Laws of Motion

JEE Main level, single correct, medium

128 of 210

Question 128 of 210

The time taken by a block of mass mm to slide down from the highest point to the lowest point on a rough inclined plane is 50%50\% more compared to the time taken by the same block on identical inclined smooth plane. Both inclined planes are at 45∘45^\circ with the horizontal. The coefficient of kinetic friction between the rough inclined surface and block is _____.

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Answer: (C) 5/95/9

Let the length of the inclined plane be ss.

For the smooth inclined plane, the acceleration of the block is a1=gsin⁡θa_1 = g \sin \theta.

The time taken to slide down is t1=2sa1=2sgsin⁡θt_1 = \sqrt{\dfrac{2s}{a_1}} = \sqrt{\dfrac{2s}{g \sin \theta}}.

For the rough inclined plane, the acceleration of the block is a2=gsin⁡θ−μkgcos⁡θa_2 = g \sin \theta - \mu_k g \cos \theta.

The time taken to slide down is t2=2sa2=2sg(sin⁡θ−μkcos⁡θ)t_2 = \sqrt{\dfrac{2s}{a_2}} = \sqrt{\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)}}.

Given that t2t_2 is 50%50\% more than t1t_1, we have t2=t1+0.5t1=32t1t_2 = t_1 + 0.5 t_1 = \dfrac{3}{2} t_1.

Squaring both sides gives t22=94t12t_2^2 = \dfrac{9}{4} t_1^2.

Substituting the expressions for t12t_1^2 and t22t_2^2:

2sg(sin⁡θ−μkcos⁡θ)=94(2sgsin⁡θ)\dfrac{2s}{g (\sin \theta - \mu_k \cos \theta)} = \dfrac{9}{4} \left( \dfrac{2s}{g \sin \theta} \right)

1sin⁡θ−μkcos⁡θ=94sin⁡θ\dfrac{1}{\sin \theta - \mu_k \cos \theta} = \dfrac{9}{4 \sin \theta}

4sin⁡θ=9sin⁡θ−9μkcos⁡θ4 \sin \theta = 9 \sin \theta - 9 \mu_k \cos \theta

9μkcos⁡θ=5sin⁡θ9 \mu_k \cos \theta = 5 \sin \theta

μk=59tan⁡θ\mu_k = \dfrac{5}{9} \tan \theta

Given θ=45∘\theta = 45^\circ, we have tan⁡45∘=1\tan 45^\circ = 1.

μk=59×1=59\mu_k = \dfrac{5}{9} \times 1 = \dfrac{5}{9}

Answer: 5/95/9

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