question_answer

The system starts from rest and \[A\] attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless everywhere and pulley and strings to be light, the value of the constant force \[F\] applied on A is

A) 50 N               

B) 75 N

C) 100 N  

D) 96 N

Correct Answer: B

Solution :

[b] \[a=\frac{{{v}^{2}}}{2s}=\frac{25}{10}=2.5m/{{s}^{2}}\]  \[For\text{ }6\text{ }kg:\text{ }-F-2T=6a...\left( i \right)\] \[For\text{ }2\text{ }kg:\text{ }-T-2g=2(2a)...\left( ii \right)\] From (i) and (ii), \[F=75\text{ }N\]