A 40kg slab rests on a frictionless floor . A 10kg block rests on top of slab . coefficient of kintic friction between the block and slab is 0.4 . A horizontal force of 100 N applied on 10 kg block .Find resulting acceleration of slab

10 * 9.81 = 98.1 N is the normal response of a 40 kilogramme slab on a 10 kilogramme block.
The static frictional force = 98.1 * 0.6 N is less than the applied force of 100 N.
=> 10 kg blck will slide on 40 kg slab and net force on it
= 100 N – kinetic friction
= 100 – 98.1 * 0.4 = 61 N
=> 10 kg block will slide on 40 kg slab with 61/10 = 6.1 m/s

Frictional force on 40 kg slab by 10 kg block = 98.1 * 0.4 = 39 N
=> 40 kg slab will move with
39/40 m/s
= 0.98 m/s.

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