In the arrangement shown in figure `m_(A)=4.kg` and `m_(B)=1.0kg`. The system is released from rest and block B is found to have a speed `0.3m//s` after it has descended through a distance of `1.m` find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take `g=10 m//s^(2)`).
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Explanation

From constant relations, we can see that `v_(A)=2v_(B)`
Therefore, `" " v_(A)=2(0.3)=0.6 ms^(-1)`
as `" " v_(B)=0.3 ms^(-1) " "`(given)
Applying `" " W_(nc)=DeltaU+DeltaK`
we get `" " -mu m_(A)gS_(A)=-m_(B)gS_(B)+(1)/(2)m_(A)v_(A)^(2)+(1)/(2)m_(B)v_(B)^(2)`
Here `" " S_(A)=2S_(B)=2m`
as `" " S_(B)=1 m " " `(given)
`:.-mu(4.0)(10)(2)=-(1)(10)(1)+(1)/(2)(4)(0.6)^(2)+(1)/(2)(1)(0.3)^(2)`
or `" " -80 mu= -10+0.72+0.045`
or `" " 80 mu=9.235` `" or " mu=0.115`