A40 kg slab rests on a frictionless floor. a 10 kg block rests on top of the slab (fig. 6-58). the coefficient of static friction µstat beltween the block and the slab is 0.70, whereas their kinetic friction coefficient µkin is 0.40. the 10 kg block is pulled by a horizontal force with a magnitude of 100 n.

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g  = 10 × 9.8 = 98 N  

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied  

10 kg block will slide on 40 kg slab and net force on it  

= 100 N - kinetic friction  

= 100 - 39.2

= 60.8 N

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4  

Frictional force on 40 kg slab by 10 kg block = 39.2 N  

40 kg slab will move with =

f=ma newton's second law 7x8=56n