InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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. |
|
Answer» Correct Answer - 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`. |
|