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For shown situation draw a graph showing...

For shown situation draw a graph showing acceleration of A and B on y-axis and time on x-axis `(g=10ms^(-2))`

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limiting friction between A & B, `f_(1)=mum_(A)g=((1)/(2))(2)(10)=10N`
Block B moves due to friction only. So maximum acceleration of B,
`a_(max)=(f_(L))/(m_(B))=10/42.5ms^(-2)`
So both the blocks move together till the common acceleration becomes `2.5 ms^(-2)` after that acceleations of B will become constant while that of A will go on increasing. Slipping will starts between A & B at `2.5 ns^(-2)`
`implies2.5=(F)/(m_(A)+m_(B))=(3t)/(6)impliest=5s`
Hence for `t le5s,a_(A)=a_(B)=(F)/(m_(A)+m_(B))=(3t)/(6)=t/2`
and for `tgt5sa_(B)=2.5ms^(-2),a_(A)=(F-f_(1))/(m_(A))=(3t-10)/(2)=3/2t-5`
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