Question

A 40 kg slab rests on a frictionless floor. a 10 kg block rests on top of the slab. the coefficient of friction between the block and the slab is 0.40. the 10 kg block is acted upon by a horizontal force of 100 n. if g = 10 m/s2, the resulting acceleration of the slab will be

Answer 1

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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.


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