Laws Of Motion Ques 28
- A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force $2 N$ down the inclined plane. The maximum external force up the inclined plane that
does not move the block is $10 N$.
The coefficient of static friction between the block and the plane is
(2019 Main, 12 Jan I)
(Take, $g=10 m / s^{2}$ )
(a) $\frac{2}{3}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{\sqrt{3}}{4}$
(d) $\frac{1}{2}$
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Answer:
Correct Answer: 28.(b)
Solution:
Formula:
- Block does not move upto a maximum applied force of $2 N$ down the inclined plane.
So, equating forces, we have;
$$ 2+m g \sin \theta=f $$
or $2+m g \sin \theta=\mu m g \cos \theta$
Similarly, block also does not move upto a maximum applied force of $10 N$ up the plane.
Now, equating forces, we have
$$ m g \sin \theta+f=10 N $$
or $m g \sin \theta+\mu m g \cos \theta=10$
Now, solving Eqs. (i) and (ii), we get
$$ m g \sin \theta=4 $$
and $\quad \mu m g \cos \theta=6$
Dividing, Eqs. (iii) and (iv), we get
$$ \begin{array}{rlrl} & \mu \cot \theta & =\frac{3}{2} \\ \Rightarrow & \mu & =\frac{3 \tan \theta}{2}=\frac{3 \tan 30^{\circ}}{2} \\ \Rightarrow \quad \mu & =\frac{\sqrt{3}}{2} \end{array} $$
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