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Question

In the figure shown below, mA=2 kg and mB=3 kg. A force F=40 N is applied on 2 kg block as shown. Find the acceleration of COM of the system. (Neglect friction everywhere and assume the string to be inextensible)

A
1.2 m/s2
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B
1 m/s2
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C
2.4 m/s2
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D
2 m/s2
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Solution

The correct option is A 1.2 m/s2
FBD's of Blocks


From F.B.D of 2 kg block
mAg+FT=mAa
2g+40T=2a ...(i)

For 3 kg block
TmBg=mBa
T=3g+3a ....(ii)

From (i) and (ii)
2g+40(3g+3a)=2a

5a=30a=6 m/s2

acom=mAaA+mBaBmA+mB=2×(6)+3×65
acom=65 m/s2=1.2 m/s2

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