Consider the situation shown in figure. The system is released from rest and the block of mass `1 kg` is found to have a speed `0.3 m//s` after it has descended through a destance of `1 m`. Find the coefficient of kinetic friction between the block and the table.
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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`