Question

please answer ! in the arrangement shown tension in the string connecting 4kg and 6kg masses is

Answer 1

HELLO DEAR,


normal reaction on the block of mass 2 kg = f1 = 2×​g = 2×​10 = 20N
Now,

the tension in the string between the block of mass 2kg and the block of mass 4 kg = 20*frictional force
= 20*f
= 20*μf1
= 20 - 4
= 16N

Now,
the friction acting on the block of mass 4 kg = f2
= μmg
= 0.2*4*10
= 8N

now the tension in the string between the mass of 2 kg and the mass of 8 kg = 16 - 8 = 8N

HENCE, Option ( A ) is the correct


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

8N

Explanation:

Maximum frictional force=12+8+4=24N∴Using 20N we can't accelerate the system Applying equilibrium condition on 2kg mass,T1+4=20⇒T1=16NSimilarly on, mass 4kgT2+8=T1⇒T2=8N