Laws of Motion

JEE Main level, single correct, medium

186 of 210

Question 186 of 210

A block of mass mm is placed on a surface with a vertical cross section given by y=x36y=\dfrac{{x}^{3}}{6}. If the coefficient of friction is 0.50.5, the maximum height above the ground at which the block can be placed without slipping is

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Answer: (A) 16m\dfrac{1}{6}\text{m}

For block remain stationary.

μ=tan⁡θ(θ\mu =\tan \text{θ}(\text{θ} is angle at that point)

∴tanθ=μ=12(slope)∴ \text{tan}\text{θ}=\mu =\dfrac{1}{2}\text{(slope)}

y=x36y=\dfrac{{x}^{3}}{6}

∴dydx=3x26=x22⇒slope=12⇒x=1∴ \dfrac{\text{d}y}{\text{d}x}=\dfrac{3{x}^{2}}{6}=\dfrac{{x}^{2}}{2}\Rightarrow \text{slope}=\dfrac{1}{2}\Rightarrow x=1

∴y=(1)36∴ y=\dfrac{{(1)}^{3}}{6} =16=\dfrac{1}{6} m

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