Question

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

Answer 1

Normal reaction from 40 kg slab on 10 kg block = 10 * 9.81 = 98.1 N 
Static frictional force = 98.1 * 0.6 N is less than 100 N applied 
=> 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.

Answer 2

right answer is

$.98m {s}^{2}$