JEE MAIN - Physics (2014 (Offline) - No. 28)
A block of mass $$m$$ is placed on a surface with a vertical cross section given by $$y = {{{x^3}} \over 6}.$$ If the coefficient of friction is $$0.5,$$ the maximum height above the ground at which the block can be placed without slipping is:
$${1 \over 6}m$$
$${2 \over 3}m$$
$${1 \over 3}m$$
$${1 \over 2}m$$
Explanation
At limiting equilibrium, $$\mu = \tan \theta $$
Equation of the surface,
$$y = {{{x^3}} \over 6}$$
Slope, $$\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$$
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Given that, Coefficient of friction $$\mu = 0.5$$
$$\therefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$
$$ \Rightarrow \,\,\,x = \pm \,1$$
Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$
Equation of the surface,
$$y = {{{x^3}} \over 6}$$
Slope, $$\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$$
Given that, Coefficient of friction $$\mu = 0.5$$
$$\therefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$
$$ \Rightarrow \,\,\,x = \pm \,1$$
Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$
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