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

A, B

Explanation

If speed of block of 1.0 kg is `0.3m//s` then speed of `4.0 kg` block at this instant would be `0.6 m//s`. Applying =,
`E_(i)-E_(f)=`work done against friction
`:. 0-[1/2 xx 1.0 xx (0.3)^(2) + 1/2 xx 4.0 xx (0.6)^(2)`
`-1 xx 10 xx 1] =mu_(k) xx 4 xx 10 xx (2)`
Solving this equation we get,
`mu_(k) =0.12`.