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A block slides down an inclined plane of angle 45° with constant velocity. If it is then projected up the same plane with velocity 10 ms^-1, then (g = 10 ms^-2)
I) The retardation while going up is 10√2 ms^-2
II) It comes to rest after moving up for $\frac{1}{\sqrt{2}} s$
III) It comes to rest after moving up for √2 s
IV) Coefficient of friction is 0.5

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I, II & IV are correct

I, III & IV are correct

I & II are correct

III & IV are correct

answer is C.

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Detailed Solution

$\begin{array}{l} θ = 45^{\circ}; μ = 1 \\ ∴ a = g (\sin θ + μcos θ) \\ v = u - at \end{array}$

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