Question

A block of mass m is placed on a Surface with a vertical height given by $y=\dfrac{{{x}^{3}}}{6}$. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
A. $\dfrac{1}{3}m$
B. $\dfrac{1}{2}m$
C. $\dfrac{1}{6}m$
D. $\dfrac{2}{3}m$

Answer 1

Since a block of mass m is placed on surface with a vertical height given by $y=\dfrac{{{x}^{3}}}{6}$, we use the formula of limiting friction $\mu =\tan \theta$ . As the coefficient of friction is $0.5$ then we have to find the value of y without slipping.

Complete answer:
A diagram can be illustrated as follows:

As we know that, a block of mass m is placed on a surface with a vertical height given by $y=\dfrac{{{x}^{3}}}{6}$First we define the limiting friction. The limiting fiction is that maximum friction that can be generated between two static surfaces in content with each other once a force applied to the two surfaces exceeds the limiting friction motion will occur. For two dry surfaces, the limiting friction is a product of the normal friction force and the coefficient of limiting friction is given by
$\mu =\tan \theta \text{ }\left( 1 \right)$
Equation of the surface$y=\dfrac{{{x}^{3}}}{6}$
Differentiate w.r.t x, on both sides we get
$\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{2}\text{ }\left( 2 \right)$
And we know$dy=\tan \theta =\dfrac{{{x}^{2}}}{2}$
From1 and 2 we get
$\mu =\dfrac{{{x}^{2}}}{2}$
$\Rightarrow 0.5=\dfrac{{{x}^{2}}}{2}\Rightarrow {{x}^{2}}=0.5\times 2=1.0$
$x=1$
So, $y=\dfrac{1}{6}$

Hence the correct option is (c).

Note:
It must remember the definition of limiting friction $\mu =\tan \theta$ and displacement is $y=\dfrac{{{x}^{3}}}{6}$ . On differentiating we get $y=\dfrac{{{x}^{3}}}{6}$ and put the value we get the maximum height at which the block is placed without slipping.