Question

The minimum value of $\mu$ for which the system remains at rest once it has stopped for the first time is:

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{1}{3}$

When the system is held with the spring at its unstretched length and then released, the blocks will start moving with positive acceleration towards right.
writing equations for two blocks
$mg-T=maT-\mu mg-kx=ma$
as the spring stretches it's force increases in magnitude and balances the pull due to mass m, net acceleration of system becomes zero,
$kx+\mu mg=mg$
however, due to acquired velocity it will keep moving to right and elongation in spring will increase, hence, spring force and the system will start decelerating, till both masses comes to rest.The elongation in the spring when system comes to rest can be found from work energy theorem.
$W_{gravity}+W_{friction}+W_{spring}=\Delta K.E.mg\times x-\mu mg\times x+\frac{1}{2}k{x}^2=0\Rightarrow x=\frac{2mg(1-\mu )}{K}$
now due to maximum elongation in spring, mass on table will have tendency to move towards left, friction will start acting to right of block. from force balance equation( as the friction force should be sufficient to keep the system stationary:
$\Rightarrow Kx=T+f=mg+\mu mg$
For the minimum value of $\mu$,
$K\frac{2mg(1-\mu )}{K}\le mg+\mu mg$
$2(1-\mu )\le 1+\mu$
$\mu \ge \frac{1}{3}$
therefore, least value of $\mu =\frac{1}{3}$