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A force F = 20 N is applied to a block (...

A force `F = 20 N` is applied to a block (at rest) as shown in figure. After the block has moved a distance of `8 m` to the right, the direction of horizontal component of the force F is reversed. Find the velocity with which block arrives at its starting point.
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The correct Answer is:
A, C

`mu N = 2,N=8,a =(16-2)/(2)=7 (rarr)`
`v_(2)=2(7) 8 rarr`....(i)
when the direction of horizontal component of the force f is reversed
`a_(1) = 9 m//s_(2) (larr)`
and distance covered by the block before it stops `= S_(1) = (16 xx 7)/(2 xx 9)`
Again `lars_(2)=8+s_(1)=8+(16xx 7)/(2 xx9)`
and `a_(2) = 7 (larr)`
`v_(2) =0+2(a_(2))(s_(2))=2(7)(8+(16xx 7)/(2 xx9))`
`rArr v=(16 sqrt(7))/(3) m//s`
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