Question

35/ Two blocks A and B of masses 4 kg and 12 kg are
placed on a smooth horizontal surface. A force of
magnitude 80 N is applied on A as shown in the
figure. The contact force between A and B is
F = 80 N->
А
90 N
(3) 30 N
(2) 25 N
(4) 60N

Answer 1

Answer:

4) 60N

Explanation:

F=(12x80)/(4+12)

=12x80/16

=12x5

=60 N

Answer 2

Explanation:

   Given Two blocks A and B of masses 4 kg and 12 kg are  placed on a smooth horizontal surface. A force of  magnitude 80 N is applied on A as shown in the  figure. The contact force between A and B is  F = 80 N->

Net force acting on the system of blocks A and B = 80 N

We need to fin acceleration of blocks

F = (m A + m B)a  

   = (4 + 12) a  

80 = 16 a

So a = 5 m / s^2

Taking block B as system we get

The only force acting on B is the contact force due to A will be

N = m B a

   =  12 x 5

  = 60 N