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Consider A situation

Block A is kept on Block B,and The coefficient of friction between them is $\mu$, the ground on which B is kept is frictionless. Block A is attached to a spring with a spring constant '$K$' What will happen if the spring and block system is placed at a distance $x$ from natural length position such that spring force on A exactly equals the maximum static friction between A and B

In this case A will be at equillibrium. Will B have any acceleration? I have certainly not such cases where lower block has greater acceleration than upper (if force is applied on upper block)

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If the inital force $Kx$ from the spring on block A is less than or equal to the maximum static friction between the two blocks then friction from block B will exert a force $F$ to the right on block A that is sufficient to prevent relative motion between the blocks. Note that no relative motion does not means that block A is at equilibrium - if block B accelerates then block A must accelerate too.

Since Block B exerts a force $F$ to the right on Block A, then Block A will exert an equal and opposite force $F$ to the left on block B. So block B will accelerate at a rate $a= \frac F {m_B}$ to the left. Since block A cannot move relative to block B due to the friction between them, block A must also accelerate at a rate $a$ to the left as well. So resolving the forces on block A we have

$\displaystyle Kx - F = m_Aa = F \frac {m_A}{m_B} \\ \displaystyle \Rightarrow Kx = F \frac {m_A+m_B}{m_B} \\ \displaystyle \Rightarrow F = \frac {Kxm_B}{m_A+m_B} \\ \displaystyle \Rightarrow a = \frac {Kx}{m_A+m_B}$

In other words, the two blocks will move as if they were a single block with mass $m_A+m_B$.

I'll let you work out what happens as the blocks start to move to the left and $x$ decreases.

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