Block A of mass 2 kg is placed over block B of mass 8 kg .The combination is placed over a rough horizonatal surface .Cofficient of friction between B and the floor is 0.5 .Coefficient of friction berween A and B is 0.4 .A horizontal force of 10 N is applied on block B .The force of friction between A and B is
(g=10 m`s^(-2))`

Options

A. 100N
B. 40N
C. 50N
D. zero

Answer

D

Explanation

Here,
`m_(A)`=2kg,`m_(B)`=8kg,`mu_(1)`=0.4,`mu_(2)`=0.5,F=10N The frictiobnal force between block B and surface is
`fmu_(2)R=mu_(2)(m_(A)+m_(B))g=0.5xx(2+8)xx10`
=50N
As applied force F(=10N)`lt` f(=50 N).the system will not move Hence,the force of friction between A and B is zero