A force of $20 N$ is applied to a block at rest as shown in figure. After the block has moved a distance of 8m to the right, the direction of horizontal component of the force F is reversed in direction. Find the velocity with which block arrives at the starting point (in m/s)

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$\frac{5 \sqrt{7}}{3}$

$\frac{\sqrt{28}}{3}$

$\frac{16 \sqrt{7}}{3}$

$\frac{8 \sqrt{7}}{3}$

answer is A.

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Detailed Solution

$R + 12 = 20 \Rightarrow R = 8 N \Rightarrow f = \frac{1}{4} \times 8 = 2 N$

$a = \frac{16 - 2}{2} = 7 m / s^{2}$

$v^{2} = 2 \times 7 \times 8 \Rightarrow v = 4 \sqrt{7} m / s$

Further distance $s_{1} \Rightarrow 2 \times 9 \times s_{1} = {(4 \sqrt{7})}^{2}$

$\Rightarrow s_{1} = \frac{56}{9} \Rightarrow {v_{1}}^{2} = 2 \times 7 \times (8 + \frac{56}{9})$

$= 2 \times 7 \times \frac{128}{9} \Rightarrow v_{1} = \frac{16 \sqrt{7}}{3} m / s$

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