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

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50 N

75 N

100 N

96 N

answer is B.

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Detailed Solution

a=v22s=2510=2.5m/s2 For 6kg:−F−2T=6a ……(1) For 2kg:−T−2g=2(2a) ……(2)From (1) and (2), F= 75 N

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