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

Options

A. `0.371 ms^(−1)`
B. `0.471 ms^(−1)`
C. `0.521 ms^(−1)`
D. `0.571 ms^(−1)`

Answer

A, C

Explanation

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`.