Question

A block of mass $m$ is attached with a massless spring of force constant $k$. The block is placed over a rough inclined surface for which the coefficient of friction is $\mu =\frac{3}{4}$. The minimum value of $M$ required to moved the block up the plane is (Neglect mass of pulley, string and friction in pulley).

• A. $\frac{3}{5}m$
• B. $\frac{4}{5}m$
• C. $\frac{6}{5}m$
• D. $\frac{3}{2}m$

Answer 1

The correct option is A $\frac{3}{5}m$

Given, coefficient of friction $\mu =\frac{3}{4}$,
mass of block $A=M$,
mass of block $B=m$
When block $A$ moves for a distance $x$, then spring will stretch to same $x$ value. And at that value of $x$, let block $B$ be just about to move.
FBD for block $B$,


Block $B$ is about to move upwards along the wedge, hence friction force will act opposite to it (downwards) along the wedge. Friction will be limiting because we want the minimum value of $M$.
So, $kx=mg\sin \theta +f$
$\Rightarrow kx=mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ }$
$\Rightarrow kx=mg\left(\frac{3}{5}\right)+\frac{3}{4}\times mg\times \frac{4}{5}$
$\Rightarrow kx=\frac{6}{5}mg.....\left(1\right)$
For Block $A$,
Applying work energy theorem,
Decrease in potenital energy of $A=$ Potential energy stored in spring
$\Rightarrow \frac{1}{2}k{x}^2=Mgx$
$\Rightarrow kx=2Mg.....\left(2\right)$

Now, from eq $\left(1\right)$ & $\left(2\right)$, we get
$M=\frac{3}{5}m$ is the minimum value of $M$ required to move the block $B$