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Two blocks mA=10 kg and mb=5 kg are connected with cord & pulley system as shown in fig

Determine the velocity of each block when system is started from rest & block B gets displacement by 2m. Take µk=0.2 between block A & horizontal surface.

enter image description here

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Consider length of string w.r.t. block A as XA &

Length of string w.r.t block B as XB

enter image description here

So, applying C.S.L.M. we get

2XA+XB=C

But as B moves down, XB increases & XA decreases

So, rewriting eqn

2XA+XB=C______(i) 

Differentiating w.r.t. t

$-2VA+VB=0............2VA=VB\text …

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