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))`

Options

A. 1
B. 2
C. 3
D. None of these

Answer

B

Explanation

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`