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A block is placed on an rough inclined plane of inclination equal to the angle of repose. If it is moving up on incline with velocity v₀ then its maximum displacement on incline will be -
Option: 1
$\frac{v_{0}^{2}}{2g\sin \theta }$

Option: 2
$\frac{v_{0}^{2}}{4g\sin \theta }$

Option: 3
$\frac{v_{0}^{2}}{2g\cos \theta }$

Option: 4
$\frac{v_{0}^{2}}{4g\cos \theta }$

Answers (1)

best_answer

Given-

$\mathrm{\mu=tan\theta}$

Initial velocity of block, $\mathrm{u=v_{0}}$

Let acceleration of block be a as shown in diagram below.

$\mathrm{ma=mg\ sin\theta+f_{k}}$

$\\ \mathrm{R=mgcos\theta}\\ \mathrm{f_{k}=\mu R=tan\theta \times mgcos\theta =mgsin\theta }$

$\\ \mathrm{ \Rightarrow ma=mg\ sin\theta+mg\ sin\theta}\\ \mathrm{ \Rightarrow a=2gsin\theta}$

Let the maximum displacement up the incline be s. At the position of maximum displacement the velocity will be zero as the block is decelerated.

Using 3rd equation of motion-

$\\ \mathrm{ v^2=u^2+2as}\\ \mathrm{ 0=v_{0}^2+2(-a)s}\\ \mathrm{s=\frac{v_{0}^2}{2a}=\frac{v_{0}^2}{4gsin\theta}}$

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