Laws of Motion

JEE Main level, numerical answer, medium

126 of 210

Question 126 of 210

A block takes tt time to slide down a plane inclined at 45°45° to the horizontal. If the surface is made smooth (frictionless), the block takes time t2\dfrac{t}{2} to slide down the plane. The coefficient of friction between the block and the inclined plane is (α100)\left(\dfrac{\alpha}{100}\right). The value of α\alpha is __________.

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Answer: 75

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 is t1=t2t_1 = \dfrac{t}{2}. Using the second equation of motion s=ut+12at2s = ut + \dfrac{1}{2}at^2 with u=0u = 0:

s=12(gsin⁡θ)(t2)2=18gsin⁡θt2s = \dfrac{1}{2} (g \sin \theta) \left(\dfrac{t}{2}\right)^2 = \dfrac{1}{8} g \sin \theta t^2

For the rough inclined plane, the acceleration of the block is a2=g(sin⁡θ−μcos⁡θ)a_2 = g(\sin \theta - \mu \cos \theta).

The time taken is t2=tt_2 = t. Using the equation of motion again:

s=12g(sin⁡θ−μcos⁡θ)t2s = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2

Equating the two expressions for ss:

18gsin⁡θt2=12g(sin⁡θ−μcos⁡θ)t2\dfrac{1}{8} g \sin \theta t^2 = \dfrac{1}{2} g(\sin \theta - \mu \cos \theta) t^2

14sin⁡θ=sin⁡θ−μcos⁡θ\dfrac{1}{4} \sin \theta = \sin \theta - \mu \cos \theta

Given θ=45∘\theta = 45^{\circ}, we have sin⁡45∘=cos⁡45∘=12\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}. Substituting these values:

14=1−μ\dfrac{1}{4} = 1 - \mu

μ=1−14=34=0.75\mu = 1 - \dfrac{1}{4} = \dfrac{3}{4} = 0.75

We are given μ=α100\mu = \dfrac{\alpha}{100}.

α100=0.75⇒α=75\dfrac{\alpha}{100} = 0.75 \Rightarrow \alpha = 75

Answer: 7575

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