Question

Three identical blocks of masses m = 2 kg are
drawn by a force F = 10.2 N on a frictionless
surface, then what is the tension (in N) in the
string between the blocks B and C ?
(a) 9.2 (b) 3.4
(c) 4 (d) 9.8

Answer 1

Answer:

T = m3 × F /(m1+m2+m3)

T = 2×10.2/(2+2+2)

T = 10.2/3

T = 3.4

Answer 2

Answer:

( b )

Explanation:

{ In this explanation, T1is represented by T' and F refers to the force acting }

For first block,

= > F - T' = ma

= > 10.2 - T' = 2a

= > T' = 10.2 - 2a

For second block,

= > T' - T'' = ma

= > 10.2 - 2a - T'' = 2a

= > 10.2 - 4a = T'' ...( 1 )

For third block,

= > T" = ma

= > 10.2 - 4a = 2a { from ( 1 ) }

= > 10.2 = 6a

= > 1.7 = a

Therefore,

= > T" = 10.2 - 4( 1.7 )

= > T" = 10.2 - 6.8

= > T" = 3.4

Hence, tension between the last blocks is 3.4 N.