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A block slides down a slope of angle $θ$ with constant velocity. It is then projected up with a velocity of $10 {ms}^{- 1}, g = 10 {ms}^{- 2} & θ = 30 °$ . The maximum distance it can go up the plane before coming to stop is

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4 m

5 m

15 m

10 m

answer is B.

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Detailed Solution

As the block is sliding down with constant velocity

$mg \sin θ = μ mg \cos θ$

When moving up total force of the block $= mg \sin θ + μ mg \cos θ$

$= 2 mg \sin θ$

Acceleration $= 2 g \sin θ$

$S = \frac{u^{2}}{2a} = \frac{u^{2}}{4 g \sin θ} = \frac{100}{40 \times \frac{1}{2}} = 5 m$

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