Answer :


Given: Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg


Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.


Actual acceleration of the blocks m1, m2 and m3 will be: a1, (a1 − a2) and (a1 + a2)


From figure 2, T − g – a1 = 0 …(i)


From figure 3, -2g-2(a1-a2)=0 …(ii)


From figure 4, -3g-3(a1+a2)=0 …(iii)


From equations (i) and (ii), eliminating T, we get:


g + a2 = 4g + 4 (a1 + a2)


5a2 − 4a1 = 3g …(iv)


From equations (ii) and (iii), we get:


2g + 2(a1 – a2) = 3g − 3 (a1 – a2)


5a1 + a2= g …(v)


Solving equations (iv) and (v), we get: a1=


a2=g-5a1



Then


And,


So, accelerations of m1, m2 and m3 are up, down and down, respectively.


Now, u = 0, s = 20 cm = 0.2 m





t=0.25 s


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