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A block B of mass 5kg rests on a rough horizontal surface (mu=0.2),A block A of mass 2kg rests on block B(mu=0.4). If a horizontal force of 21 N be applied to the block B, what is force ("in N") of friction acting between the blocks A and B ? (g=10ms^(-2))

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Solution :Let friction between blocks be `f_(1)` and on ground be `f_(2)`
`because 21Ngtf_(2"MAX")=14Nimplies5kg"moves on ground"`
`THEREFORE ""f_(2)=f_(k)=14N`
`a_("Amax")=(f_(1"max"))/(2)=(0.4xx20)/(2)=4m//s^(2)`
If both are to move together, COMMON acceleration,
`a=(21-14)/(7)=1m//s^(2)lta_(Amax)`
`therefore` both will move together with `1m//s^(2)implies f_(1)=f_(s)=2a=2N`


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