now making FBD of small mass m you will get following forces on it 1- tension upward=T 2-weight downward=mg 3- normal reaction due to block of mass M=ma towards right 4-friction force upward=F"=k1ma ma will provide acceleration a towards right equating vertical component to mass*acceleratin(plz note that the acceleration of this block will be twise as that of block of mass M ) u will get this eq mg-T-F"=2ma mg-T-k1ma=2ma T=mg-k1ma-2ma................4
Moderation Team
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