Question

In the figure shown below, a $4kg$ block is compressed by $2m$. The coefficient of friction between the block and the plane is $\mu =0.8$. Then the frictional force (in $\text{N}$) acting on the block is given by

Answer 1

The FBD of the block is as shown in the below figure.


From the FBD, at limiting condition we have
$\sum F_y=0$ (perpendicular to the plane)
$\Rightarrow N=mg\cos {37}^{\circ }=40\times \frac{4}{5}=32\text{N}$
$\sum F_x=0$ (along the plane)
$\Rightarrow mg\sin {37}^{\circ }+f=kx$...........(1)
Limiting friction
$\left(f_{max}\right)={\mu }_{s}N=0.8\times 32=25.6\text{N}$

Let the maximum compression in spring for which block will not slide be $x_{max}$
From eq (1),
$k{x}_{max}=mg\sin {37}^{\circ }+f_{max}=40\times \frac{3}{5}+25.6=49.6$
$\Rightarrow x_{max}=\frac{49.6}{k}=\frac{49.6}{20}=2.48m$

As per the question, the spring is compressed by a length $2m$, so it will not slide ($\because 2m<2.48m$)

$\therefore$ from equation (1), the friction force acting on the block is given by
$mg\sin {37}^{\circ }+f=kx\Rightarrow f=kx-mg\sin {37}^{\circ }$
$\Rightarrow f=20\times 2-40\times \frac{3}{5}=16\text{N}$