A block of mass m is placed on a surface with a vertical cross-section given by y = x³/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

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$\frac{2}{3} m$

$\frac{1}{2} m$

$\frac{1}{3} m$

$\frac{1}{6} m$

answer is A.

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Detailed Solution

$\begin{matrix} \tan θ = \frac{dy}{dx} = \frac{d (\frac{x^{3}}{6})}{dx} = \frac{x^{2}}{2} \\ μ = \tan θ \\ 0 .5 = \frac{x^{2}}{2} \Rightarrow x = 1 \\ y = \frac{(1)^{2}}{6} \end{matrix}$

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