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Question

# Two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley as shown in figure m1=3kg and m2=6kg . The system is released from rest (a) Find the distance traveled by the first block the first two seconds. g=10m/s2

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Solution

## Step 1: Drawing FBD [Ref. Fig.]T= Tension in stringStep 2: Applying Newton's Second LawOn block 2 (Taking downward positive) ∑Fy=ma −T+m2g=m2a ....(1)On block 1 (Taking upward positive) ∑Fy=ma T−m1g=m1a ....(2)Adding Equations (1) and (2), we get m2g−m1g=m1a+m2a⇒ (m2−m1)g(m1+m2)=a⇒ a=(6−3)×106+3m/s2=103m/s2Step 3: Applying equations of motionSince acceleration is constant, therefore applying equation of motion for block 1(Taking upward positive) S=ut+12at2 u=0;t=2s S=12×103×(2)2m =6.66mHence, the distance traveled by the first block is 6.66m in the first two seconds.

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