Question

For the situation shown in the figure, if the spring is elongated by ${}^{\prime }{x}^{\prime }$. The condition for no slipping on the inclined plane is given by

• A. $\tan \theta >{\mu }_s$
• B. $\tan \theta \le {\mu }_s$
• C. $\tan \theta >\frac{1}{{\mu }_s}$
• D. $\tan \theta \le \frac{1}{{\mu }_s}$

Answer 1

The correct option is B $\tan \theta \le {\mu }_s$

Let the spring be elongated by a length $x$.
The FBD of the block is as shown for the elongation.
Here, $R$ is the normal reaction force on the block by the inclined plane.

From the FBD, we have
$R=mg\cos \theta$
$f_s={\mu }_{s}R={\mu }_{s}mg\cos \theta$
(limiting value of static friction)

And, $mg\sin \theta +kx=f_s$
$\Rightarrow mg\sin \theta +kx={\mu }_{s}mg\cos \theta$
$\Rightarrow kx={\mu }_{s}mg\cos \theta -mg\sin \theta$
$\Rightarrow x=\frac{mg}{k}({\mu }_s\cos \theta -\sin \theta )$
Since, $x\ge 0$
${\mu }_s\cos \theta -\sin \theta \ge 0$
$\Rightarrow {\mu }_s\cos \theta \ge \sin \theta$
$\Rightarrow \tan \theta \le {\mu }_s$