The potential energy of a particle of mass 1 kg free to move along x-axis is given by `U(x) =(x^(2)/2-x)` joule. If total mechanical energy of the particle is 2J then find the maximum speed of the particle. (Assuming only conservative force acts on particle)

Explanation

Potential energy `U = x^(3)/2 - x`
For minimum `U ,(dU)/(dx) = (2x)/(2) - 1 = 0` & `(d^(2)U)/(dx^(2)) = 1 = Positive`
So at `x = 1`, U is minimum Hence `U_(min) = - 1/2J`
Total mechanical energy `= Max. K.E + Min P.E Rightarrow Max K.E = 1/2 mv_(max)^(2) = 2 -(1/2) = 5/2`
`Rightarrow V_(max) = sqrt((2)/(1) xx(5)(2)) = sqrt 5 ms^(-1)`