A particle oscillates back and forth in a frictionless bowl whose height is given by h(x) = 0.22x2 where h and x are meters. (a) Show graphically how the potential and kinetic energies of the particle vary with x. (b) Where does the particle have maximum kinetic energy? (c) If the maximum speed of the particle is 0.4 ms-1, find the x-coordinates at which the particle has maximum potential energy. Well, for (b) the particle has maximum kinetic energy at the bottom of the bowl, where h=0 at x=0. for (c) v=.04ms-1 so K=1/2mv^2 K=.08m.. U(potential energy)=.08m also U=mgh, .08m=9.8mh, therefore h=.008m The x-coordinates must be (h=.22x^2) .008=.22x^2 .036=x^2 x= 0.19, -0.19 i'm just making sure i did this right..