Question

a block of mass 5 KG is placed on another block A of mass 10 KG which rests on a smooth frictional surface if coefficient of friction is 0.2 and force of 40 N is applied on A the acceleration ofAis

Answer 1

Answer: a=1 m/s^2

Explanation:

coefficient of frictio(u)=0.2

frictional force (f)= uN

normal force N=mg=(10+5)*10=150

therefore f= 0.2*150=30

therfore F(net)=40-30=10N

therefore acceleration = F(net)/Mass=10/10 =1m/s^2