Question

21. Two blocks of masses 2 kg and 4 kg are hanging with the help of massless string passing over an ideal pulley
inside an elevator. The elevator is moving upward with an acceleration . The tension in the string connected
between the blocks will be (Take g = 10 m/s).​

Answer 1

Acc. To me,

Your answer is T =(8a+80)/3 (a is accleration of elevator)

-----------------------------------------------------------------------------

Suppose your diagram like below and

M=4kg,m=2kg with any accleration (a°)

And elevator goes up with accl= a

In mass 4kg accl is downward with (a°-a) ----becz mass go down(a°) and elevator accl(a) oppose it and net accl in a° direction.

In mass 2kg accl is upward with (a°+a)----becz elevator accl helps mass 2kg.

By FBD

For 2kg

T-mg=m(a°+a)         ByFnet=ma

T-20=2(a+a°)-----------1

For 4kg

mg-T=m(a°-a)

40-T=4(a°-a)-------------2

Just multiply eq 1 by 2and add eqn 1 and 2

then you find all terms cancel and left

T=(8a+80)/3

--------------------------------------------------------------------------

I hope you understand my answer