In the arrangement shown in figure `m_(A)` = 4.0 kg and `m_(B)` = 1.0 kg. The system is released from rest and block B is found to have a speed 0.3 m/s after it has descended through a distance of 1m. Find the coefficient of friction between the block and the table. Neglect friction elsewhere.
(Take g = 10 m/`s^(2)`)

Explanation

From constraint relations, we can see that
`v_(A) = 2 v_(B)`
Therefore, `" " v_(A) = 2(0.3) = 0.6 ` m/s
as `" " v_(B) = 0.3 `m/s (given)
Applying `" "W_(nc) = Delta U + Delta K`
We get `" " - mu m_(A) g S_(A)`
= `- m_(B) g S_(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 )
`therefore " " - 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