InterviewSolution
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InterviewSolution
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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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Answer» 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` |
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