \cos 	heta}}$

$\mu,$ applicable here, is kinetic friction as the block moves over the inclined plane. $\mathrm{n}=\frac{1}{\sqrt{1-\mu_{\mathrm{k}}}}\left(\because \cos 45^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}ight)$

$\mathrm{n}^{2}=\frac{1}{1-\mu_{\mathrm{k}}} \quad$ or $1-\mu_{\mathrm{k}}=\frac{1}{n^{2}}$

Or $\mu_{\mathrm{k}}=1-\frac{1}{\mathrm{n}^{2}}$

$\geqslant \mu\left(2 \sin ^{2} \frac{	heta}{2}ight)$

$	herefore \cot \frac{	heta}{2} \geqslant \mu$

A force $f$ is acting on a block of mass $m$. Coefficient of friction between block and surface is $\mu$. The block can be pulled along the surface if :-

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