Three-blocks A,B and C of mass 4kg, 2kg and 1kg respectively are in contact of a frictionless surface as shown. If a force of 14N is applied on the 4kg block, then the contact force between A and B will be
A
6N
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
8N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
18N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2N
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A6N
Writing force equations (assuming common acceleration to be a), 14−N=4a N−N1=2a N1=1×a
Solving, we get N=6N
Alternate solution:
Since all the three blocks are going to move with common acceleration, we can consider them as one system. a=Net external forceTotal mass=147=2m/s2
Now if you consider blocks B and C together as a system, the net external force acting on it is contact force N.