Question

A block of mass $10kg$ is kept on an inclined plane having an angle of inclination ${37}^{\circ }$ attached with a spring of force constant $20N/m$ as shown in the figure. If the coefficient of friction between block and incline is $0.8$, then the maximum value of elongation for which the block is in contact with the inclined plane and has no slipping is

• A. $0.3\text{m}$
• B. $0.156\text{m}$
• C. $2\text{m}$
• D. $6.66\text{m}$

Answer 1

The correct option is B $0.156\text{m}$

The FBD of the block can be shown as below.
Here, $R$ is the normal reaction force


Since the block will not slide, the maximum elongation can be obtained at the limiting condition.
From the FBD, we have
$\sum F_y=0$ (perpendicular to the plane)
$\Rightarrow kx\sin {37}^{\circ }+R=mg\cos {37}^{\circ }$
$\sum F_x=0$ (along the incline)
$\Rightarrow mg\sin {37}^{\circ }+kx\cos {37}^{\circ }=f$

Since limiting friction $f=\mu R$, we get
$mg\sin {37}^{\circ }+kx\cos {37}^{\circ }=\mu (mg\cos {37}^{\circ }-kx\sin {37}^{\circ })$
$\Rightarrow kx(\cos {37}^{\circ }+\mu \sin {37}^{\circ })=mg(\mu \cos {37}^{\circ }-\sin {37}^{\circ })$
$\Rightarrow kx=\frac{mg(\mu \cos {37}^{\circ }-\sin {37}^{\circ })}{\cos {37}^{\circ }+\mu \sin {37}^{\circ }}$
$\Rightarrow x=\frac{mg}{k}\left(\frac{\mu \cos {37}^{\circ }-\sin {37}^{\circ }}{\cos {37}^{\circ }+\mu \sin {37}^{\circ }}\right)$
$\Rightarrow x=\frac{\frac{100}{20}(\frac{4}{5}\times \frac{4}{5}-\frac{3}{5})}{(\frac{4}{5}+\frac{4}{5}\times \frac{3}{5})}$
$=5\left[\frac{\frac{16}{25}-\frac{3}{5}}{\frac{4}{5}+\frac{12}{25}}\right]=\frac{5}{32}=0.156\text{m}$