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`(sqrt(3)g)/(23) ms^(-2)``(3sqrt(3)g)/(23) ms^(-2)``(3g)/(23) ms^(-2)``(g)/(23) ms^(-2)`

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BSolution :

If initial acceleration of M towards right is A, then we can shown that acceleration of m w.r.t. M down the incline is <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_E01_169_S01.png" width="80%"> <br> `a=A(1+cos theta)(3A)/(2) ( because theta = 60^(@))` <br> FBD of block m (w.r.t. M) is shown below: <br> FBD of M <br> Equartion of motion: <br> for m: `mg sqrt(3)/(2) + mA xx 1/2 -T =m 3/2 A` ..(i) <br> `N+mA (sqrt(3))/(2)=mg 1/2` ...(ii) <br> For M: `T+N(sqrt(3))/(2)=MA` ..(iii) <br> From eqs. (i), (ii) and (iii) `A=(3sqrt(3)g)/(23) ms^(-1)`. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C06_E01_169_S02.png" width="80%">