Particle oscillation and energy (potential/kinetic)

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SUMMARY

The discussion focuses on the oscillation of a particle in a frictionless bowl defined by the height function h(x) = 0.22x². The particle achieves maximum kinetic energy at the bottom of the bowl, specifically at x = 0, where potential energy is zero. Given a maximum speed of 0.4 m/s, the maximum potential energy occurs at x-coordinates of approximately 0.19 m and -0.19 m, derived from the equations for kinetic and potential energy.

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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..
 
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Totally correct dude
 

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