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Asked in NEET 2024 · Applying the second law and free-body diagrams
Given: F=10 N on A, m_A=2 kg, m_B=3 kg, frictionless surface.
Idea: take the two blocks together first to get the shared acceleration, then look at B alone.
Whole system: a=F/(m_A+m_B)=(10)/(2+3)=2 m s⁻².
Now B on its own: the only horizontal force on it is the contact push N from A.
N=m_Ba=3×2.
N=6 N.
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