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(A) The condition for a block to remain at rest on an inclined surface without slipping is that the angle of inclination $	heta$ must be less than or

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$1/4 \ m$

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(A) The condition for a block to remain at rest on an inclined surface without slipping is that the angle of inclination $	heta$ must be less than or equal to the angle of repose $\phi$,where $	an \phi = \mu$.Given the surface equation $y = x^2 / 4$,the slope at any point $x$ is given by the derivative $\frac{dy}{dx} = 	an 	heta$.$\frac{dy}{dx} = \frac{d}{dx} (x^2 / 4) = \frac{2x}{4} = \frac{x}{2}$.For the block to be at the maximum height without slipping,the slope must equal the coefficient of friction $\mu = 0.5$.So,$\frac{x}{2} = 0.5$,which gives $x = 1$.Substituting $x = 1$ into the equation of the surface $y = x^2 / 4$,we get the maximum height $y = (1)^2 / 4 = 1/4 \ m$.

$A$ block of mass $m$ is placed on a surface having a vertical cross-section given by $y = x^2 / 4$. 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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