A block of mass $$m$$ is placed on a surface having vertical cross section given by $$y = \frac{x^2}{4}$$. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

Surface is given by
$$y=\frac{x^2}{4}$$

At any point, slope:
$$\frac{dy}{dx}=\frac{x}{2}$$

So,
$$\tanθ=\frac{x}{2}$$

For no slipping:
$$\tanθ\leμ$$

Given $$μ=0.5$$

So:
$$\frac{x}{2}\le0.5$$

$$x\le1$$

Now find maximum height:

$$y=\frac{x^2}{4}$$
$$=\frac{(1)^2}{4}$$
=

$$\frac{1}{4}$$

Final answer:

$$\frac{1}{4}$$

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