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A block slides down a rough inclined plane of angle of inclination ‘ $θ$ ’ with a constant velocity. If it is pushed up the same plane with an initial velocity ‘v’, then the distance travelled by the block up the plane before coming to rest is

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$\frac{v^{2}}{4 g \sin θ}$

$\frac{v^{2}}{2 g \sin θ}$

$\frac{v^{2}}{g \sin θ}$

$\frac{4 v^{2}}{g \sin θ}$

answer is A.

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Detailed Solution

when it is sliding down

$f = m g \sin θ,$ net acceleration is zero as velocity is constant.

when it is projected up, retardation

$a = \frac{f + m g \sin θ}{m} = \frac{m g \sin θ + m g \sin θ}{m} = 2 g \sin θ$

From $v^{2} - u^{2} = 2 as$

$0 - v^{2} = - 2 as \Rightarrow v^{2} = 2.2 gsin θ . s \Rightarrow s = \frac{v^{2}}{4 g \sin θ}$

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