Question

solve it please asap. No spamming​

Answer 1

$\sf \Large Solution!!$

Let x be the compression in the spring when the bar $\sf m_{2}$ is about to shift. Therefore at this moment spring force on $\sf m_{2}$ is equal to the limiting friction between the bar $\sf m_{2}$ and horizontal floor. Hence,

$\sf k_{spring}x=km_{2}g\dots (1)$ where k is the coefficient of friction.

The spring has to be extended by x to make the block $\sf m_{2}$ move. Notice that the bar2 is just about to move, it hasn't started moving. The work done by the constant force

F is a sum of work done on the spring to stretch it by x and moving the bar $\sf m_{1}$ by x.

$\sf{\therefore Fx=\dfrac{1}{2}k_{spring}x^{2}+km_{1}gx}$

$\sf{\implies F=\dfrac{1}{2}(km_{2}g)+km_{1}g=kg\bigg(m_{1}+\dfrac{m_{2}}{2}\bigg)}$

where we have substituted

$\sf k_{spring}x=km_{2}g\:from\:1$

Comparing with $\sf{F_{min}=kg\bigg(m_{1}+\dfrac{m_{2}}{2}\bigg)}$