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An atwood machine is setup in an elevator moving upward at `5 m//s` and slowing down at `2m//s^(2)` The initial velocity of block B is `2m//s` upward and the acceleration of block A is `3 m//s^(2)` downwards Find the time (in sec) at which block B will return to its initial position. Assume the string remains taut and the acceleration of the elevator does not change during the required time interval |
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Answer» Correct Answer - `4 sec` `v_(B)+v_(A)=2v_(P)=2xx5` `2+v_(A)=10 Rightarrow v_(A)=8 uparrow` `a_(B)+a_(A)=2a_(P)=2xx-2` `a_(B) -3=-4` `a_(B)=-1` `s=0=2xxt-1/2xx1xxt^(2)` `t=4 sec` |
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