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In the figure, the distance BQ=3m, BP=14...

In the figure, the distance `BQ=3m, BP=14m` at time `t=0`. The system of blocks is released from rest at time `t=0`. The string connecting `B` and `C` is suddenly cut at time `t=2s`. Calculate the velocity of `B` at the instant when it hits the pulley `Q`. The coefficient of friction between `B` and the horizontal surface is `mu_(s)=mu_(k)=0.25`. Take `g=9.8 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
7
initialy `8g-T = 8a`
`T-T'-(1)/(4) xx 4g = 4a , T'-2g = 2a`
solving we get a `a_(B) = (5g)/(14)` leftward.
Now `T = 0` at `t = 2s , v_(B) = (5g)/(7), s_(B) = (5g)/(7)`
`2g-T_(0) = 2a' , g + T_(0) = 4a'`
`2g = 6a rArr a' = (g)/(2) , v = (5g)/(7)-(g)/(2)t = 0`
`(t)/(2) = (5)/(7) rArr t = (10)/(7)s 0- ((5g)/(7))^(2) = 2 xx (g)/(2)t = 0 , S = (25 g)/(49) m`
Now `f` gets reserved. `2g - T_(0) = 2a`
`T_(0) = T_(k) = 4a , g = 6a rArr a = (g)/(6)`
`v^(2) -0^(2) = 2 xx (g)/(6)((25g)/(49)+(5g)/(7)+3)`
`v^(2) = (g)/(3) xx 15 rArr v^(2)-5 xx 9.8`
` = 49.0 rArr v = 7 ms^(-1)`
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