Question

A block of mass $4\text{kg}$ is kept on an inclined plane having an angle of inclination ${37}^{\circ }$ attached with a spring of force constant $20\text{N/m}$ as shown in the figure. If the coefficient of friction between the block and incline is $0.8$, find the range in which the block can be at rest without slipping from its equilibrium position?

• A. $2.56\text{m}$
• B. $5\text{m}$
• C. $10\text{m}$
• D. $1.25\text{m}$

Answer 1

The correct option is A $2.56\text{m}$

Case 1: Spring is Compressed:
Let the spring be compressed by ${}^{\prime }{x}_1^{\prime }$ length.
$R$ is the normal force acting on the block applied by the inclined plane on it.
At equilibrium,
$\sum F_y=0$
$\Rightarrow R=mg\cos {37}^{\circ }$
$\sum F_x=0$
$\Rightarrow mg\sin {37}^{\circ }+f_s=k{x}_1$
$\Rightarrow k{x}_1=mg\sin {37}^{\circ }+{\mu }_{s}mg\cos {37}^{\circ }$
($\because f_s={\mu }_{s}R={\mu }_{s}mg\cos {37}^{\circ }$ is limiting value of friction)
$\Rightarrow 20x_1=40\times \frac{3}{5}+0.8\times 40\times \frac{4}{5}$
$\Rightarrow x_1=\frac{6}{5}+\frac{8}{10}\times \frac{8}{5}$
$\Rightarrow x_1=2.48\text{m}$

Case 2: Spring under elongation:

Let the spring be elongated by ${}^{\prime }{x}_2^{\prime }$ length.
At equilibrium,
$\sum F_y=0$
$\Rightarrow R=mg\cos {37}^{\circ }$
$\sum F_x=0$
$\Rightarrow mg\sin {37}^{\circ }+k{x}_2=f_s$
$\Rightarrow k{x}_2={\mu }_{s}mg\cos {37}^{\circ }-mg\sin {37}^{\circ }$
($\because f_s={\mu }_{s}R={\mu }_{s}mg\cos {37}^{\circ }$ is limiting value of friction)
$\Rightarrow 20x_2=0.8\times 40\times \frac{4}{5}-40\times \frac{3}{5}$
$\Rightarrow x_2=\frac{2}{5}(3.2-3)=0.08\text{m}$
$\therefore$ the range in which the block can be at rest without slipping from its equilibrium position is $x_1+x_2=2.48+0.08=2.56\text{m}$