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Let $\displaystyle M_1$ be displaced by a small distance $\displaystyle x$

Using the Work energy theorem,
$\displaystyle Fx - \mu_1M_1gx - \frac{1}{2}Kx^2 = \Delta K.E.$

For the situation that $\displaystyle M_2$ just moves,
$\displaystyle \Delta K.E. = 0$

Thus,
$\displaystyle F - \mu_1M_1g - \frac{1}{2}Kx = 0$

$\displaystyle F = \mu_1M_1g+\frac{1}{2}Kx$

For $\displaystyle M_2$ to move slightly,

$\displaystyle \mu_2M_2g = Kx$

Hence,
$\displaystyle F = \mu_1M_1g + \frac{1}{2}\mu_2M_2g$

$\displaystyle F = \left(\mu_1M_1+\frac{\mu_2}{2}M_2\right)g$

Therefore,
$\displaystyle F = \left(0.4\times 3 + 0.3\times 5\right)\times 9.8 = \boxed{26.46N}$

Solution without using energy.

let $x$ be a displacement for $M_{1}$ . Using Newton 2nd law for $M_{1}$ we get, $a + \frac{k}{M_{1}} x = \frac{F-\mu_{1} M_{1}g}{M_{1}}$ with initial condition $x(0) = 0$ and $v(0)=0$ .

Those are equation for simple harmonic oscillation, then we could write the solution as $x(t) = \frac{F-\mu_{1}M_{1}g}{k}(1 - \cos(\omega t))$ and the velocity, $v(t) = \omega \frac{F- \mu_{1} M_{1}g}{k} \sin(\omega t )$

For minimum $F$ , the second mass will move when $M_{1}$ at maximum compression, i.e. the velocity is zero. It occurs at $t = \pi / \omega$ and the maximum compression is $x_{m} = 2 \frac{F - \mu_{1}M_{1}g}{k}$ .

Since $M_{2}$ just about to move, then $kx_{m} = \mu_{2} M_{2}g$ , or $F = \frac{2 \mu_{1}M_{1}g + \mu_{2} M_{2} g}{2}$

Hence, $F = 26.46 N$ .

force greater than U1M1g will start move first block and it would cross mean position and stop at other exreme..spring will be most compressed when this block reaches other extreme. hence second block can move with least force at this moment.. F-U1M1g = Kx, 2kx=U2M2g.. solving it we get F=26.46 N