Question

The potential energy of a particle of mass 1 kg moving along x-axis given by U (x)= $[\frac{x^2}{2}-x]J$. If total mechanical energy speed of particle is 20 J, then find its maximum speed .

Answer 1

$U=[\frac{x^2}{2}-x]$

For minimum potential energy-

$\frac{dU}{dx}=x-1=0$

$⟹x=1m$

$⟹U_{min}=\frac{-1}{2}J$

$E=K+U$

$K_{max}=E-U_{min}$

$⟹K_{max}=20.5J$

$⟹\frac{1}{2}m{v}_{max}^2=20.5$

$⟹v_{max}^2=41$

$⟹v_{max}=6.4m/s$