Solving Mechanics Question: Show Energy of Motion in x-z Plane

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SUMMARY

The discussion focuses on deriving the energy of motion for a particle sliding under gravity along a specified path in the x-z plane, defined by the equation z = b - (x^2/a). The total energy expression is established as E = 0.5m*xdot^2*[1+(2x/a)^2] + mgb[1-(x^2/ab)]. Key equations include the conservation of energy principle T(A) + V(A) = T(B) + V(B), where T represents kinetic energy. The participant expresses clarity on potential energy calculation but seeks guidance on determining the kinetic energy term, specifically the relationship between velocity components.

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  • Familiarity with kinetic and potential energy equations.
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Homework Statement


A particle slides under gravity along a smooth path in the x-z plane by z=b-(x^2/a).
Show that the energy of this motion can be expressed as E=.5m*xdot^2*[1+(2x/a)^2] + mgb[1-(x^2/ab)]


Homework Equations


T(A)+V(A) = T(B)+V(B), where T is kinetic energy


The Attempt at a Solution


I understand how to get the potential energy term, but I am confused on how to calculate the kinetic term.
 
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[tex]v^2=\dot z^2+\dot x^2[/tex]

[tex]\dot z=-\frac{2x\dot x}{a}[/tex]
 

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