Laws Of Motion Ques 6
- A block of mass $m$ is placed on a surface with a vertical cross-section given by $y=x^3 / 6$. If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is
(2014 Main)
(a) $\frac{1}{6} m$
(b) $\frac{2}{3} \mathrm{~m}$
(c) $\frac{1}{3} \mathrm{~m}$
(d) $\frac{1}{2} \mathrm{~m}$
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Answer:
Correct Answer: 6.( a )
Solution:
Formula:
$ \tan \theta=\frac{d y}{d x} \frac{d\left(\frac{x^3}{6}\right)}{d x}=\frac{x^2}{2} $
At limiting equilibrium, we get
$ \begin{aligned} & \mu=\tan \theta \\ & 0.5=\frac{x^2}{2} \\ & \Rightarrow \quad x^2=1 \\ & \Rightarrow \quad x= \pm 1 \\ & \end{aligned} $
Now, putting the value of $x$ in $y=\frac{x^3}{6}$, we get
$ \begin{array}{l|l} \text { When } x=1 & \text { When } x=-1 \\ y=\frac{(1)^3}{6}=\frac{1}{6} & y=\frac{(-1)^3}{6}=\frac{-1}{6} \end{array} $
So, the maximum height above the ground at which the block can be placed without slipping is $1 / 6 \mathrm{~m}$.
BETA
