Acceleration of Slab and Block: 58N Force

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A slab of mass 18 kg is on a frictionless floor with a 12 kg block on top, and a horizontal force of 58 N is applied to the block. The static friction coefficient is 0.30 and the kinetic friction coefficient is 0.20. The first attempt to calculate the acceleration of the block correctly accounted for the force and kinetic friction. However, the second attempt to find the slab's acceleration incorrectly assumed that the kinetic friction force was the only factor acting on it. The discussion highlights the importance of accurately applying friction concepts in multi-body problems.
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Homework Statement


A slab of mass m1 = 18 kg rests on a frictionless floor, and a block of mass m2 = 12 kg rests on top of the slab. Between block and slab, the coefficient of static friction is 0.30, and the coefficient of kinetic friction is 0.20. A horizontal force ,F of magnitude 58 N begins to pull directly on the block. What are the accelerations of (a) the block and (b) the slab?

Homework Equations


The Attempt at a Solution


Attempts for (a)
m1*a= F - fk
Solved for a.
I got it right.

Attempts for (b)
It's sliding (F > Fs) and there is no friction between floor and the slab.
I assumed the fk is pulling the slab. I put it in this way :
m2*a = fk.
Solved for a. But my answer is wrong.

fk = magnitude of kinetic friction between block and slab.
I would guess the equation for my (b) or my assumption is wrong.
 
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Looks right to me. Did you check your math and value for fk?
 
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PhanthomJay said:
Looks right to me. Did you check your math and value for fk?
I just checked it. It was right. Thank you for checking, phantomjay.
 
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