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:

Question

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: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b2828cac183c3ce64df6b8459b0b2a95-.png" /> (A) 50 N (B) 75 N (C) 100 N (D) 96 N

Tardigrade Admin · Accepted Answer

a=(V2/2s) =(25/10) =2.5 m/s2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5dd8e194aa92121bc8c9d88b5ad7400a-.png" /> For 6 kg :-F-2T=6a For 2 kg: -T-2g=2(2a) From (1) and (2) F= 75 N

Q. 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:

$50 N$

$75 N$

$100 N$

$96 N$

$a = \frac{V ^{2}}{2 s} = \frac{25}{10}$

$= 2.5 m / s^{2}$

For 6 kg : $- F - 2 T = 6 a$

For 2 kg : $- T - 2 g = 2 (2 a)$

From (1) and (2) $F = 75 N$