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The system is released from rest with th...

The system is released from rest with the spring initally stretched `75 mm`. Calculate the velocity `v` of the block afrer it has dropped `12 mm`. The spring has a stiffness of `1050 N//m`. Neglect the mass of the small pulley.

Text Solution

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The correct Answer is:
A, C
If block drops 12 mm, then sping will further stretch by 24 mm. Now,
`E_(i) =E_(f)`
`:. 1/2 xx 1050 xx (0.075)^(2) =-45 xx 10 xx 0.012`
`+(1)/(2) xx 45 xx v^(2) + 1/2 xx 1050 xx (0.099)^(2)`
Solving we get, `v=0.37 m//s`.
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