Question

Three blocks of masses 3 kg, 2 kg and 1 kg are placed side by side on a smooth surface as she
in figure. A force of 12 N is applied on 3 kg block. The contact force between 3 kg and 2 ka bloc
3kg|2kg 1kg
(1) 2N
(2) 4 N
(3) 6N
(4) & N​

Answer 1

First look at the whole system:  A 12 N force is applied to a total mass of 6 kg.  

Since F = ma, 12N = 6 kg x a, we see the whole mass will accelerate at 2 m/s/s.

The 2 kg must push on the 3 kg block hard enough to accelerate it at 2m/s/s.  Again, F = ma, F = 3kg x 2m/s/s, F = 6N

Same thing for the second part:  The 1kg block is pushing 5kg ahead of it hard enough to accelerate the 5kg @ 2m/s/s... a force of 10N.  That's the force between the two blocks.  I think of it as the 1kg pushing on the 2kg - because that's the direction they're moving, etc.  However, if you put a miniature bathroom scale between them the scale would read 10 N regardless of which way it was facing.  Block 1 pushing on block 2?  Fine.  Block 2 pushing back on block 1?  Also fine.

Another Method-

find the acceleration :

12 = (1+2+3) a

=> a = 12 / 6 = 2 m/s^2

a) F23 = m3 x a

.........= 3 x 2 = 6 N <<<<<<

b) the total force on the 1 kg block would be 12 - F21 , so :

12 - F21 = 1 x a

=> 12 - F21 = 2

=> F21 = 12 - 2 = 10 N <<<<<<<<

Mark Brainliest if it is correct...

Answer 2

m=3+2+1=6

f=12

a= 2 ms

so easy