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A block of mass $m$ is pressed against a vertical surface by a spring of unstretched length $l$ . If the coefficient of friction between the block and the surface is $\mu $. Choose the correct statement.
if spring constant $k = \frac{{2mg}}{{\mu d}}$, block will not be in equilibrium.
if spring constant is $k = \frac{{2mg}}{{\mu d}}$, the normal reaction is $\frac{{mg}}{\mu }$
in the part $(2)$ force of friction is $2mg$
minimum spring constant $k_{min}$ to keep the block of mass $m$ in equilibrium is $\frac{{mg}}{{\mu d}}$
Solution
For equilibrium $:-$ $\quad \mathrm{f} \geq \mathrm{mg}$
$\mu \mathrm{Kd} \geq \mathrm{mg} \quad \mathrm{K} \geq \frac{\mathrm{mg}}{\mu \mathrm{d}}$
Hence $\mathrm{K}_{\min }=\frac{\mathrm{mg}}{\mu \mathrm{d}}$
