The potential energy of a particle oscillatin | vedclass

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(C) The potential energy function is $U(x) = 20 + (x - 2)^2$.The minimum potential energy $(U_{min})$ occurs when the term $(x - 2)^2$ is zero,which h

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$16$

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(C) The potential energy function is $U(x) = 20 + (x - 2)^2$.The minimum potential energy $(U_{min})$ occurs when the term $(x - 2)^2$ is zero,which happens at $x = 2 \, m$.Thus,$U_{min} = 20 + 0 = 20 \, J$.The total mechanical energy $(TE)$ of the particle is the sum of its kinetic energy $(KE)$ and potential energy $(U)$: $TE = KE + U$.To find the maximum kinetic energy $(KE_{max})$,we use the relation: $TE = KE_{max} + U_{min}$.Substituting the given values: $36 = KE_{max} + 20$.Therefore,$KE_{max} = 36 - 20 = 16 \, J$.

The potential energy of a particle oscillating along the $x-$axis is given as $U = 20 + (x - 2)^2$,where $U$ is in $joules$ and $x$ is in $meters$. The total mechanical energy of the particle is $36 \, J$. The maximum kinetic energy of the particle is ............... $J$.

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