Three identical blocks of masses m=2kg are drawn by a force 10.2 N on a frictionless surface. What is the tension (in N) in the string between the blocks B and C?

Options

A. 9.2
B. 8
C. 3.4
D. 9.8

Answer

C

Explanation

( c) Given, `m_(1)=m_(2)=m_(3)=2 kg,` F=10.2 N in a frictionles surface as shown in free body diagram

For body A, `F-T_(1)=ma` ….(i)
For body B, `T_(1)-T=ma`…..(ii)
For body C, `T=ma`………(iii)
Adding these three equations, F=3 ma
`a=(10.2)/(3xx2)=1.7 ms^(-2)`
Alternately acceleration can be found as net acceleration of a system, i.e.,
`a=("Total net force")/("Total mass")`
`=(10.2)/((2+2+2))=(10.2)/(6)=1.7 ms^(-2)`
So, net tension in string between the blocks B and C is
`T=mxxa=2xx1.7=3.4N`