Saved Bookmarks
| 1. |
The potential energy of a `4kg` particle free to move along the x-axis varies with x according to following relationship : `U(x) = ((x^(3))/(3)-(5x^(2))/(2)+6x+3)` Joules, where `x` is in meters. If the total mechanical energy of the particle is `25.5` Joules, then the maximum speed of the particle is `x m//s`, find `x` |
|
Answer» Correct Answer - 3 The potential ………….. `U(x) = (x^(3))/(3)-(5x^(2))/(2)+6x+3` `(dU)/(dx) = x^(2) - 5x+6=0` `x = 2, 3` `(d^(2)U)/(dx^(2)) = 2x -5` At `x = 2m, (d^(2)U)/(dx^(2)) = -1`, which is a negative value this `U` is maximum At `x = 3 m, (d^(2)U)/(dx^(2)) = +1`, which is a positive value thus `U` is minimum `U_(min) = U(x=3)=7.5J` `T.M.E = 25.5J` `K_(max)+U_(min)= 25.5J` `K_(max) = 25.5-7.5=18J` `(1)/(2)(4)V^(2)max=18` `implies V_(max)^(2) = 9 implies V_(max) = 3m//s` |
|