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 $0.5$. $M$ is released from rest when the spring was unstretched. The minimum value of $M$ required to move the block $m$ up the plane is (neglect mass of string and pulley and friction in pulley)

• A. $\frac{m}{2}$
• B. $\frac{m}{3}$
• C. $\frac{m}{4}$
• D. None of these

Answer 1

The correct option is A $\frac{m}{2}$

Let the extension produced in spring be $x$. Then $M$ will move by the same distance $x$.
FBD of the block $m$ as it just starts to move.


Here, $f=\mu N=\mu mg\cos {37}^{\circ }$
(Limiting value of friction)
$kx=\mu mg\cos \theta +mg\sin \theta$
$\Rightarrow kx=0.5mg\times \frac{4}{5}+mg\times \frac{3}{5}=mg$

For minimum value of $M$, extension produced in the spring will be just enough to move the block $m$.
According to law of conservation of energy,
Loss in gravitational potential of $M$ = Potential energy stored in spring
$Mgx=\frac{1}{2}k{x}^2\Rightarrow Mg=\frac{1}{2}kx$
$\Rightarrow Mg=\frac{1}{2}mg\Rightarrow M=\frac{m}{2}$