Question
Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is
-
2 N
-
6 N
-
8 N
-
18 N
Solution
B.
6 N
Given, mA = 4 kg
mB = 2 kg
=> mC =1 kg
So total mass (M) = 4+2+1 = 7 kg
Now, F = Ma
14 = 7a
a=2 m/s2
F-F' = 4a
F' = 14-4x2
F' = 6N
