A) \[\frac{g+a}{3}\]
B) \[\frac{g-a}{3}\]
C) \[\frac{4m(g+a)}{3}\]
D) \[\frac{m(g-a)}{3}\]
Correct Answer: C
Solution :
Let mass \[2m\] come downwards with acceleration \[x\] relative to pulley. The mass m rises up with acceleration \[x\] with respect to pulley. The entire system is lifted upwards with acceleration\[a\]. So, the mass m moves up with acceleration\[a+x\] and mass 2m moves up with acceleration\[a-x\]. Now, free body diagram of each mass is as shown: So, \[T-mg=m(a+x)\] and \[T-2mg=2m(a-x)\] or \[T-mg-ma=mx\] ? (i) and \[T-2mg-2ma=-2mx\] ? (ii) Dividing equations (i) by fu^ we have \[\frac{T-2mg-2ma}{T-mg-ma}=-2\] or \[T-2mg-2ma=-2T+2mg+2ma\] or \[T=4\frac{m}{3}(g+a)\]You need to login to perform this action.
You will be redirected in
3 sec