A constant force F=m2g2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m2 .
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a
m1g2(m1+m2)
b
m2g2(m1+m2)
c
3 m2g2(m1+m2)
d
3 m2g2 m1
answer is B.
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Detailed Solution
From the FBD of the two blocks, T-F=m1a and m2 g-T=m2a Adding these equations we get a=m2g-Fm1+m2 ⇒a=m2g−m2g/2m1+m2=m2 g2 m1+m2 towards right