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Question

A pulley-block system as shown in the figure is released from rest. What is the magnitude of acceleration of COM of pulley-block system? (Take g=10 m/s2)


A
7 m/s2
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B
5 m/s2
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C
3.77 m/s2
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D
10.5 m/s2
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Solution

The correct option is C 3.77 m/s2

Tension T is same throughout string length and tension in string connecting block 4 kg will be 2T


Applying string constraint relation:
l1+l2+l3= constant
Differentiating twice w.r t time gives:
aa+a=0

Rate of decrease of length=ve
Rate of increase of length=+ve
a=2a ....(i)

From the FBD of blocks, we get the equation of dynamics in direction of acceleration as:


9gT=9×2a....(ii)

2T4g=4a ....(iii)
Solving the equations (ii) and (iii) we get:
a=14040=3.5 m/s2.
Hence acceleration of 9 kg is 2a=2×3.5=7 m/s2

Applying formula for COM of system:
aCM=m1a1+m2a2m1+m2
Considering the downward direction as +ve:
a2=+7 m/s2
a1=3.5 m/s2
aCM=(4×3.5)+(9×7)4+9
aCM=4913=+3.77 m/s2
Hence COM of system will move downwards with magnitude of acceleration as 3.77 m/s2.

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