Question

A 8 kg block (A) is placed on 2 kg block (B) which rests on a table. Coefficient of friction between (A) and (B) is 0.5 and between (B) and table is 0.2. A 60 N horizontal force is applied on the block (A), then the friction force between the blocks (A) and (B) is

Answer 1

Answer:

ANSWER

First calculate the static friction force.

At bottom surface friction coefficient is 0.5

F 1

=μN 1

N 1

=(2+8)g=10×10=100N

Thus,F 1 =μN 1

=0.5×100N

=50N

Static friction force for motion of lower block we have to apply minimum 50 N force there is no motion until we apply 50 N force.

Thus, there is no motion hence there is no friction force between block A and B.

solution

Answer 2

Answer:

answer attached in enclosed Photo