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 every where and pulley & strings to be light, the value of the constant force F applied on A is​

Answer 1

Answer:

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Explanation:

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Answer 2

The value of the constant force F applied on A is 75N.

Explanation:

$u=0,v=5\frac{m}{s^{2}}$

$s=5m$

$V^{2}=2as$

$5^{2}=2*a*5$

$25=10a$

$a=\frac{25}{10}$

$=2.5\frac{m}{s^{2}}$

$A=F-T$

$F=T+6(2.5)$

$=60+15$

$=75.$