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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 & string to be light, the value of the force F applied on A is (take $g = 10 {ms}^{- 2}$ )

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50 N

75 N

100 N

96 N

answer is B.

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Detailed Solution

From the given diagram

Question Image

$F - 2 T = 6 a \to (1)$

$T - 20 = 2 \times 2 a$

$\Rightarrow 2T - 40 = 8a \to (2)$

$(1) + (2), a = \frac{F - 40}{14}$

But given $v = 5 m/s, s = 5m$

$∴$ From equation of motion

$v^{2} - u^{2} = 2as$

${(5)}^{2} - {(0)}^{2} = 2 \times \frac{F - 40}{14} \times 5$

$25 = \frac{F - 40}{7} \times 5$

$35 = F - 40$

$F = 75 N$

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