Question

A block of mass $\sqrt{2}\text{kg}$ is kept on an inclined plane having an angle of inclination ${45}^{\circ }$ attached with a spring of force constant $10\text{N/m}$ as shown in the figure. If the coefficient of friction between block and incline is $0.4$, then which of the following statements is true.

• A. If spring is elongated by $x>2\text{m}$, then the block has no slipping.
• B. If spring is compressed by $x<1.4\text{m}$, then the block has no slipping.
• C. If spring is compressed by $x>1.4\text{m}$, then the block has no slipping.
• D. If spring is elongated by $0<x<2\text{m}$, then the block has no slipping.

Answer 1

The correct option is B If spring is compressed by $x<1.4\text{m}$, then the block has no slipping.

Case 1: When the spring is elongated

For no slipping, at limiting condition
$\Sigma F_y=0$
$\Rightarrow N=mg\cos \theta$
$\Rightarrow f=\mu N=\mu mg\cos \theta$
and, $\Sigma F_x=0$
$\Rightarrow f=k{x}_1+mg\sin \theta$
$\Rightarrow k{x}_1=mg(\mu \cos \theta -\sin \theta )$
As $\theta ={45}^{\circ }$
$\Rightarrow k{x}_1=mg(\mu \cos {45}^{\circ }-\sin {45}^{\circ })$
$\Rightarrow k{x}_1=mg(\frac{\mu }{\sqrt{2}}-\frac{1}{\sqrt{2}})$
$\Rightarrow x_1=\frac{mg}{\sqrt{2}k}(\mu -1)$
as $\mu =0.4<1$, if string is elongated, there always will be slipping.

Case 2: For compression
For no slipping, at limiting condition
$\Sigma F_y=0$
$\Rightarrow N=mg\cos \theta$
$\Rightarrow f=\mu N=\mu mg\cos \theta$
and, $\Sigma F_x=0$
$\Rightarrow k{x}_2=mg\sin \theta +f$
$\Rightarrow k{x}_2=mg(\sin \theta +\mu \cos \theta )$
As $\theta ={45}^{\circ }$
$\Rightarrow k{x}_2=mg(\mu \cos {45}^{\circ }+\sin {45}^{\circ })$
$\Rightarrow k{x}_2=mg(\frac{\mu }{\sqrt{2}}+\frac{1}{\sqrt{2}})$
$\Rightarrow x_2=\frac{mg}{k\sqrt{2}}(1+\mu )$
$\Rightarrow x_2=\frac{10\sqrt{2}}{10\sqrt{2}}(1+0.4)=1.4\text{m}$
Hence, If spring is compressed by $x<1.4\text{m}$, then the block has no slipping.