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A 40kg slab rest on a frictionless floor...

A 40kg slab rest on a frictionless floor as shown in . A 10kg blocks rests on the top of the slab. The static coefficient of friction between the block and the slab is `0.60`, while kinetic friction is `0.40`. The 10kg block is acted upon by a horizontal force of 100N. What is the resulting acceleration of the slab ?

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The correct Answer is:
`0.98m//s^(2)`
`a=F//M=mu mg//M=0.4xx10xx9.8//40=98ms^(-2)`
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