Question

A 40kg slab rests on frictionless floor. A 10kg block rests on the top of the slab. The static coefficient of the friction between the block and the slab is 0.60 while the kinetic friction ids 0.40 the 10kg block is acted upon by a horizontal force F. What is the resulting acceleration of the slab if F=10N and F=100N

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.