Saved Bookmarks
| 1. |
A partical of mass `2kg` is moving on the `x-`axis with a constant mechanical energy `20 J`. Its potential energy at any `x` is `U=(16-x^(2))J` where `x` is in metre. The minimum velocity of particle is `:-`A. `2m//s`B. `4m//s`C. `6m//s`D. zero |
|
Answer» Correct Answer - A From expression of potential energy Maximum `PE=16J` so minimum `KE=4J` Therefore `(1)/(2)xx2xxv^(2)=4rArrv=2m//s` |
|