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

Hint: 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.