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In the arrangement shown in figure m(A)=...

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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Text Solution

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