Question

If acceleration of A is 2 m/s^2 which is smaller than acceleration of B then value of frictional force applied by B on A is

Answer 1

As A is sliding on the block B with an acceleration of 2 m/ s2, there is relative motion between the blocks so the block a is under the action of a maximum frictional force which is providing it the acceleration of 2 m/s^2. so the frictional force on A is

 F = 5 x 2 = 10 N

Now, this is also the frictional force acting on the 10 kg block in the backward direction. (Action-Reaction pair)

Answer 2

Answer:10N.

Explanation:f=mass of A × acceleration of A

= 5kg× 2m/s^2=10 kgm/s^2= 10N.