Forces on a slab of mass Question

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A slab of mass 40 kg rests on a frictionless floor with a 10 kg block on top, subject to a 100 N horizontal force. The static friction coefficient between the block and slab is 0.60, allowing a maximum static friction force of 60 N, which is exceeded by the applied force, leading to slipping. Kinetic friction, with a coefficient of 0.40, then acts on both masses, causing them to accelerate differently. The kinetic friction force calculated is 40 N acting backward on the block and forward on the slab. This results in distinct accelerations for both the block and the slab until they separate.
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Homework Statement


A slab of mass m_{1} = 40 \mbox{kg} rests on a frictionless floor and a block m_{2} = 10 \mbox{kg} rests on top of the mass. Between the block and the slab, the coefficient of static friction is 0.60 and the coefficient of kinetic friction is 0.40. The block ontop of the mass is pulled by a horizontal force of 100 N. What are the resulting accelerations of the block and the slab?


Homework Equations





The Attempt at a Solution


What I'm having trouble is identify the forces. My friend says that with the force being applied on m_{2}, there would be friction resisting the motion. Consequently, a force of the same magnitude of the mentioned friction force would occur on m_{1}in the direction of the applied force. How does that work?
 
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The max value of static friction possible between the two is \mu_sm_2g =0.60*10*10=60 N.
Since the applied force is 100 N, definitely there will be slipping between the block and the slab and ultimately, they will get separated.

Until they get seperated, Kinetic friction will act and both of them will have different accelerations.

f_k=\mu_km_2g=40 N acting on m_2 in the backward direction and on m_1 in forward direction.

so a_1=<< solution deleted by berkeman >>



and a_2=<< solution deleted by berkeman >>
_________
I may be wrong.
 
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