Force needed to move blocks against friction 
Two blocks of masses 3kg and 5kg are connected by a spring of stiffness k. The coefficient of friction between the blocks and surface is 0.2 . Find the minimum constant horizontal force F to be applied to 3kg block in order to just slide the 5kg block.

My work : For 5kg block to just slide, the spring force should be equal to friction experienced by it which is equal to 10N. Now the external force F applied on 3kg block should be greater than or equal to the spring force + friction due to surface which is equal to 10N + 6N = 16N . So my result is that the minimum force should be equal to F = 16N. But this is not the answer. I want to know where am I wrong.
Ans - F =11 N
 A: 
The force causes acceleration on $m_1$ (right hand block) and work is done on the spring:
$$(F-\mu m_1g)x=\frac12 kx^2$$
The maximum spring extension before $m_2$ starts to move is:
$$x_{max}=2\Big(\frac{F-\mu m_1g}{k}\Big)$$
This causes the tension in the spring to reach a maximum:
$$T_{max}=kx_{max}=2(F-\mu m_1g)$$
Which has to exceed the friction provided by $m_2$, to cause motion of $m_2$:
$$2(F-\mu m_1g)>\mu m_2g$$
$$\implies F>\mu(\frac{m_2g}{2}+m_2g)$$
With the provided numerical values this gives:
$$F>11\:\mathrm{N}$$
If this condition isn't met then $m_1$ will simply come to a halt and $m_2$ won't start moving.
A: For 3kg block:-
By work- energy theorem, Work by external force + work by non-conservative force = change in K.E + change in P.E
Let required extension of spring be x metres
So, Fx - mu(=0.2) × 3 × g(=10) × x = change in P.E + change in K.E
As from the equation, eveything except change in K.E is fixed , so for minimum value of F , K.E should be equal to zero. So assuming change in K.E. is very negligible
Fx - 0.2 × 3 × 10 × x = 0.5kx^2 + 0
F - 0.2 × 3 × 10 = 0.5kx
Since to move the 5 kg block, kx minimum magnitude is 10 N. So-
F - 6 = 5
So, F = 11 N
